Applications of Force

We are now ready to look at some applications of Newton's laws. There are three types of forces that we will deal with in this course. The first is applied forces. The other two are natural forces and contact forces.

Applied Forces

Applied forces are those that arise due to the interaction of the object with some other object. Usually the two objects are in contact for a brief period. An example of this type of force is the force exerted by a person pulling a block or the tension in a line. These forces always depend on the situation being examined.

Natural Forces

Natural forces are those that arise from fundamental interactions in nature. In reality, both direct action forces and contact forces are applications of natural forces. There are four known natural forces. From our knowledge of particle physics we know that all of the natural forces can be characterized as a force field times a charge. Each charge is associated with a particle.

The first of the natural forces to be studied was the gravitational force. This is the force that attracts to bodies together. It creates weight and keeps the planets in orbit around the sun. It has an infinite range. The equivalent charge is the mass of the body, and the associated particle is called the graviton. It is the only one we will look at this semester.

The second force that was known classically is the electromagnetic force. This is more commonly recognized as electric forces and magnetic forces. As with gravity, it has infinite range. The charge is the electric charge, and the associated particle is the photon. It is studied in the second semester of this course.

The third force is the strong nuclear force. This is the force that holds an atomic nucleus together. It has a limited range (roughly the diameter of a nucleus). The charge is the strong nuclear charge, or color charge, and the associated particles are called gluons.

The last known natural force this the weak nuclear force. This is another short ranged nuclear force. It is responsible for the decays of certain particles. The charge is the weak nuclear charge, or hypercharge, and the associated particles are called the gauge bosons.

Since the early 1970's there has been considerable effort to unify these four forces into a single theory that describes the forces as a single force. To date, only the electromagnetic and weak forces have been unified, although there are various theories, called grand unified theories, that unite electromagnetic, strong and weak forces. Attempts to unify these with gravity are found in the various superstring theories, or the theories of everything, that are being postulated.

Contact Forces

The last class of forces we will deal with are the contact forces. These forces arose due to the contact of one object with another. They can be decomposed into two components. The normal force, which we encountered earlier, acts perpendicular to the contact surfaces. The parallel component is the friction force and acts to either hold the objects at rest or to resist their motion. In general, the true nature of the friction force is complex and difficult to describe. However, there are certain general properties which can be described and serve as an easy means to describe the force. The main property is that the magnitude of the friction force is proportional to the magnitude of the normal force. The constant of proportionality depends on whether the objects are moving relative to each other or not.

If the objects are moving relative to each other, then the magnitude of the friction force is given by

 

 

f = mkN

(6.1)

 

where mk is called the coefficient of kinetic friction. The coefficient of kinetic friction is a property of the object and must be determined based on its composition. Don't forget that the friction force acts parallel to the contact surfaces and in the direction opposite the motion.

If the objects are at rest, then the friction force is approximately proportional to the normal force. Its magnitude can range from zero up to a maximum of msN. Thus the friction force is given by

 

 

(6.2)

 

where ms is the coefficient of static friction. The equality holds only when the object is about to begin motion. In most cases, when sliding begins, the friction force decreases. This means that, in general, the coefficient of kinetic friction is less than the coefficient of static friction.

Examples with Zero Acceleration

Example:

A toboggan loaded with vacationing students slides down a long, snow-covered slope having a coefficient of kinetic friction mk. The slope has just the right angle to make the toboggan slide with constant speed. Find the angle of the slope.

 

 

Since the toboggan is sliding at constant speed, the acceleration is zero. We chose a frame of reference that is parallel to the slope. In this frame of reference, the force equations are

 

|| component: mg sinq - mkN = 0

 

^ component: N - mg cosq = 0

 

Look at the || equation first

 

 

Substituting this into the ^ equation and solving for q yields

 

 

Thus, a constant velocity is achieved only if the tangent of the angle is equal to the coefficient of kinetic friction, regardless of the number of people on the toboggan.

 

Example:

A 75 kg man hangs from the middle of a tightly stretched rope so that the angle between the rope and the horizontal direction is 5. What is the tension in the rope?

