www.lsmsa.edu teacher Robert Dalling's Physics Lectures (see also www.ushumans.net)

Force

If you push on a mass then it will accelerate. This is Newton's Second Law of Motion which he published in the year 1687. Using F to represent force, which is a push or a pull, and m to represent mass, Newton's motion equation is written

F = ma.

Force = mass x acceleration.

If you put a force on a mass then it will accelerate.

If you put a force on a mass then its velocity will change.

If you put a force on a mass then its speed or its direction of motion will change.

            This is really the only equation in physics. All other equations simply describe special cases occurring in specific situations.

            Force is the cause of acceleration that was studied in the previous chapters. We feel force as a push or a pull. The harder we push the greater the acceleration. Mass resists the acceleration. Smaller masses are easier to accelerate.

CQ: Try to shake these large and small masses (being passed around the classroom) and feel their differing resistance to acceleration. Which of these numbered masses matches that of a coin? Name an item whose mass is about 1 kilogram, about 10, and about 100.

            Mass is measured in kilograms (kg) units. Force is measured in units of mass times acceleration, which is kg m/s². The combination kg m/s² is also a unit of one Newton (Isaac not fig). There are 0.2248 lb/N.

            We know that an object’s speed changes when it accelerates. We feel acceleration when we speed up in our car and are pressed back into the car's chair or when riding in an elevator that is starting or stopping its motion.

            Force is a vector quantity because the sum of two forces depends on the angle between them. We can believe that two persons pushing forward equally hard on a rock will realize a different result than two persons pushing equally hard against the rock but in opposite directions. In the x-direction, we have Σ F_x = ma_x, in the y-direction, we have Σ F_y = ma_y, and in the z-direction we have Σ F_z = ma_z.

CQ: What force is needed to accelerate a 100 kg object by 9.8 m/s²?

F = ma = ( 100 kg )( 9.8 m/s² ) = 980 kg m/s² = 980 Newtons. This force produces the weight of a 100 kg person.

            Notice that if F = 0 then a = F/m will be zero. This is Newton’s First Law of Motion. It is a statement of Galileo's inertia: an object at rest will remain at rest and an object in motion will remain in motion at constant speed until a force acts to change its speed. Notice also that if two or more forces act on an object but have a combined force of zero, then the acceleration will be zero. This happens when you push a box forward across the floor while ground friction pushes equally hard against the movement of the box.

CQ: What is the net force on a bridge?

            Notice also that if F = constant, then a = F/m will be constant and the resulting motion will be described by the constant acceleration equations we studied in the previous chapters. Only a few force relations are found in nature. In the next chapters, we will study forces that are proportional to distance, speed, speed squared, and to one over the distance squared. Nuclear forces vary as inverse exponentials. Interatomic forces vary with separation distance in more complicated ways but are readily modeled with a computer.

            You exert a force on some object once every few seconds. You exert a force when you grasp a pencil, turn a page, stand up, walk, lift a spoon, open a door, press a button, or turn a knob.

CQ: List the objects on which you exerted a force in the last few hours. How many did you find? Which student found the greatest number?

            Force comes in action-reaction or “equal and opposite” pairs. To understand this, consider that your fingers are a force meter. You know there is a force on your fingers whenever you see that they are bent backwards. As you push on a chair, you can feel that you are pushing and you know the chair is pushing back because your fingers are bent. The push of your finger on the chair and the force of the chair on your fingers, are equal and opposite, action-reaction pairs. Action-reaction forces always have the same magnitude but point in opposite directions. When you push on the wall with your fingers, you know that the wall is pushing back because your fingers are bent.

            You have experienced action-reaction forces in numerous situations. Action-reaction forces occur when a bat hits a ball. The bat slows from the ball's force and the ball speeds up from bat's force. These two items experience equal but opposite forces and undergo differing accelerations since they have differing masses. You feel the force of the ball as a kick against the bat held in your hands. When any two objects collide, equal and opposite action-reaction forces occur on the two objects. Surprisingly, this means that the force of a car on a bug has the same magnitude (but opposite direction) as the force of the bug on the car. Action-reaction forces occur when you kick a ball. The harder you kick, the greater the force of your foot on the ball and the greater the force of the ball on your foot. You feel the reaction force when you use a hammer to hit a nail. When an airplane wing pushes downward against air, the air pushes upward on the wing. This upward reaction force is what lifts an airplane into the air. Action-reaction pairs also occur when jet engines push air backwards and in reaction, the exhausting air pushes the jet forward. When swimming, you feel the force that you exert with your hand against the water. The water pushes on you with that same force and propels you forward. You throw an object by pushing on it with your hand, and you feel the reaction force as the object being thrown pushes back against your hand. The reaction force from a thrown object would push back and accelerate an astronaut floating in space or on a person who has jumped into the air or is falling through the air. While water skiing, we feel the pull of a rope while we pull on the rope. While riding in a car, you wouldn't move forward but for the push of the car chair or seat against your back. You feel that you are being pressed into the chair back because it is pushing forward on you. You push equally hard backwards against the seat. When the car is braked to a stop, you feel the force of the seatbelt slowing you or the force of your hand pressed against the dash. The car and road are pushing on each other with equal and opposite forces. While walking, we extend our foot backwards as we push against the ground. The ground’s reaction force pushes us forward. While walking on ice, we can not push as hard against the ground. You have experienced these swimming and throwing forces but may never before have had any reason to look carefully at their nature.

CQ: Picture one of the above forces or pushes in your mind and describe the sensation you experience while pushing against an object.

CQ: A car that is moving at 100 km/hr collides with a bug traveling at 30 cm/s.

a) The force of the bug on the car is less than the force of the car on the bug

b) The force of the bug on the car is greater than the force of the car on the bug

c) The force of the bug on the car is equal to the force of the car on the bug

CQ: A large truck that is moving at 100 km/hr collides with a small car traveling at 30 km/hr.

a) The force of the truck on the car is less than the force of the car on the truck.

b) The force of the truck on the car is greater than the force of the car on the truck.

c) The force of the truck on the car is equal to the force of the car on the truck.

CQ: A train that is moving at 100 km/hr collides with a bottle that is sitting on the track..

a) The force of the train on the bottle is less than the force of the bottle on the train.

b) The force of the train on the bottle is greater than the force of the bottle on the train.

c) The force of the train on the bottle is equal to the force of the bottle on the train.

            When we look more closely, we refine our notion of motion. We may not believe it at first, but measurements show that these action-reaction forces are always equal and opposite.

            Newton’s Third Law of Motion is the statement of his realization that force occurs in action-reaction pairs. We sometimes hear that for every force, there is an equal and opposite reaction. This also means that a force does not exist by itself but always occurs in action-reaction pairs. A force doesn’t push unless the object it pushes, pushes back. To be a force, a force has to push or pull something. Otherwise, its not a force.

            Mass is the amount of matter in a material and is measured in kilograms. It is determined by the total number of atoms comprising the material. Mass is also the resistance to acceleration. We all know that larger masses are harder to accelerate than are smaller masses.

            Mass is not the same thing as volume or density. The volume of matter is the amount of space it occupies and is measured in cubic meters. Density is the amount of mass per unit volume, measured in kg/m³. A block of material can be compressed, changing its mass density but not its number of atoms, which is its mass. For example, compressing a foam cushion will change the density of the material but not the number of atoms within the material.

CQ: Name some things you can compress and some that you can not compress.

            Mass is not the same thing as weight. The mass of an object does not change when it is taken to other planets but its weight does change. While on the Earth, the weight of a person results from the Earth’s gravitational force pulling him or her against the scale and toward the ground. Bathroom scales measure the attractive force of the Earth’s mass on your mass. This force can be envisioned as the Earth’s “invisible hands” pulling you downward against the scale. The more massive the planet, the more it pulls you against the scale. Just as the bathroom scale reading would increase as increasing numbers of your friends pushed downward on your feet, the bathroom scale reading also changes when you stand on the scale on increasingly massive planets. Since the gravitational force of each planet or moon differs, the weight of an object differs on each planet or moon. You may have heard that a person weighs one-sixth as much on the Moon than on the Earth. When a person sits in a swing, it is easy to accelerate that person forward by supplying a small push against their back. This push is parallel to the ground. The mass of the person resists the acceleration. When that person is on the moon, it is no easier to accelerate that person but it is much easier to get under that person and lift him or her upward. The masses that were passed around the classroom, that you held and shook, are easier to lift on the moon but they are just as hard to shake.

            Mass is measured in kilogram units, while weight is measured in units of Newtons. The U.S. is still using “pounds” as a unit of weight. A pound is not a unit of mass. The weight of one kilogram is 2.2 pounds.

CQ: What is the weight in pounds of 100 kg? ( 100 kg )( 2.2 pounds/kg ) = 220 pounds.

            A mass density is measured in units of kg/m³, while weight densities are measured in units of pounds/ft³. Some weight densities in lbs/ft³ are water = 64, person = 58 (0.9 water), Aluminum = 170, iron = 500, gold = 1200 (gold bars in the movies are never so heavy). Since water is defined to be 1 g/cm³, divide each of these numbers by 64 to convert to SI mass density in g/cm³.

            Physicists have studied millions of systems and have found that Newton’s motion equation, F = ma, describes each of them. Newton’s Laws are used to write equations that quantitatively specify the position, speed, and acceleration of any system ever observed, including pendulums, oscillating springs, weight-lifting pulleys, cars and their internal parts, rocket propulsion, rocket trajectories to the Moon or to Saturn, space shuttle orbits, tornadoes and hurricanes, baseballs, footballs, basketballs, hockey pucks, spinning tops, forces on muscles and bones inside animals or bridges or skyscrapers, propulsion of fish through water, motions of air past airplane or bird wings that lifts them into the air, and billiard balls (which the philosopher George Bush describes as "a sign of a misspent youth"). Each of these motions–and the motion of every object you have ever seen–are described by Newton’s equation. Each is just a special case.

            Newton's law of motion was published in 1687. At that time, nobody could imagine the endless uses of Newton’s equation. In fact, Newton’s equations were seen to also describe the motions of the planets around the sun. In 1687, people were very surprised that a planet would follow an equation just as did objects on the ground, and they were surprised mere mortals could write down the equation. Newton’s equation (if you push on an object, its speed will change) is a fundamental truth of nature. It will be used by humans for centuries to come. It is really the only equation in all of physics because all others are obtained from it. See the power that you have in your hand to be able to numerically predict the details of motion for any system.

            Through millions of experiments conducted in the last three centuries, physicists have studied millions of phenomena, including heat, light, motion, gravity, electricity, magnetism, gases, fluids, solid materials, sound, radiation, atoms, and nuclei. At first it seemed like the world was full of millions of unconnected phenomena, but we now understand how each of these is just one of multiple aspects of a few more-fundamental phenomena. Richard Feynman says "The adventure of our science of physics is a perpetual attempt to recognize that the different aspects of nature are really different aspects of the same thing." Physicists have found that only five forces exist in nature. And they suspect that each of these five are simply different aspects of a single, more-general force. It may be possible to explain all physical phenomena with a single law of nature. As we see in this chapter that the countless situations involving motion are all described by the single equation F=ma, we will begin to see that at its most fundamental level, nature is simple.

            In the following chapters, we will see how a few but fundamental laws govern natural phenomena, including every chemical or biological process that occurs. We will see that the fundamental aspects of nature are simple, that nature exhibits endless variety, and that each complicated end product–for example, a human–consists of a large number of these more-simple underlying phenomena. Five forces form the fundamental basis of nature. We will see that these forces are in action behind every natural phenomenon.

            Newton's motion equation describes the motion that results from applying forces to masses. The five forces found to exist in nature are the gravitational, electric, magnetic, and the strong and weak nuclear forces. We'll be studying each of these.

            Gravity is the attractive force that pulls two masses toward each other. In this way, it keeps the Moon in orbit around the Earth, the Earth in orbit around the Sun, and it also holds us to the surface of the Earth. In the year 1680, Newton published the equation of describing the Universal Law of Gravitation: Gm₁m₂/r², where G = 6.67 x 10^-11 Nm²/kg², that describes this gravitational force.

            Electricity and magnetism are the forces between charges and magnets. In contrast to gravity, these forces can be either attractive or repulsive. The magnetic force of a current carrying wire goes as 1/x. In the year 1785, Coulomb found the equation kq₁q₂/r², where k = 9 x 10⁹ Nm²/coul², that relates the electrical force between two charges q₁ and q₂. The electric force is zillions of times stronger than the gravitational force. The electrical force is behind the chemical bonds occurring between electrically charged atoms and multi-atom molecules–including the molecules of life, even the billion-atom DNA molecule. All of chemistry results from the electrical force. (Through millions of experiments, chemists have found many aspects of atomic interactions.) Static electricity is used in xerox machines, provides the cling of plastic food-wrap, keeps you from falling through the floor, and is often encountered as a sparking jolt when touching metal. A lightning bolt is a flow of electrical charges. There are countless manifestations of the electric force between neighboring atoms, including the tension as a string or spring is pulled. In the year 1864, Maxwell wrote a set of four equations that describe all electric and magnetic phenomena and showed that they are different aspects of a single, electromagnetic force. We will see that light is an aspect of electromagnetism. Every electromagnetic machine that has ever or will ever be invented, is described by Maxwell’s equations. Nobody in the year 1864 could imagine the endless uses of this equation.

             The short-ranged, weak and strong nuclear forces were found only in the last century as we became able to measure processes on an atomic scale. The strong nuclear force holds together the particles, called protons and neutrons, found to makeup the nucleus of an atom, and it also holds together the three quarks that are the particles forming a proton or neutron. The equation describing this force is not yet known. Physicists use particle accelerators to study this force. There is no telling what sort of “magical” machines will result. The weak nuclear force is involved in the process of radioactive decay in which the nucleus of an atom ejects electrons, light, or Helium nuclei.

            Since the phenomenon of radioactive decay involves both the weak and electromagnetic forces, those forces must be related. In the year 1987, Weinberg and Shalom wrote an equation showing that the electric, magnetic, and weak forces are different aspects of a single electroweak force. Since the Big Bang phenomenon involved all forces, they must all be related. There have not yet been any measurements of the gravitational interactions of atomic matter (you might like to make these measurements), so we do not yet know how to describe gravity on an atomic scale. Many physicists hope to unite gravity with the other four forces, resulting in a single equation that describes the interactions of all the particles of the universe in a “Theory of everything.” This is called the Grand Unified Theory. One such attempt is given by string theory.

The frictional force between two objects sliding past each other

We've all pushed a box across the floor. You may have noticed that it is harder to begin the motion of the box then it is to keep the box moving once it is in motion. At a microscopic level, surfaces of manmade objects are a bit rough, with occasionally prominent protuberances. By increasing the normal force, the two surfaces are pressed together such that even smaller protuberances are forced into contact. It turns out that while sitting in place, a small portion of the molecules of the box and of the floor electrically bond, as do any two materials when in contact. It takes a bit of extra force to break the more numerous electrical bonds that create this “static frictional force” to get the box to begin moving. Before motion begins, you are pushing harder and harder–increasing your force–and the static frictional force is pushing back against you with an action-reaction force that is identically increasing. Soon the maximum possible resistance of static friction is reached, attaining a "breaking point" in the bonds. Once moving, bonds between molecules are less able to form, so they occur in lower numbers than when the two adjacent materials are stationary. There are two forces acting on the box: your push and friction's resistance. When these two forces are equal, the box moves at constant speed. Since the frictional force is smaller once the box is moving, a smaller push by you is needed to keep the box moving at constant speed. If we measure closely, the acceleration of the box might be seen to cycle between slightly negative, positive, and zero. If you get the box moving and then stop pushing on it, friction quickly stops the box. Friction has two forms: static friction F_S, and kinetic friction F_K.

Here is a plot of frictional force versus applied force.

            |breaking point

           .

          . . . .. . . . . . . .

        .

      .

    .

  .

.

<--------><--F-kinetic—>

 F-static

            The exact details of friction are not yet known. Scientists are trying to more accurately understand its cause. In the last few decades, a new type of microscope, the scanning tunneling microscope, has been developed that sort of “feels” out a surface one atom at a time. Instead of a photograph, a map is made of the location and height of each atom on a surface. Here is a sheet of gold atoms.

            As an object presses against a material, it is squeezing together the molecules that comprise that material. The squeezed material pushes back with equal force unless a breaking point is reached. When you squeeze a cushion, it pushes back. Solid materials do the same, they are just more difficult to squeeze and compress by tiny distances. When a block sits on the top of a table, the weight of that object is pressing downward against the desk. In response, the desk pushes upward. Since this upward force is perpendicular or normal to the surface, it is referred to as the “normal force” and is written F_N. In this case the normal force is perpendicular to the ground. If a block is pressed against a vertical wall, the wall pushes back with a normal force. This normal force is still perpendicular to the wall but in this case, it is parallel to the ground. If a block is pressed against a board that is inclined at an angle, the normal force is still perpendicular to the surface, so in this case it is neither parallel nor perpendicular to the ground.

Demonstration:

Press a bathroom scale against the wall and read the resulting normal force.

            Why don't we fall through the floor, because the floor pushes up with a force equal to our weight. This force results from the electrical repulsion of the atoms squeezed within the floor. It means that the upward, electrical force from a small volume of atoms under your feet is as great as the downward gravitational force from all the atoms of the earth.

             It is found experimentally that the static and kinetic frictional forces are proportional to the normal force. The proportionality constant is called the coefficient friction. For static friction, we have

F_S = μs F_N,

The static (not moving) frictional force = the coefficient of static friction x the normal force.

            For kinetic friction, we have

F_k = μ_k F_N.

The kinetic (moving) frictional force = the coefficient of kinetic friction x the normal force.

It always occurs that

0 < Fs < F_k.

            The coefficient of kinetic friction is a measure of the slipperiness of a material. We have all noticed that some objects are more difficult than others to push across some materials. For example, it is easier to push ice across wood tables than it is to push wood across concrete. We also know by experience that cars slip more easily when driving on wet roads–even more so when driving on icy roads. In the study of “science friction” or tribology, researchers have measured the static and kinetic coefficients for thousands of materials moving across other materials. Friction coefficients have no dimension. Frictional resistance is less for boats moving through water than it is for vehicles moving across roads. The coefficient of kinetic friction for a tire rolling across concrete is 0.9. This means that the frictional force is about 90% as large as the weight of the car. Synovial fluid in bone joints have the very low coefficient of friction of about 0.003. Friction in biological systems occurs as moving bones, lungs, and hearts scrape past adjacent materials. This is the field of biotribology. Bone growth may occur through a piezoelectric effect in which the more bones are squeezed by forces, the more rapidly growth will occur.

            Friction often grabs and slips in a repeating, stick-slip, stick-slip motion. This happens when violin bows scrape across strings, when chalk screeches across a chalkboard, or as a finger scrapes the rim of a wine glass that then vibrates and sings in response. The stick-slip movement of tires rolling across dirt roads results in the washboard surfaces that develop on those roads.

            Friction has some surprising characteristics. It is proportional to normal force, which is often the weight of an object, but it is not proportional to the area of the object. For example, the frictional force is the same when sliding a book on its face or on its edge. This means that friction barely decreases even after an object has been polished to greatly reduce its roughness. When the speed of the object is greatly increased, friction barely increases.

Example:

A box with a mass of 20 kg (44 lbs) is fairly hard to push across a floor. When the coefficients of static and kinetic friction are μs = 0.8 and μ_k = 0.6, with what force P must you push to get the box to begin moving, and once it is moving, what force is needed to keep it in motion at constant speed?

       F_N

        ↑ 

P →██← F_S

        ↓

       W = mg

            Let’s choose a coordinate system such that x lies along the ground with positive x toward the right and positive y upward. While motionless, the box sits on the floor and is subject to four forces. Since the box is not moving, its acceleration is zero in both the x and y directions. The two forces in the x direction add to zero and the two forces in the y direction add to zero.

In the y direction, we have

+F_N - mg = 0.

This can be rearranged as F_N = mg. The normal force here is equal to the weight of the box.

The normal force is pointing in what we’ve chosen to be the positive direction, so F_N is written here with a positive sign. The weight of the box is mg and this force points in what we’ve chosen to be the negative direction, so we have written -mg here.

In the x direction we have,

+P - F_s = 0.

This can be rearranged as P = F_s.

The push P points toward the positive x direction, so P gets a positive sign. The frictional force resists the push, so F_s points in the negative x direction and is written with a minus sign.

While pushing against the motionless box, the static friction force pushes back equally until the force of your push reaches the limit of the static frictional force given by

F_S = μs F_N = μs mg.

Since the y-forces showed that P = F_s, we have

P = μs mg = ( 0.8 )( 20 kg )( 9.8 m/s² ) = 157 N.

This force is ( 157N )( 0.2248 lb/N ) = 35 lb. Does that amount of force agree with your past experience in pushing heavy boxes across the floor?

To keep the box moving at constant speed requires that the push P just balance the opposing, kinetic frictional force. In this case,

P = F_K = μ_k F_N = ( 0.6 )( 20 kg )( 9.8 m/s² ) = 118 N

A smaller push is needed to keep the box moving at constant speed than was required to begin the motion of the box. If we push with a force larger than this, the box will accelerate. For example, if we continue pushing with the earlier force of 157 N, what will be the acceleration of the box?

In this case, we have

Σ F_x = ma_x,

or

P - F_K = ma_x.

Since F_K = μ_k F_N = μ_k mg, this equation becomes

P - μ_k mg = ma_x.

Solving for a_x we have

a_x = P/m - μ_kg = ( 157 N )/( 20 kg ) - ( 0.6 )( 9.8 m/s² ) = 2 m/s².

You would have to run faster and faster, accelerating at a rate of 2 m/s² to keep pushing against the box. After just three seconds your speed would be

v_x = v_xo + a_x t = 0 + ( 2 m/s² )( 3 s ) = 6 m/s,

which is 14 mph and faster than most of us can run. We see that once the box is in motion, a push of magnitude F_s < P < F_K results in an acceleration of zero, while a larger push produces an acceleration.