 

 

 

Since the two sections of the rope are symmetrical with respect to the man, the tensions must have the same magnitude. Thus, summing the forces in the y direction yields

 

 

Examples with Acceleration

How do we go about solving force problems with acceleration? As always, we follow the steps outlined in the first chapter, taking special attention to the magnitudes and directions of the various forces involved. We then set up the vector equations using F = ma, and solve for the acceleration and subsequent motion.

 

Example:

An elevator and its load have a total mass of 800 kg. Find the tension T in the supporting cable when the elevator, originally moving downward at 10 m/s, is brought to rest with constant acceleration in a distance of 25 m.

 

 

We first need to find the acceleration. This can be found from the equation

 

 

All of the forces are in one dimension, so we get

 

 

Notice that the tension is greater than the weight. This is the cause of the upward acceleration. Also note that the relative signs on the acceleration and gravity were accounted for in setting up the equation, so we use the positive values when substituting.

 

Example:

A block of mass m1 moves on a level surface with a coefficient of kinetic friction m. It is connected by a light (massless), flexible cord passing over a small, frictionless pulley to a second hanging block of mass m2. What is the acceleration of the system, and what is the tension in the cord connecting the two blocks?

 

 

We treat each block as a separate system. Starting with the block on the table we get

 

 

x component:

y component:

 

Solving for the acceleration we get

 

 

The forces acting on the second block are

 

 

x component: 0

y component: T m2g = -m2a

 

Substituting the previous result for a, we get

 

 

The acceleration can be written in terms of the masses by substituting this back into the acceleration equation.

Circular Forces

As continued examples of the applications of Newton's laws, we now turn to circular motion. Recall that we saw that if an object was undergoing circular motion at a constant speed, it experienced centripetal acceleration which had a magnitude of

 

ac = v2/r

 

By Newton's laws, this acceleration is related to a force. This "force" is called the centripetal force and is given by

 

F = mv2/r

 

pointing towards the center of the circle. Notice that the centripetal force is not a physical force. The actual force used on the left hand side of Newton's law must be supplied by some external force.

 

Example:

A 2 kg rock is spun at the end of a 3 m rope in a horizontal circle. If it has a constant speed of 5 m/s, what is the tension in the rope?

 

 

The only true force acting on the rock in the horizontal direction is the tension in the rope. This must be the source of the centripetal acceleration, so

 

 

Example:

A car of mass m is rounding a flat curve with a coefficient of static friction m. If the curve has a radius of r, what is the maximum speed that the car can go before it begins to slip?

 

 

If there were no friction on the road, the car would continue to move in a straight line. The friction between the tires and the road cause the car to turn. Hence, we see that the centripetal force is caused by the friction force. Since the road is flat, the normal force is just equal to the weight of the car, as so

 

 

Example:

At what angle should the road be banked so that friction is not needed to keep the car on the road?

 

 

Splitting the normal force into components, we get

 

x component: N sinq = mv2/r

y component: N cosq = mg

 

dividing the first equation by the second one, we get

 

tanq = v2/gr

 

If the curve has a radius of 100 m and is part of the interstate (with a legal speed of 29.0563 m/s), then the banking angle should be

 

 

For horizontal circular motion it is possible to have the speed be a constant. However, if the circle is vertical, we must also include the tangential acceleration due to gravity.

 

Example:

What is the minimum speed that a roller coaster can go in order to stay on the tracks at the top of a loop if the loop is 20 m tall?

 

 

The forces acting on the car at the top of the loop are its weight mg, and the normal force of the tracks against the car N. These must equal the centripetal force.

 

mg + N = mv2/r

 

If the velocity is to be the minimum, then the normal force would be zero, so this becomes

 

 

Example:

A rock is whirled at the end of a rope in a vertical circle. Find a general expression for the tension T. What is the magnitude of the total acceleration?

 

 

Let the rock have a mass m, the rope have a length r, and measure the angle from the bottom of the circle. Then for an arbitrary angle q, the forces are

 

tangential:

normal:

 

We can find the tension from the normal component,

 

T = m(v2/r + g cosq)

 

The magnitude of the total acceleration is

 

 

Notice that the tension is greatest at the bottom of the loop, and is smallest at the top.