            Our initial push gets the box in motion. We might stop pushing once the box reaches a speed of 1 m/s. In this case, the box is acted on in the x direction by only the frictional force, F_k. In agreement with our past acceleration, the box quickly slows to a stop. While slowing, we have

Σ F_x = ma_x,

or

- F_K = ma_x.

Since F_K = μ_k F_N = μ_k mg, this equation becomes

- μ_k mg = ma_x.

Solving for a_x we have

a_x = - μ_kg = ( 0.6 )( 9.8 m/s² ) = - 5.9 m/s².

The acceleration points in what we chose to be the negative x direction. The box will stop in a distance given found from v_o² = v² - 2a( x - x_o ), where x will now be measured from the point at which the box is released with a speed of 1 m/s. Since the final speed of the box will be zero, we have

x = x_o - ½ ( v_o² - v² ) / a = 0 - ½ [ ( 1 m/s )² - 0 ] / ( - 5.9 m/s² ) = 0.08 m.

            Does this distance of 8 cm ( 3 inches ) agree with your past experience in shoving 20 kg (44 lb) boxes?

            When we push a box for a moment and then discontinue pushing it, the box quickly stops. While growing up, we acquire a vague notion of friction, but we might not of thought about the way in which friction is opposing our push or the way that friction alone brings the box to a halt. When we push the box, we sort of think that our own push is the only force acting on the box.

            Projects are underway to measure the coefficient of friction continually on airport runways so that pilots will know the value before landing. A fleet of GPS-equipped taxis could measure the coefficient of tire friction throughout town on a continual basis, resulting in a real-time map. Larry Strawhorn calculates that worn brake lining reduces its frictional braking force by 22%. Dag Anders Moldestad discusses efforts to reduce friction on skis, canoes, and kayaks and other boats and has developed a high-friction sock for car tires. He says that one-third of the energy used in by civilization does work against friction.

Inertia revisited

            In our daily lives, everything we see moving soon stops due to friction and then sits motionless on the ground. Our everyday experience tells us that the natural state of an object is to be at rest on the ground. But, our experience is limited to the special case of case of air and ground friction and ever-present gravity. Can you improve the accuracy your past, unmeasured notions of motion? When we observe and measure the motion of objects apparently unsubjected to frictional and gravitational forces, such as occurs for “floating” astronauts, we find that nature behaves differently than we expect due to our limited experience. The astronauts seem to undergo unnatural motion because it is unlike the motions seen in everyday experience. But movement under gravity or frictional forces are special cases of motion. We want to go beyond our everyday notion of motion and discover the nature of motion in its most general form throughout the universe. We want to look beyond the single example of motion always stopping because of friction. Friction is not always present. We don't want to have one equation for motion with friction and another equation for motion without friction. We seek the most general description of motion that includes all cases with or without friction. Inertia is a more fundamental property of motion than is friction.

            Newton’s Law of Motion, F=ma, shows that if the force is zero then the acceleration will be zero. An acceleration of zero does not mean that the speed is zero, it means that speed is constant. In our experience limited to motion along the friction-covered ground, we have built up the expectation that the speed of an object can only be zero and that it is nonzero only while we are pushing on it. Our everyday but limited experience has led us to believe that a continual force is needed to maintain an object's motion and we believe that moving objects always stop. But Newton’s Law tells us that a continual or constant force produces a constant acceleration not a constant speed. And it tells us that after a force is removed, the object will continue in motion at a constant speed forever. The absence of force is the reason objects have inertia. We saw that inertia is the name given (by Galileo) to the observed property of matter to remain in its state of motion at a constant speed until acted upon by an outside force. Do you believe that once in space, a spaceship will move at a constant speed forever without having to run its engine? What about a space-walker outside the ship? Does this person need a rocket engine to keep up with the ship? The answer is no. Both continue moving at the same speed because no outside force is acting on either of them.

            Since our limited everyday experience has led us to believe that objects always stop moving unless they are being pushed, we vaguely feel that after having tossed a rock upwards, something must keep pushing on it as it continues its upward motion. After an upward-tossed rock has left your hand at a speed v, your hand no longer influences the thrown object. What keeps it moving upward? Its inertia does. (Aristotle “sat in his arm chair” trying logically to guess how nature behaves and incorrectly explained that the air must somehow be rushing in behind the object and forcing it onward. In contrast, the understanding of inertia resulted from repeatable experiments.) The rock would continue in the tossed direction at speed v forever but the external force of gravity, which is equal to its weight mg, causes a downward acceleration. The speed of the tossed object decreases to zero, stops for an instant, and then accelerates downward. In the same way we think that after using our hand to start a ball rolling up a hill, there must be some force that keeps it in motion. Now we understand that no force is needed to keep the ball in motion and that gravity slows its upward speed and then accelerates it downward.

            Galileo came to understand that objects in motion naturally continue in motion. He allowed a ball to roll down one inclined plane, cross a short level region, and then roll up a second inclined plane.

∙                                                                                  ∙          -

            ∙                                                          ∙                      h

                        ∙          ∙          ∙          ∙                                  _

            He released the ball from a height h and found that the ball would rise nearly to the same height on the second ramp before coming to a stop. He reasoned that if friction were not present, that the ball would return exactly to its starting height. (While moving in a swing, your downward motion begins at a certain height and then subsequent upward motion ends at that same height.) When he inclined the second ramp at a lower angle, the ball still returned to its starting height, no matter how far it had to travel.

∙                                                                                                                                  ∙          -

            ∙                                                                                  ∙                                              h

                        ∙          ∙          ∙          ∙                                                                                  _

            He observed that the second incline was slowing the speed of the ball. He then reasoned that if the second ramp was not inclined at all (an angle of 0°) then the ball would continue moving forever without being slowed down. Galileo concluded that objects in motion stay in motion until a force stops them. We call this inertia.

            Consider again, the examples of inertia given above in the introduction to projectile motion. When you drop an object within a moving car, the object falls straight to the floor right under your hand. From the point of view of an observer sitting on the sidewalk, that object undergoes horizontal motion at a constant speed while falling with a constant acceleration–just as was the case for our frictionless projectile. When Apollo astronaut Alan Shepard hit a golf ball on the moon, where there is no air or air friction, the flying ball underwent motion with constant horizontal speed and constant vertical acceleration.

            We have all been traveling in a car at 100 km/hr (60 mph) while we tossed a coin straight up into the air and found that the object falls straight back down into our hand. It does not smash through the back window at 100 km/hr and land in the road behind the car. Since your past experience has taught you this, do you believe that a rock dropped from the mast of a moving ship will fall straight to the deck below the mast–that is, it will not land in the water behind the moving ship? If you toss a coin while riding in an airplane moving at 1000 km/hr (600 mph), will the coin crash into the back of the aircraft at 1,000 km/hr? If an object is launched straight upwards from the back of a moving truck, the object will later land in the same spot in the back of the truck from which it was launched. Astronauts travel inside a space shuttle moving at 20,000 mph and can step out of the ship without being left behind. If any of these tossed objects are in mid-air when the car, boat, airplane, or shuttle accelerates, then the back of the vehicle will move forward and can meet the mid-air object. Galileo used the term inertia to describe the property of objects in motion to remain in motion and for objects at rest to remain at rest. While driving in a car and making a turn, the stuff on our dash board wants to keep moving forward and so the side of the turning car collides with it. In a similar situation, Mr. Schlangen of Two Harbors High School demonstrates a Hewitt Inertia Hat. If no force acts, then the speed of an object will not change. If there were no air friction, an object dropped from an airplane would continue in motion directly under the airplane just as does the object dropped under your hand while moving in the car.

Forces in various situations

Forces add like vectors. For example, we can believe that two equally strong pushes will combine either to a value between nothing or double that of one force acting alone. In which direction will be the net force in these situations?

A) P = F.

P →███← F

B) P = F

P → ████→ F

C) P = 2F

   F

   ↑ 

███← P

            When solving problems involving forces, you might like to use the following procedure. Draw a picture, show and label every force, show the coordinate system and indicate the positive x and y directions so that they point consistently throughout the solution, list known and unknowns, and then apply Newton’s law to the system. In the x-direction, we have Σ F_x = ma_x, in the y-direction, we have Σ F_y = ma_y, and in the z-direction we have Σ F_z = ma_z. Solve the equations for the unknown value, put numbers into the equation, and then decide if the answer is reasonable. Check that forces point in the right directions. Sometimes, three forces add up to nothing. To increase the accuracy of our notion of force, let’s analyze the forces occurring in many systems. Your physical understanding begins after solving ten problems and becomes more accurate after solving fifty.

Example:

What will be the acceleration when a force of 10 N is placed on a 2 kg mass?

We are given F = 10 N, m = 2kg, and asked to find a. From F = ma we have

a = F/m =10 N / 2 kg = 5 m/s².

If the force is doubled, how will a change?

Let the new force be P, where we have P = 2F. Let a_F = F/m and a_P = P/m. The ratio

a_P / a_F = ( P/m )( m/F ) = P / F,

but since P = 2F this becomes,

a_P / a_F = 2F/F = 2.

We see that the acceleration is doubled when the force is doubled.

CQ: What will be the change in a if F is not changed but m is instead doubled? (a is halved)

            Let’s draw the forces that occur when a person with a mass of 50 kg stands on a scale.

O

/\

–

The weight mg of the person results from the downward pull of gravity. Gravity pulls the person downward against the scale and in response, the scale pushes upward with an equal and opposite normal force N, whose magnitude is read on the scale. The two forces can be drawn by themselves in a “free body diagram” and the coordinate system can be drawn as shown.

y

┼ x

N

↑

█

↓

W = mg

There are no forces in the x direction. Since there is no acceleration in the y direction, we have

+ N - mg = ma_y = 0.

This gives

N = mg = ( 50 kg )( 9.8 m/s² ) = 490 Newtons as the scale reading. (European scales measure mass in kilograms while scales in the U.S. measure weight in pounds.)

CQ: What will each scale read when a 50 kg person stands on two scales? (Half)

Demonstration:

Hang a 1 N weight from a string and see that the scale reads 1 Newton. The tension in either string is 1 Newton.

____ bar

    | upper string

   O scale

    | lower string

   █ weight

A free body diagram for the weight shows that the two forces acting on it are the tension T in the string holding it up and the force of gravity W=mg pulling downward. As with the scale, we have

y

┼ x

N

↑

█

↓

W = mg

In the y direction,

+ T - mg = ma_y = 0,

or

T = mg.

The tension in the lower string is equal to the weight of the object.

CQ: What is the tension in the upper string? T is the same as W = mg = 1 Newton.

Example:

Instead of attaching the upper string to the bar, loop the upper string over a pulley and attach it to a wall. This has no effect on the tension in the string. The tension T is the same throughout the entire length of the string and is given by T = W = mg = 1 Newton. The string is always pulling (remember the engineering rule that “you can’t push with a string”) along its current direction. The upper, horizontal portion of the string is pulling toward the wall, while the vertical portion of the string is pulling upward. The equation tells us the tension in the string. We usually omit the scale from the drawing. The scale does not need to be there unless we wanted to verify the value. If the scale were used here, it would still read a value of 1 Newton.

wall

  |─┐ string over pulley

     │ lower string

     █ weight

Example:

Instead of attaching the upper string over a pulley to a wall, replace the wall with another mass that has the same weight as the first. This has no effect on the tension in the string. The tension T is the same as W = mg = 1 Newton. Both scales read the same, 10 Newtons, and the tension is 10 Newtons in each of the three sections of the single string.

 ┌──┐ string over two pulleys

 │....│ lower portions of the two strings

 █ .. █ two weights

            This means that the tension is the same when two teams of horses pulling a box both left and right or one horse pulling a box tied to wall. In the coming chapters, we will see the similar result that two cars colliding head produces an identical force to one car colliding with wall.

Example:

Hang one mass of weight W = mg from two strings. What will be the tension in each string?

__

| |

█

The free body diagram shows the tension T in each string and the weight W acting on the suspended object.

y

┼ x

T T

 ↑↑

 █

 ↓

W

In the y direction, we have

+ T + T = W,

or

T = W/2.

We see that each string supports half the weight of the object. If bridge consists of n wires supporting a roadway of weight W, the tension in each wire wile be W/n.

Example:

Here are two strings tied at an angle to two masses held together by a third string.

Example:

Two people are pulling on Greg who has mass m = 50 kg. Kari is pulling with a force K = 100 N and is directed at angle θ = 60° above the positive x axis and Bryan applies a force of B = 200 N directed at angle α = 30° below the positive x axis. What will be the direction and magnitude of the resulting acceleration?

y

               K

            ●

          ●

       ● θ

  ┼ ------------>x

         ● α

                ●

                   B

Notice that the y-component of force B points in the negative y direction.

Σ F_x = ma_x = K cos( θ ) + B cos( α ) = 100 cos( 60 ) + 200 cos( 30 ) = 50 N + 173 N = 223N

Σ F_y= ma_y = K sin( θ ) - B sin( α ) = 100 sin( 60 ) - 200 sin( 30 ) = 86.6 - 100 = -13.4 N

The magnitude of the resultant force is | F | = √( F_x² + F_y² ) = 224 N.

The magnitude of the resultant acceleration is then a = | F | / m = 224 N / 50 kg = 4.5 m/s².

The resultant force has angle β = tan^-1( F_y / F_x ) = tan^-1( 13.4 / 223 ) = -3.4 degrees, which is below the positive x axis. The resultant acceleration also points in this direction.

Example:

A force F = 30 N is applied to a block of mass m = 5 kg that in turn pushes on a second block of mass M = 10 kg. What will be the acceleration of the blocks, and with what force P does the first block push on the second?

          M

F →■██

       m

            The first block feels both the force F pushing it to the right and the resisting force P of the block ahead of it. P is called the contact force and it pushes m toward the left. Block M is pushed by block m with a force R acting toward the right. The two forces P and R are equal and opposite action-reaction forces. Imagine yourself being this first block, being pushed from the left into block M that doesn’t want to move. This causes you to be squeezed, as sometimes occurs when being pushed in line. The second block M is pushed only by the first. Both blocks will move together toward the right with the same acceleration. For that reason, let’s choose the positive direction to be toward the right.

We first write Newton’s Second Law ( F = ma ) for the combined system, because F will accelerate both blocks. This gives

F = ( m + M )a.

Solving for a, we get

a = F / ( m + M ) = ( 30 N ) / ( 5 kg + 10 kg ) = 2 m/s².

We can also write Newton’s Second Law for each individual block.

Block m is subject to two forces. We have

F →■← P

      m

What is the contact force P between m and M? Let’s write Newton’s Second Law for block m. We have

F - P = ma,

 or

P = F - ma = 30 - ( 5 kg )( 2 m/s² ) = -20 N, which points to the left.

The only force acting on block M is the contact force R. Writing Newton’s Second Law for block M, we get

       M

R →██

R = M a = ( 10 kg )( 2 m/s² ) = +20 N,

which is equal and opposite to force P.

Example:

Make free body diagrams for each of these systems.

Example:

After jumping out of a perfectly good airplane, a parachuter of mass m = 40 kg is being pulled toward the ground by the force of gravity, which is W = mg, while also being pushed by the force of air friction R. (We might mention in passing that R is often written as R = ½C_dρv². The drag coefficient C_d depends on the shape of the object but may also very by a few orders of magnitude as the speed changes. The density of the medium is given by ρ.) As the downward speed of the parachuter increases from zero, the upwards frictional force increases as the square of the speed. When the upward, frictional force grows to be as large as the downward gravitational force, the parachuter then falls at constant speed. What is the force of the air friction at that time? The forces on m are R and W.

R

↑

█ m

↓

W

            We choose the usual coordinate system

y

┼ x

but have motion only in the vertical or y direction. Newton’s equation is then

Σ F = ma,

or

+R - W = ma = 0.

Since the parachuter is moving at constant speed, the acceleration is zero, and we get

R = W = mg = ( 40 kg )( 9.8 m/s² ) = 390 N.

By the way, did you hear about the lucky parachuter who was caught for 45 minutes in thunderstorm air that was circulating up and down.

Example:

What are the forces on a moving car?

            While moving, the car tires push backwards against the ground, just as you feel your own muscles pushing backwards against the ground. In an equal and opposite reaction, the ground pushes forward on the car tires (or your foot while walking). It is this forward force F that moves the car (or walker) forward. In addition, ground and air friction R resist the motion of the car . When moving toward the right, the two forces F and R act on the car.

F →██ ← R

            When the acceleration is zero, we have

F = R

            The car engine burns gasoline to do work against the ground and air frictions. A similar situation occurs when an airplane is pushed forward by the force F of the propellers or jets and pushed backwards by the air resistance R. While moving at constant speed, we have F = R.

Example:

A 2000 kg (4400 pound) car accelerates from a speed of zero to 30 m/s (68mph) in 30 seconds. What is its acceleration?

a = Δv / Δt = ( 30 m/s ) / (30 s ) = 1 m/s².

What force F is needed to produce this acceleration when the coefficient of friction between the tire and the road has the value μ_K =0.9?

Let’s have the car accelerating in the positive direction toward the right. The only horizontal force acting on the car is the force of friction F which pushes the car forward in reaction to the car pushing backwards against the ground.

       F_N

        ↑ 

F →██

        ↓

       W = mg

            In the vertical or y direction, the y forces are F_N - W = 0, so that once again the normal force is equal to the weight of the car. We have F_N = W = mg. In the x direction, we have

F = μ_K F_N = μ_K mg = ( 0.9 )( 2000 kg )( 9.8 m/s² ) = 17,600 N.

Notice that the higher the friction coefficient, the higher can be the acceleration of the car. This is true whether the car is taking off or braking to a stop. When μ_K goes to zero, as occurs on icy roads, then a goes to zero.

            Let’s practice our physical reasoning involving force. Indicate the forces occurring in the following systems.

Example:

A water skier is being pulled to the right by a boat. The forces on the skier are downward gravity W, upward normal force F_N, forward rope tension T, and backward friction F_K. The greater the difference T - F_K, the greater will be the forward acceleration. The skier feels the pull T in her hands and arms and feels the resistance F_K in her feet and legs.

         F_N

          ↑ 

F_K ←██→ T

          ↓

         W = mg

CQ: What are the forces on a windsurfer? What are the forces on a dogsled?

Example:

What are the forces on a launching space shuttle? The forces are the upward engine thrust T, the downward weight W, and air drag D. The upward acceleration is determined by the difference between these forces.

y

┼ x

L

↑

█

↓

D+W

            Rocket propulsion can also be used to travel along the ground.

Example:

As air flows past a moving ball, the air takes symmetric paths above and below the ball. We can believe that the upward and downward force of the air are matched and that the net vertical force is zero. There will be a backwards frictional force on the ball.

Example:

Now suppose we pinch the sphere so that it is shaped like this. Now that it has an unsymmetrical shape, the flow of the air past it will no longer be symmetrical and a net vertical force will result. This wing still has drag but it now also has lift. The faster the wing moves, the greater will be the lift, as you may have noticed when holding your hand out of the window while in a moving car. The lift also increases when the wing moves at an angle to the wind. When an airplane is stopped on the ground, no air is flowing past the wing so its lift is zero. As the speed of an airplane increases down the runway during take off, the lift of the wing is increasing. At a certain speed, the lift force matches the weight of the plane and it can begin flying. This is the take-off speed of the plane.

Example:

The forces on an airplane moving to the right are downward gravity W, upward lift force L, forward thrust T, and backward air drag D. These forces change while the aircraft maneuvers. The engine thrust T must work against the drag D.

       L

        ↑ 

D ←██→ T

        ↓

       W = mg

CQ: What are the forces on a moving butterfly or on a jellyfish propelling itself through the water?

            We are able to swim because the resisting force of water on an object increases with the speed of the object through the water. Without realizing what we are doing while swimming, we push our hands rapidly backwards but slowly forward. As we push backwards against the water, in reaction we are pushed forward. As we next move our arms forward, in reaction, the water is pushing us backwards–but with a smaller force than occurred with the more rapid backwards motion of our hand. As we move our hand in alternating the backward and forward directions, the water is pushing us much forward and then less backwards in an alternating manner. The forward push of the water exceeds the backward push of the water so that overall, we move forward. If the resistance of water did not increase with speed, we would not be able to move within water.

            Cislunar Aerospace, Inc. explains the propulsion of an octopus, jellyfish, clam, seal, turtle, and sea-lion. Jellyfish push their water-filled umbrella backwards and in reaction are propelled forward. Squid squeeze a water-filled, muscular mantle. As water is expelled backwards from its mantle, the squid is propelled forward. Squid choose the direction in which they will move by varying the direction of the expelling water. Scallops and clams avoid predatory sea stars by clapping together their two shells so that the expelling water will propel them away. The airplane-wing-shaped cross section of sea turtle and sea lion flippers are used to fly through water. The flipper is tilted 25° below the direction of the animal’s forward movement so that lift is generated on downstrokes. Seals instead press backwards against the water with their hind flippers and tail to propel themselves forward.

Example:

What are the forces on a skier sliding down a hill that is sloped at an angle θ relative to the horizontal? (This problem is also done here, here, here, and here, click “Skier goes down a slope.”) The forces are downward gravity W, a normal force F_N that is perpendicular to the hill, and a friction force F_K that points up the hill. For this problem, it is convenient to choose the positive -axis to point down the hill. The force W then has a component Wsin( θ ) that points down the hill. The net force down the incline is Wsin( θ ) - F_K and it causes the skier to accelerate down the hill. If Wsin( θ ) = F_K then the acceleration is zero.

                            F_K

                           .

                         . .

                       . .

F_N . . .

     . . .

        . . .

                . .

             . .

          . . .

        . | . .

      . \/ .

    . θ W .

  . . . . . . . . . . . . .

Example:

What are the forces on a plank that is supported at both ends while also holding up a person?

   O

   /\

┌────┐

│.........│

            Drawing just the plank, the forces are the weight w of the plank, which is located at its center point, the weight W of the person, the supporting force R on the right edge, and the supporting force L on the let edge. If you and a friend took the place of the two supports at L and R, each of you would feel squeezed by the downward forces w + W.

 ____________

↑....↓.....↓..........↑

.L..w...W.........R

Example:

What are the forces on a leaf hanging from the edge of a wind-blown branch? The forces are the tension T in the stem of the leaf, the downward weight W, and the force F of the wind on the leaf. As the force of the wind increases, the tension increases until a breaking point is reached.

F ←██→ T

         ↓

        W = mg

Example:

What are the forces on a tree frog that is clinging to a vertical tree trunk? In addition to the weight of the frog, there is a normal force F_N and a frictional force f on each foot. The sum of the vertical forces and the sum of the horizontal forces are zero. Here is one foot.

   f

   ↑ 

██→ F_N

   ↓

  W = mg

Example:

What are the forces on nearly-standing toddler being partially supported by his hands? In addition to his weight W, there is an upward normal force L and R under his left and right feet and an upward tension l and r in his left and right hands due to the holding parent. The two forces L and R are reaction forces to the force with which the baby pushes against the ground. We push by contracting muscles composed of long proteins. When the baby is not moving, the sum of the vertical forces and the sum of the horizontal forces are zero. The normal ground’s, upward force increases in reaction to the baby’s increased downward push in his leg muscles. Sometimes the baby stops pushing downward with his leg muscles. The baby accelerates downward if

L + R + l + r < W.

l....r

↑...↑

███

↑.↓.↑

LWR

Student:

Run the Phet, Motion: The Ramp and Forces in One Dimension simulations and then write paragraphs explaining what each does, what you have learned from it, and how it helps you understand this physical phenomenon. Include numerical examples.

Student:

Run one of the ActivPhysics Forces and Motion units and then write a paragraph explaining what it does, what you have learned from it, and how it helps you understand this physical phenomenon. Include numerical examples.

Online Example:

Richard Vawter of Western Washington University has an animation showing the motion of a block pressed against a frictional, vertical wall.

Example:

Let’s look at what happens when we accelerate in an elevator while standing on a scale. As the elevator accelerates upward to begin moving, we feel an increase in our apparent weight as our knees slightly buckle. While cruising at constant speed our weight feels normal again. We feel less weighty in our knees and feet as the elevator accelerates downward.. In the previous pages we saw that, as a person stands on a scale, the person is pulled downward by gravity. In reaction to this constant force W = mg, the scale pushes upward with a normal force N. The forces on the person are N and W.

y

┼ x

N

↑

█

↓

W

            Choosing the positive direction to be upward, Newton’s equation Σ F = ma becomes

N - mg = ma.

            Notice that we list the forces on this person and not the forces of this person on other things. (It often helps to draw a circle around the person and then include only those forces that poke through the circle.) The scale measures the normal force, N. Solving for N we have

N = m( a + g ).

            The elevator, scale, and person all accelerate together. They share the same value of a. When the acceleration is upward, then the normal force N = m( a + g ) will increase as if the person suddenly weighs more than usual. When the acceleration is downward, then a is negative and the normal force N = m( a + g ) will decrease as if the person suddenly weighs less than usual. This agrees with our experience riding in elevators or in helicopters.

            Our apparent weight is a multiple of our usual or resting weight.

N / W = m( a + g ) / mg = 1+a/g.

Or,

N = ( 1+a/g )W.

            Let’s calculate our apparent weight N when riding in an elevator that accelerates upward from a speed of 0 m/s to 5 m/s in 5 seconds. The acceleration is

a = Δv / Δt = ( 5 m/s ) / 5 s = 1 m/s².

            Our apparent weight, which is the scale reading, is then

N = ( 1+a/g )W = ( 1 + 1/9.8 ) = 1.1 W.

            We feel 10% heavier than usual. While accelerating downward at this same rate,

a = -1 m/s²,

            and our apparent weight is

N = ( 1+a/g )W = ( 1 - 1/9.8 ) = 0.9 W

Demonstration:

While standing on a scale, take on ride in the school’s elevator.

Example:

To jump straight up into the air, you bend your knees and then push against the ground. You feel the force in your legs and knees. The harder you push against the ground, the higher you will jump. If this is done while standing on a scale, then the scale will read the force with which you press on the ground as in this video clip. Biologists measure the force under each foot of walking insects. Dancers, athletes, and physical therapists measure forces on feet in the science of biomechanics.

            How high can you jump? A student might like to demonstrate by jumping up and touching a spot on the wall. We can then measure or calculate the following jump parameters. An athletic person jumps vertically about 0.6 meters, while a professional might jump 1.25 meters. For how many seconds will a professional be in the air? To fall from a height of 1.25 meters, we have x_o = 1.25 m, x= 0, v_o = 0, v = ?, a = -9.8 m/s², and t=? The motion equation x = x_o + v_ot + ½at² gives

t = √( 2x_o / g ) = 0.5 second

for the time needed to fall from the top of the trajectory. Since the upward and downward portions of the jumps last for equal amounts of time, the person is in the air for 1.0 seconds. An athlete’s “hang time” is less than one second no matter how much flapping is done, nor does it matter if the person is moving horizontally.

            What is the jumper’s initial speed? From v = v_o + at we have

v_o = v - at = 0 - ( -9.8 m/s² )( 0.5 s ) = 4.9 m/s.

This speed occurs as the jumper leaves the ground and again as the jumper lands on the ground. Both speeds have the same value but the person is moving in opposite directions. Suppose the jumper has a mass of 75 kg, which corresponds to a weight of 165 pounds. To make the jump, the person presses his or her feet against the ground for a period of about 0.3 seconds. If the jumper is to attain a launch speed of 4.9 m/s during a 0.3 second push, what will be his or her acceleration? We have

a = Δv / Δt = ( 4.9 m/s ) / 0.3 s = 16.3 m/s².

Since this is an acceleration of 1.7 g, a scale placed under the feet of the jumper would rad 170% of his or her weight. To develop this acceleration, one must push against the ground. With what multiple of your weight must you push? We have

F = ma = ( 75 kg )( 16.3 m/s² ) = 1225 N.

This a multiple of

F / W = ma / mg = a / g = 16.33 / 9.8 = 1.7

times the weight of the person. Does that agree with your past experience?

Example:

A 0.2 kg arrow gains a speed of 40 m/s as bowstring pushes it through a distance of 40 cm. Its velocity increases at an unknown instantaneous rate but we can find the average acceleration throughout the entire launch using the motion equations. What is the average acceleration?

            Let’s put the origin at the place where the arrow begins moving toward the right. Then we have x_o = 0, x = 0.4 m, v_o=0, v = 40 m/s, t = ? and a = ? The speed of the arrow increases while the string is pushing on it. Its velocity vector increases in length. The acceleration is found from

v² = v_o² + 2a( x - x_o ),

or

a = ( v² - v_o² ) / [ 2( x - x_o ) ] = ( 40 m/s )² / (2 ⋅ 0.4 m ) = 2000 m/s².

Then

F = ma = ( 0.2 kg )( 2000 m/s ) = 400 Newtons = ( 400 N ( 0.22 lbs/N ) = 88 lbs.

We know it takes a strong arm to do this.

Example:

Two rams compete by colliding heads in the manner of two Hollywood stars “as if the one who thinks of it doesn't get hurt." As a ram collides with another, his speed drops from 1 m/s to zero as soft bone and vertebrae-gaps compress by just 1 cm. What is the force on a ram whose mass is 200 kg? In the old English units, the weight of 200 kg ram is 440 lbs (“lbs” comes from the old pilot's abbreviation for "land before stopping"). Combining F = ma and v² = v_o² + 2a( x - x_o ), we have

F = ma = m{ ( v² - v_o² ) / [ 2( x - x_o ) ] }.

Putting the origin at the point of contact, then x = 0.01 m, the final speed is zero, and we have

F = ma = mv² / ( 2x ) = ( 400 kg )( 1 m/s )² / (2 ⋅ 0.01 m) = 20,000 N = 4400 lb.

            This force is ten times the weight of the animal. Each ram experiences this same force because the force of the left ram on the right is equal and opposite to the force of the right ram on the left. Since each ram has a different mass, each experiences a different acceleration–but an equal force.

            Male Stegoceras dinosaurs had thick bones on the top of their skulls that must have been used for colliding. This indicates that they had a social system of competing males. For two colliding 20-kg Stegoceras animals running at 3 m/s and stopping in 1 cm, the stopping force is

F = ma = mv² / ( 2x ) = ( 20 kg )( 3 m/s )² / (2 ⋅ 0.01 m) = 9000 N.

            Woodpeckers, skiers, and skateboarders feel similar forces when they hit trees. The shorter the stopping distance, the greater the force. Without realizing it, this is the reason you bend your knees upon contacting the ground after jumping down from a height. It may be that relaxed, drunk people less often suffer broken bones in falls because their loose limbs are not flexed. Stiffened necks and backbones suffer larger stopping forces during a collision because the stopping distance is shorter than would occur with relaxed bodies.

            When toddlers try to stand or walk they often fall onto their butt. They aren't hurt because their center of mass falls only 20 cm. Heavy horses, cows, and elephants can break bones simply by falling. People are not quite as susceptible, cats less so. Animals over 100 kg (220 lbs) can be hurt when falling through a distance equal to their own height. For animals having a mass in the range 100 grams down to 100 milligrams, their terminal velocity is low and are uninjured by a fall. A mouse is not hurt when falling five stories as is occasionally demonstrated.

            Smaller creatures with a mass less than 100 milligrams do not fall through the air but are carried with the wind wherever it might go. In their physics course, they are not interested in un-experienced "free fall," only in hurricane force winds. While our lack of experience with unforced inertia makes us assume that all objects “stop as soon as pushing ceases,” these small creatures believe that motion does not stop unless the wind stops. Who is right, us or them? Rather than checking to see if equal masses fall at the same rate, they might want to know if equal masses are blown by the wind at the same rate?

Example:

Lets look at the forces occurring as a demolition team uses dynamite to bring down a building in an implosion. What would happen if one floor is severed from its support so that the now-unsupported, upper portion of the building falls down onto the floor below, collides with it, and then stops in distance h. That is, the upper portion drops a distance h, hits a floor below, and stops in a distance d.

            While dropping a distance h from an initial sped of 0 at the origin, the speed of the falling portion is found from v² = v_o² + 2a( x - x_o ) to be

v = √( 2gh ).

While falling, velocity increases at a constant rate due to the constant force of gravity. After having fallen that distance h, the falling portion of the building hits the floor below while moving downward at speed v = √( 2gh ). During the collision, it comes to a stop in a distance s, experiencing an acceleration a. The acceleration is found from v² = v_o² + 2a( x - x_o ). To solve this equation, let’s choose a coordinate system at the stopping level, positive upward, so that x_o = s, now it is v_o = √( 2gh ), v = 0, and x = 0. The acceleration is then

a = ( v² - v_o² ) / [ 2( x - x_o ) ] = ( √2gh )² / ( 2s ) = gh/s..

Then the force on the stopping floor is

F = ma = mgh / s,

but mg is the weight W of the floor(s), so that

F = Wh/s.

            While stopping, the floor feels a force of h/s multiplied by the weight of the portion of the building that is falling. The falling portion and the floor below each feel an equal but opposite force. This means that the floor stopping the falling, upper-building feels a force equal to the weight of the portion of the building above that stopping floor, multiplied by h/s. If h/s = 1 then the stopping force equals the weight that each floor of the building is designed to support. Like a rubber floor bending down and rebounding back. For the case that h = 5m and s = 0.5 m, we have F = 10W, which means that the floor has to support 10 times the weight of the building above it. Engineers and architects use safety factors of 5 or 20 or so. If the building's floors give like rubber then the impact is spread in time and occurs over a larger distance s, which makes the force F smaller. If the building's floors are very stiff and have less give, then the force is increased. For the case of s = 0.05 m = 5cm, we get F = Wh/s = 100W. In this case, the floor that tries to stop the falling portion of the building, has to support 100 times the weight of the building above it.

            By the way, Boston's 60-story, 800-foot tall (250 m) John Hancock Building was built with too little steel and so had a nauseating sway on windy days. Nobody would rent any room on the upper half of the building. Windows would also pop out and fall to the street below. To put an end to the sway, two 300-ton blocks were placed on the 58^th floor. As wind begins to move the building in one direction, one of the masses is moved in the opposite direction to keep the building's center-of-mass straight up. Purposefully moving the masses on a clam day will get the building swaying by one meter. After this adjustment was found to work, a few buildings were purposefully designed with too little steel so that they would cost less; they use a moving mass to keep the building upright.

Conclusion:

Notions of force, speed, and acceleration are valid in any and all systems. Nobody in the year 1687 would have imagined it would be used in systems from bones and skulls to skyscrapers and space ships.

Two-D forces and motions

For repeated use in identifying equivalent angles, let’s review that geometry thing of two parallel lines cut by a transversal.

                a . b

-----------.--------

         b . a

       a . b

-----.------------

b . a

The angles a are all equal, the angles b are equal, and we have a + b = 180.

In the case of a right triangle, the sum of the angles is 90°.

              N

                  ·

                         ·

                                  ·

                                                                              ●

                                            ·

                                                                 ●

                                                  ·

                                                                               ●

                                            ●

                                                 b·c

                                                                ·

                                                                               ●

             ●a

                                                                               ●

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

            We see that b + c = 90 and a + b = 90. These mean that c = 90 - b and b = 90 - a, so we conclude that c = a. We are often given angle a, and care only to know that angle c = a.

            Next we solve a number of problems. The Physics Classroom discusses inclined plane problems. Dr Finkenthal solves many problems from Tipler. Active Online Physics solves many problems.

Example:

A car sits on a hill inclined at an angle θ. (Like this coin on this tilted book ) What is the normal force? Here we choose to put the coordinate system such that the x axis is positive down the hill and the y axis is positive upward from the hill, perpendicular to the incline. The force of gravity points straight down to the center of the Earth but is an angle θ away from the positive y axis and so has x-component Wcos( θ ) and y component - Wsin( θ ). The normal force N is perpendicular to the surface and so has no x-component. The frictional force f points only in the negative x direction. Since the isn’t accelerating

Σ F_x = ma_x = 0,

or

+ Wsin( θ ) - f = 0.

This tells us that

f = Wsin( θ ) = mg sin( θ ).

We also have

Σ F_y = ma_y = 0,

or

+ N - Wcos( θ ) = 0.

This gives

N = Wcos( θ ) = mg cos( θ ),

which is reduced from the normal force that occurs on level land. The frictional force is

f = μ_S N = μ_S Wcos( θ ) = mg μ_S cos( θ ).

Comparing f in the previous line to f found above for the x-forces, we have

f / f = 1 = mg sin( θ ) / [ mg μ_S cos( θ ) ] = tan( θ ) / μ_S .

This tells us that μ_S = tan( θ ). A way to measure μ_S is to place the object on an incline and then increase the angle until the object begins to move. To measure μ_K, one measures the angle at which the object slides downhill at a constant speed. Here is another online animation. 

Example:

When the angle is increased until the object slide downhill, what will be the acceleration down the incline?

            From the previous problem, the y motion still gives

f = mg μ_S cos( θ ).

            The x motion is now

+ mg sin( θ ) - f = ma_x.

Inserting f gives

mg sin( θ ) - mg μ_S cos( θ ) = ma_x.

Or,

a_x = sin( θ ) - μ_S cos( θ ).

            Various parameters in this motion can be plotted online here.

            Remember that for a car driven on level ground, N = mg, so f = μ N = μmg. But the normal and the frictional force change when the car is driven on a hill. This means that as a car is driven from level ground onto an uphill section of road, the car is sort of pressed into the road, increasing N and hence f, while a downturn in the road means the car is slightly falling downhill and both N and f are decreased. Both N and f are increased going uphill and decreased going downhill. Picture that in your mind. Also picture that a high N will make the block move up the inclined plane.

Example:

A sign is hung from two ropes. Since the angles are equal, the tensions T are equal. The weight W points straight down.

T.......................................T

  .

                                  .

       .

                            .θ

             .

                  .

                 █

                 ↓

                W

For W = 400 N, find T for angles θ = 3°, 0.1°, and 10°.

Let’s use the usual x-y coordinate system,

y

┼ x.

We have,

Σ F_x = ma_x = 0.

Or,

Tcosθ - Tcosθ = 0.

This verifies the both tensions are the same. We also have,

Σ F_y = ma_y = 0.

Tsinθ + Tsinθ - W = ma_y = 0.

Or,

T = W/(2sinθ).

For θ = 10° we get T = 400/(2sin10) = 1151 N.

For θ = 3° we get T = 400/(2sin3°) = 3820N.

For θ = 0.1° we get T = 400/(2sin0.1) =115000 N.

            In your mind, feel the tension holding the sign and imagine how you would have to pull more tightly on a rope to try to make the rope more horizontal. We see that T→∞ for θ → 0, which means that the ropes can not be pulled to a fully horizontal position because the ropes would tear.

            This same system occurs when two ancient Egyptians use ropes to pull a rock across the ground. (How many person are needed to pull a large rock?) This system might also describe a hiker who crosses a ravine by throwing a rope across it and then sliding across or it might describe a person who ties a rope between a car bumper and a tree and then pushes at the rope's midpoint. This produces a huge force on the car that might pull it out of the mud.

            Biologists measure the drag force on fish while they are swimming in a laboratory channel through which water is moving at a controllable, constant speed.. Two wires are attached to the body of the fish. As the fish is pushed by the water, the wires become taught. The free body diagram given here for the sign then applies to the fish where W is replaced by the drag force D. The measured tension in a wire enables the drag force to be calculated.

CQ: Describe your mental image of string tension

Demonstration:

Hewitt suggests that we go outside, tie a rope between a large tree and the bumper of a car, and then push perpendicularly against the rope. Does this move the car more easily than occurs when pushing directly on the back of the car?

Demonstration:

The acceleration of a mass will always be in the direction of the net force. This surprises us sometime, as when pulling a spool of thread.

Example:

As you accelerate forward in your car, the lucky charm hanging from the rear-view mirror tilts back at a constant angle. Can you picture in your mind that the greater the acceleration, the greater will be the angle. The forces on the hanging mass █ are its weight W and the tension T in the string ●●●●.

T

.●......|

|.●.....|

|..●.θ.|

|.θ.●..|.

|......●

|.......█

|........|

|.......▿

|......W

We can tell that the acceleration is toward the left because the mass is hanging toward the right. Let’s take positive x to be toward the left.

..y

x┼ .

If the line representing the W is extended vertically upward, then we have two vertical lines | being cut by the transversal representing the string. This tells us that the angle θ also occurs between T and that new line. While accelerating horizontally along the road at a constant rate, this angle will be constant. There is no y-acceleration.

The free body diagram is

       T_y

        ↑ 

T_x ←██

        ↓

       W = mg

We have,

Σ F_x = Tsinθ = ma_x.

Or,

We also have,

Σ F_y = ma_y = 0.

Or,

Tcosθ - W = 0.

Taking T = W/cosθ = mg/cosθ from this and putting it into the x equation, gives

ma_x = (mg sinθ) / cosθ

or

tanθ = a_x / g.

For a_x / g = 1, we would have θ = 45° but that rate is not possible in a car. We typically accelerate at g/3, producing 18°. Try this in your own car. If your lucky charm has about the same mass as your vehicle, then the motion of the lucky charm will affect the motion of your vehicle.

Example:

While walking, we extend a foot forward and then place a heel on the ground. We are nearly vertical; there is typically an 81° angle between the ground and our leg. As our heel lands on the ground, it would slip if it were not for the friction of the ground.

   \ \

  θ \ \ / /

.......\\ / /

Since θ = 81°, cosθ=0.15.

Here is a free body diagram. The frictional force points to the left because it is stopping the foot.

           \

Friction █<---

            |

            W

When the foot is landing (or propelling), its horizontal force component is 0.15 W. So we need μ>0.15 or we will slip. Rubber on dry concrete has μ=0.7 to 1.0, but on icy roads it is μ = 0.1 to 0.2.

Student:

Describe the forces involved while a skateboarder performs an Ollie maneuver.

Student:

Describe the physics of sailing, parachuting, or swimming, gymnastics, or another sport.

            In your next physics course involving statics and dynamics, you’ll be ready to solve more complicated problems.

Student:

Discuss the forces in these situations and show the solutions.

Centripetal force

We move in circular arcs when driving around corners, making turns in airplanes, or taking amusement park rides. The centripetal force spins the water out of our clothes during the spin cycle of a washing machine. The centripetal force is at work in chemical and biological centrifuges, it throws mud off spinning tires, and results in the display of hands, feet, hands, feet, hands, feet of motorcyclist during an unplanned tumble at 50 mph.

            The geometry of similar triangles for Δr and Δv vectors tells us that Δr/r = Δv/v. Dividing each by Δt gives Δr/(rdt) = Δv/(vdt) or v/r = a/v which is rearranged as a_c = v² /r. Whenever an object moves in a circular motion, its acceleration has a magnitude a_c = v² /r that points radially inwards toward the center of the circle. Multiplying by the mass of the object gives the centripetal force

F_c = mv²/r.

The centripetal force = mass x tangential speed squared / distance from axis.

            Since objects want to continue moving in a straight line at constant speed forever, it takes a force–or a net force–to cause the object to move in a circle. (See also here and here.)

            Remember that speed equals distance divided by time. When moving with a period T in a circle of radius r, we have v = 2πr / T. The period is the time needed to travel in one complete cycle or revolution. The frequency is the number of revolutions per second. Frequency and period are inverses of each other, T = 1/f.

            Remember that an arc length s of a circle is given by s = θr. The circle here has a constant radius. Changes in arc length and related to changes in angle by Δs = rΔθ. Dividing this by Δt gives Δs/Δt = rΔθ/Δt. The term Δs/Δt is a tangential, linear speed, v = Δs/Δt, while Δθ/Δt is an angular velocity. We use the symbol ω = Δθ/Δt for angular velocity. The tangential and angular velocities are related by v = rω. The tangential and angular accelerations are related by a = rα. We then have for centripetal acceleration, the equivalent expressions

a_c = v²/r = ( rω )²/r = rω² = 4π²r/T² = 4π²rf².

            This means that centripetal force has the equivalent expressions

F_c = ma_c = mv²/r = mrω² = 4π²mr/T² = 4π²mrf².

            Whenever an object moves in a circle, the centripetal force is involved and the sum of the radial forces add up to ma_c. Such circular motion occurs when we swing a pail of water in a vertically-oriented circle without spilling, travel upside down in a roller coaster, or being thrown to one side while making a turn in a car. Can you estimate speed, tension, weight, and centripetal force in Jay Gorh’s video.

Example:

When we suddenly realize we are driving toward a wall, we have to choose whether to turn the car to a side or to slam on the brakes to stop the car. Which should we do?

            When we slam on the brakes, how far do we travel before friction stops our car? Given the values x_o = 0, x = ?, v = 0, v_o = given, a = F\m =μ_Kg, t = ? We use v² = v_o² + 2a( x - x_o ) to obtain

x = v_o² / (2μ_Kg).

            If we instead choose to turn the car then the frictional force has to supply the centripetal force. We have

F_c = μ_Kmg = mv²/r.

            Solving for r, we get

r = v² / (μ_Kg).

            Comparing r and x, we see that x = ½ r and so conclude that it is better to stop than to turn.

Example:

When riding in a car that is turning a corner, you feel a force that seemingly pushes you to one side. Depending on whether you are turning right or left, this force throws you either inward or against the wall of the car. With what speed must you drive so that the centripetal force matches your weight? We have

F_c = mv²/r = mg.

            Or,

v²/r = g.

            For a speed of 20 m/s (45 mph), we have F_c = mg when r = 41 meters. Our cars are unable to do this but a large cylinder rotating in space can have a_c = g, as in the next example.

Example:

Suppose we make a cylinder that has a radius of 1000 meters (name an object or building that is one kilometer away from your classroom), lift in into space, and then cause it to spin about its long axis to make "artificial gravity" for people living on the inside surface of this cylinder. At what speed should it rotate to give an acceleration of 1 g? How many revolutions per second is this? We have v²/r = g, or

v = √(gr) = √( 9.8 m/s² )( 1000 m ) = 100 m/s (225 mph).

            The cylinder will be rotating at a large speed. The period T is the time taken to make one revolution, which is

T = distance / speed = ( 2πr ) / v = (2π)(1000 m)/(100 m/s) = 63 s.

            The cylinder is spinning at a rate of one rev per 63 sec or about 1 rev / min. The people would stand with their feet on the inner surface of the cylinder and have their head toward the center of the cylinder. They would walk around the inner surface and feel exactly the same “weight” as occurs when walking on the surface of the Earth. However, imagine throwing a ball toward that center line. When the ball leaves your hand it will move in a straight line due to its inertia, but the cylinder would rotate below it. The ball would appear to curve behind as it travels toward the center of the cylinder. Like a ball tossed on a merry-go-round or the Earth rotating under an orbiting satellite. In the 1890s, there was a bit of a fashion to play billiards on rotating tables.

Example:

A Millisecond Pulsar star rotates at such great speed that its surface moves at more than 10% of the speed of light. He pulsar Terzan 5 has a radius of 16 km and a rotational frequency of 716 per second. What is its period? With what speed does its surface move? Calculate the ratio v/c. What is the centripetal force on a 1 kg object placed on its surface?

            The period is T = 1/f = 1/716 s. The surface speed is v = 2πr / T = 2πrf = 7.2 x 10⁷ m/s. Since the speed of light is 3 x 10⁸ m/s, we have v/c = ( 7.2 x 10⁷ m/s ) / ( 3 x 10⁸ m/s ) = 0.24. The centripetal force on a 1 kg object is humongous, F_c = mv²/r = 3.2 x 10¹¹ N.

Example:

A rock of mass 0.1 kg is tied to a 0.8 meter long string and then whirled once per second around in a horizontal circle of radius one 0.75 meter. What is the frequency? What is the centripetal force? What is the tension in the string? The only horizontal force acting on the rock is the tension in the string. This tension is supplying the centripetal force that is making the rock travel in a circle. In a snapshot of the motion, choose a coordinate system in which x is horizontal, with positive toward the person.

█· · · · · · · · ·

m

|-------r------|

The free body diagram is

█→ T.

We have

Σ F_x = T.

Since the object is in circular motion, we know from similar Δr/r and Δv/v triangles that the sum of the forces must add up to F_c = mv²/r. The frequency is given as f = 1 /s. Since the speed of the rock is v = 2πrf, we have

T = mv²/r = ( 0.1 kg )[ ( 2π )( 0.75 m )( 1 /s )]² = 2.2 N.

            Since the tension is supplying the centripetal force, we have F_C = T = 2.2 N. The faster the rock moves, the greater will be the tension. You will feel this tension if you whirl a rock in such a circle.

Example:

A car of mass 1000 kg is traveling at a speed v on a circular path of radius 150 m. If the coefficient of static friction between the tire and the road is 0.8, what can be the maximum speed of the car before it flies off the road? What is the centripetal force? The forces on the car are friction and its weight. We choose to place a coordinate system at the location of the car, with positive x inward. Instead of the string tension in the previous example, static friction supplies the inward force along the x axis. Since the car is moving in the y direction, kinetic friction occurs in that direction. But the car is not moving in the x direction, so static friction is involved in that direction.

█· · · · · · · · ·

m

|-------r------|

            The free body diagram is

N

↑

█→ F_S

↓

W = mg

            Once again, Σ F_y = 0, so N = W. In the x direction, we have

Σ F_x = F_S .

            Since the object is in circular motion, we know that the sum of the x forces must add up to F_c = mv²/r. Equating the centripetal force with the sum of the x forces gives

F_S = mv²/r.

            We also know that the frictional force F_S = μ_SN = μ_S mg. Equating these two expressions for F_S, we get

mv²/r = μ_S mg.

            Or,

v = √[ μ_S gr ] = √[( 0.8 )( 9.8 m/s² )( 150 m ) ] = 34 m/s.

            The centripetal force is then

F_C = μ_S mg = ( 0.8 )( 1000 kg )( 9.8 m/s² ) = 7800 N.

            If the speed of the car is greater than 34 m/s, the frictional force will not be able to keep the car moving in a circle of radius 150 meters. The car will travel outward. Another way to force a car into a circular motion, is to tilt the roadway so that a component of the normal force supplies the centripetal force, as done in the next example.

Example:

At what angle should a roadway curve be banked so that a car moving at a speed of 30 m/s (68 mph) will stay on a road having radius r = 800 meters?

            Orient the x-y coordinate system so that the centripetal force points entirely in the positive x direction, which here is toward the left.

                y

                |

                            ●

                          ●

                            ●

                        ●

                             ●

· N

              θ|

- x ◃· · · · ┼●· · · · · · · · · · · · · ▹ +x

                             ·

                             ●

                                                  ·

             ●

                     ·

                             ●

                     ↓

         ●

                     W

                             ●

     ●

        θ

                             ●

●●●●●●●●●●●●

            In this case, we choose to orient the x axis so that the positive direction points inward toward the center of the circle. This is convenient because the centripetal force Fc then lies entirely along the x axis and is positive. The car is not accelerating in the y direction so we have

 Σ F_y = 0.

Or

Ncosθ = W.

            The sum of the x forces are supplying the centripetal force.

Σ F_x = mv²/r .

Or,

Nsinθ = mv²/r.

            The ratio of these two results gives

tanθ = sinθ/cosθ = mv²/( Wr ) = v²/( gr ).

            So we get θ = tan^-1[ 302 /(9.8x800)] = 6.5°.

            Banking the roadway by 6.5° enables the inward or radial component of the normal force to supply the centripetal force need to move the car in a circle of radius 800 meters. No frictional force is needed in this case. Even in icy conditions where μ_S is almost zero, the car will stay on the road. The banking angle must be increased if the speed is increased or if the radius is decreased.

            The Volunteer Speedway in Bulls Gap, Tennessee has a 0.40-mile-long oval track that is banked at a 32° angle. Drivers make a loop in 12 seconds at 110 mph. Is friction needed in this case?

            Search the internet for video clips of motorcyclists traveling on the inside surface of spherical, “Globe of Death”cages that are a few meters in diameter. If the sphere has a radius of 2 meters and 65 kg drivers have a period of 1.3 seconds, what will be the centripetal force on the bottom of the tire?

F_c = ma_c = 4π²mr/T² = ( 4π² )( 65 kg )( 2 m ) / ( 1.3 s )² = 3000 N.

            The driver’s head and feet travel in circles of differing radii and so experience differing centripetal accelerations What does that feel like?

Example:

Make free body diagrams for each of these systems.

Example:

In one circus ride, people stand against the inside wall of a cylinder having a radius of 4 meters. There is friction between a person’s back and the inside wall. The cylinder begins spinning. When it is going fast enough, the floor is removed. The riders remain held against the wall like clothes spinning in a drier. For μ_S = 0.9, at what speed v can the floor be removed?

            The forces on the rider are downward weight W, upward frictional force f_S, and normal force N. The normal force provides the centripetal acceleration. The frictional force f_S = μ_SN keeps the rider from falling downward. The greater the rate of spin, the greater will be the force with which the rider is thrown against the wall and greater will be the normal force. In turn, as the normal force increases, so does the upward frictional force. If the wall were suddenly removed, ex-riders would continue moving in a straight line.

            At one “snapshot” instant, the free body diagram is

f_s

↑

█→

↓

W = mg

            We choose a coordinate system in which positive y is upward along the wall and positive x points inward toward the center so that the centripetal force lies entirely along the x axis.

            The sum of the x forces are supplying the centripetal force.

Σ F_x = mv²/r .

Or,

N = mv²/r.

            The rider is not accelerating in the y direction so we have

Σ F_y = 0.

Or,

f_S = W.

            Which can be written as

μ_SN = mg,

or N = mg / μ_S.

            Equating the two expressions for N gives,

mg / μ_S = mv²/r,

or

v = √[ rg / μ_S ] = √[( 4 m )( 9.8 m/s² ) / ( 0.9 ) ] = 6.6 m/s = 15 mph.

            Luckily, this speed is independent of the mass of the rider, otherwise some people would fall and some would not. How much time T needed for one revolution and what is the frequency f = 1/T? We have,

T time = distance / speed = 2πr / v = ( 2π )( 4 m ) / ( 6.6 m/s ) = 3.8 s.

            It is spinning at a frequency f = 1/ T = 1 / 3.8 s = 0.26 rev /s.

Example:

Roller coasters always include looping sections of track. If the speed of the car is too low, it will not travel a complete circle. At a higher speed, the car will make a complete loop. At a certain speed, the car will just barely reach the top of the loop; the normal force will be zero at that instant. Due to its inertia, the car wants to travel in a straight line but the normal force of the track causes it to instead travel in a circular path. (Roller coaster riders experience free fall at that instant.) If a section of the looping track was missing, the car would then move in a straight line. A NASA astronaut in space, once demonstrated the normal force that a circular track exerts on a looping car and the equal and opposite reaction force of the car on the track.

            A car travels in a vertically oriented, circular loop that has a radius of 20 meters. At what speed will the rider experience free fall for an instant? Since it is the normal force that makes the car move in a circle, we find the answer by setting N = 0 at the top of the loop. The free body diagram at the top of the loop is

█

↓

W+N

            With the positive axis y pointing downwards, we have

Σ F_y = mv²/r.

            Or,

N + W = mv²/r.

            Putting N = 0 and W = mg, we have

mg = mv²/r,

            or

v = √( gr ) = √[ ( 9.8 m/s² )( 20 m ) ] = 14 m/s =32 mph.

            As the car travels around the circle, the force W always points downward and the normal force always points inward. The sum of these two vectors continually points in a changing direction, as the does the net acceleration vector a_net = a_c + g. The car slows down while moving up the loop, and it speeds up while moving down the loop.

            In a looping airplane, the lift of the wing makes the plane move and in a circle and so provides the centripetal force. Discuss the force, accelerations, and vectors of a looping biplane, of water within a glass being spun in a vertical circle, or of a rock tied to a string and spun in a vertical circle.

CQ: Describe your mental image of the normal force.

Demonstration:

In this video clip, Harry Schroedinger’s cat named Rebound demonstrates the centripetal force, the Doppler effect, tension in a string, projectile motion, wall friction, the normal force, relativistic motion, buoyancy, radioactivity, torque, gravitational acceleration, and the conservation of momentum and energy.

Harmonic, periodic, or oscillatory motion

Oscillatory motion occurs in springs, pendulums, swings, and a point on the edge of a spinning wheel. There is an equivalence between rotational and simple harmonic motion. To see how this is true, paint a spot a spot on the edge of a rotating disk, spin the disk, and then view it from edge on. The painted spot will appear to move up and down in time as a cosine or sine function because by definition, the these trig functions are the x and y components of a spot moving along the edge of a circle. Harmonic motion follows cosine or sine functions.

            Hold and stretch these springs being passed around the classroom. You can tell that some springs are easier to stretch than are others. The stiffness of the spring is measured in terms of a spring constant k. You can also tell that the more you stretch the spring from its relaxed length, the harder it is to stretch it yet further. The force exerted by a stretched spring is found to be

F = - kx.

The force developed by a stretched or compressed spring = its stiffness or spring constant times the distance stretched or compressed from the equilibrium position of the spring.

            The minus sign indicates that the spring develops a force that opposes the stretch and points in the opposite direction of the stretch. The spring constant has units of force/distance or N/m.

CQ: What is the spring constant if a force of 10 Newtons is needed to stretch a spring 10 cm from its equilibrium position? k = F / x = 10 N / 0.1 m = 100 N/m.

            When F = -kx, Newton’s motion equation is

F = ma = - kx,

or

a = -(k/m)x = -ω²x.

            The resulting motion x(t) is oscillatory with angular frequency ω. This form of the acceleration equation occurs in every oscillatory system, but the expression relating ω to other physical parameters will vary from system to system. For the periodic motion of a spring, we have ω = √(k/m).

            When an object undergoes periodic motion, its position, speed, and acceleration are naturally described by the trigonometric cosine and sine functions. (This is what such functions sound like.)

            The cosine and sin functions oscillate between the extremes of positive and negative 1.0 or between ±A when the amplitude A is not equal to 1. The cosine function begins at the positive extreme, while the sin function begins at zero. If the position of an object through time is given by a cosine function, then its speed follows a negative sin function. This can be seen by drawing a cosine function across a paper and then drawing a -sin function below that. The slope of each point in the cosine function is seen to develop the -sin function. The slope of each point in the -sin function is seen to develop the original cosine function. We can then see that whenever we have F = -kx, or equivalently, a = -ω²x, we then have

x( t ) = A cos( ωt )

v( t ) = -A ω sin( ωt )

a( t ) = -A ω² cos( ωt )

and

F( t) = ma = -mA ω² cos( ωt ),

where

ω = 2πf = 2π/T and the amplitude is A. Visit here or here for animations and plots of x, v, and a. The animations clarify how the position, velocity, and acceleration each follow a sine function but are out of phase with each other.

            These graphical relations do produce a( t ) = - ω² x( t ) and describe oscillatory motion having angular frequency ω and period T. Such oscillations occurs in sound waves, musical instruments, a ball rolling back and forth within a bowl, springs, pendulums, and rattling metal strips.

CQ: Name something that undergoes periodic motion.

Example:

The position of a 2 kg mass oscillating oscillating on a spring is given by x( t ) = 0.25 cos( 4t ).

            What is its amplitude A, angular frequency ω, frequency f, period T, and spring constant k? Comparing the given expression for x( t ) to the general form x( t ) = A cos( ωt ), we see that the amplitude is A = 0.25 meters, angular frequency ω = 4 rad/s, frequency f = ω/(2π) = 0.64 /s, period T = 1/f = 1.6 s, and spring constant k = mω² = 32 N/m².

            What is the speed of the mass at t = 1.5 seconds? We have

v( t ) = -A ω sin( ωt ) = - ( 0.25 meters )( 4 rad/s ) sin( 6 radians ) = 0.28 m/s.

Be sure to calculate the sin function using radians and not degrees.

            Pendulums have been observed for centuries. In 1583, Galileo used pendulums to time the duration of events. He first got the “clocking” idea while watching the swinging motion of church chandeliers. To time the duration of events he also counted heartbeats or counted water drops emerging from a small hole in the bottom of a container. In 1657, Christian Huygens first made a clock out of a pendulum.

            The force acting on a pendulum when at the bottom of its swing are downward gravity W and inward string tension S. Using a coordinate system at one instant in time in which positive x points to the left and positive y points inward along the string, the centripetal force then lies entirely along the positive y axis.

The forces on the hanging mass █ are its weight W and the string tension S in the string ●●●●. The length of the string is L.

S

.●......|

|.●.....|

|..●.θ.|

|.θ.●..|.

|......●

|.......█

|........|

|.......▿

|......W

The sum of the y forces is

Σ F_y = ma_c = mv²/r .

Or,

S - mg cos( θ ) = mv²/r.

The speed v here is actually the tangential speed, which is in the x direction. By the way, this tells us that the tension in the string varies with θ as

S = mg cos( θ ) + mv²/r.

            (See also here.) Put yourself in that place, tugging harder near θ = 0 than at other angles. The sum of the x or tangential forces is

Σ F_x = ma_x,

or

-mg sin( θ ) = ma_x.

            Since sin( θ ) = x/L, we have

-mg sin( x/L ) = ma_x.

            When θ is less than about 10°, the common approximation is used in which sin( θ ) ≈θ or sin( x/L ) ≈x/L. This gives

-gx/L = a_x.

            Which is another case of the general oscillation equation a = -(k/m)x = -ω²x in which

ω² = g/L.

            The period of pendulum motion is seen to be

T = 2π √(l/g).

            The period increases with string length. Have you noticed that this happens? We see that long pendulums move more slowly than do short ones. The period also depends on the acceleration due to gravity, which changes from planet to planet.

            What do you suppose would be the motion of a pendulum swinging on a rotating platform? Its inertia keeps it moving in one plane as the platform spins under it. This also happens as the spinning Earth rotates under swinging pendulums, as explained by Foucalt.

Example:

On the Moon, the acceleration g_M due to gravity is 1/6 the value that occurs on the Earth, which is g_E = 9.8 m/s². How does the period of a pendulum of length L compare on the Moon and Earth?

            These sort of questions are always answered by forming a ratio and then cancelling all the identical factors and constants. We have

T_M / T_E = [ 2π √(l/g_M) ] / [ 2π √(l/g_E)] = √(g_M / g_E ) = √( 1/6 ) = 0.4.

            NASA has information and video of a pendulum while in motion on the surface of the moon. Just to wet your appetite for your next physics course that contains additional mathematical techniques, here is the motion of one pendulum hanging from a second pendulum, here is the motion of a pendulum whose support is undergoing circular motion, here is a multi-part pendulum hitting a cube of jello, and here is a the motion of a pendulum that has an elastic string. Here are some others. This is what a pendulum sounds like when the frequency of two instruments vary with its angles. Here is the motion of two masses connected by a spring. These systems are described by Newton’s equations.

            We mentioned above that Newton’s Universal Law of Gravitation Gm₁m₂/r², where G = 6.67 x 10^-11 Nm²/kg², gives the force between two particles of mass m₁ and m₂ that are separated by a distance r. Once you’ve calculated the force between two masses, there is nothing new to learn before you can calculate the net force that 100 or one million particles exert on a mass. When a system has 100 particles, the net force on each particle is found from the vector sum of 99 gravitational forces. It saves time to use a computer to add these vectors. Today’s computers can handle a system of 100,000 particles. (Five decades ago, computers could handle a system of only 10 particles, four decades ago they could handle 100, three decades ago it was 1,000, and one decade ago it was 10,000.) Modeling complex systems does not involve any additional physics beyond what you already know. In this course, the student learns about such things as mass, acceleration, force, vectors, and gravity so that he or she can go on to the next course to model and solve more complicated systems.

            Newton’s laws of motion and the conservation of momentum, energy, mass, and charge are all that is needed to make computer models of systems ranging from fluid flow past airplanes to beating hearts. The volume comprising the modeled system is dividing into small cubes containing the material of the system and then Newton’s laws and the conservation laws are applied to each individual cube and to its neighbors. Each cube affects its neighbors. For example, suppose fifty people stand side-by-side but one-step apart while holding the out-stretched hands of their two neighbors. If one man in the middle tugs on the hand of the neighbor to his right, he necessarily tugs also on the neighbor to his left and this tug will propagate down the line from neighbor to neighbor. Two- or three-dimensional systems are modeled using a mesh or grid of interacting neighbors, as shown in these NCHC models (Click ‘Results’) of a car hitting a wall, of an airplane, and of a tanker. Here is a 2D system of springs. Movie and game animations of people and animals apply Newton’s laws to each element of a grid, as is done in surgery simulations. After calculating the pressure, temperature, speed, force, and energy and such of each cube within the grid, those physical parameters can be visualized using color scales–for example, blue might represent a value between eight and nine while green represents a value between seven and eight. Here is colored grid of the NCHC model of a tanker. The math department of the Hamburg University has many animations of vehicle motion.

Review of force and inertia

Let’s discuss again the list of examples of inertia and the examples of action-reaction force pairs.

            All of physics is obtained from Newton’s Equation, F = ma. (Today, this is written in a relativistic, quantum-mechanical form.)

            Though our everyday experience has led us to believe that motion ceases when the driving force is removed–for example, when pushing a box across the floor–nature has objects continue in motion in the absence of a force. No force pushes a tossed object up an incline or pushes a tossed ball upward.

            We might push a ball so that it rolls up an incline. To do this, we apply a force with our hand to increase the speed of a ball so it can begin rolling upward. No continual force pushes the ball up the inclined plane. Consider again how v, a, and f vary as the ball moves up, slows, stops, and then moves back down the incline. It slows with a constant deceleration as gravity alone pulls it downward.

            We might toss a ball upward by applying a force with our hand to increase the speed of the ball so it can begin moving upward. No continual force pushes the ball up the inclined plane. Consider again how v, a, and f vary as the ball moves up, slows, stops, and then moves down again. It undergoes a constant, downward acceleration as gravity alone pulls it downward.

            We also found that forces occur in equal and opposite, action-reaction pairs. When a ball collides with a high speed car, the force of the ball on the car is equal and opposite to the force of the ball on the car.

            The class might like to discuss some of the motions described on the Illinois Institute of Technology website.