Chapter 3: Forces and Linear Motion

D. Mass and Weight

E. Standard Units of Measurement

F. Applications of Newton's Second Law

G. The Normal Force

H. Motion on a Frictionsless Incline Plane

I. Frictional Forces

Examples

Problems
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Our discussions in the first two chapters focused on kinematics: the study of motion exclusive if what causes the motion. In this chapter we turn our attention to the relationship between motion and the forces affecting motion. This study is called kinetics and begins with a statement and discussion of Newton’s three laws of motion.

A. Newton’s First Law of Motion

Newton’s first law of motion can be stated simply as:

A body continues to move at constant velocity unless acted on by a force.

It is important that we remember what it means when we say that a body is moving with constant velocity: it is either at rest or moving in a straight line with constant speed.

B. Newton’s Third Law of Motion

We state the third law of motion in terms of two objects and the forces they exert on each other. In these terms we say:

When one object exerts a force on a second object, the second object exerts an equal but opposite force on the first.

The third law is often called the law of action-reaction.

C. Newton’s Second Law of Motion

The first and third laws are rather simple statements of behavior which are familiar to all of us. They seem quite reasonable, even to the uninitiated. The second law is not so obvious. This is because it is a precise mathematical statement of relationships between quite different physical quantities. In addition, this statement introduces the concept of mass. Though not really complex in itself, the concept of mass frequently confuses students as being the same as weight of an object.

The second law is simple to illustrate qualitatively. If one pushes a cart it is obvious that the harder one pushes, the larger the acceleration if the cart. Quantitatively it is easy to see that this acceleration is directly proportional to the force. We can write

a ~ F

where “~” is read “is proportional to.”

We also recognize that the acceleration of an object depends upon the massiveness of the object. Each object possesses a quality we call inertia. All objects tend to remain at rest unless an unbalanced force acts on them. Moreover, an object in motion tends to remain in motion unless an unbalanced force acts on it. We say each object possesses inertia. The inertia of an object is related to its massiveness. For example, a football coach chooses very massive players for the line since they are not easily knocked out of position. The coach knows that massive objects have more inertia than less massive objects. They are more difficult to set in motion if they are at rest, and they are more difficult to stop if they are already in motion.

This property of massiveness or inertia must enter into determining the acceleration produced by a force. We see that the acceleration produced by a given force varies inversely with the massiveness of the object. We write

a ~ 1/m

where m is a measure of the massiveness of the object. We call m the mass of the object.

We have found that a ~ F when m is constant and a ~ 1/m when F is constant. These two proportionalities can be combined into one relation:

a ~ F/m

If mass and force are measured with consistent units, the proportionality can be removed and we can write Newton’s second law as an equation.

(3.1)

This is a mathematical statement of Newton’s second law. In writing this we should note that F is the resultant force on the object whose mass is m. Recall, also, that m is a measure of the inertia of the object. It measures how hard it is to set the object into motion or to stop it if it is already moving.

D. Mass and Weight

At this point you are likely to associate the mass of an object with its weight. There is a relation between the two, of course, since massive objects are heavy. But weight and mass are not the same even though they are related.

Mass, as a measure of the inertia of a body, tells us how much matter is contained in a body

The weight of a body is the force of gravity acting on the body.

We can find the relationship between weight and mass by considering the simple example of free-fall. The object shown is falling freely on the earth under the effect if the earth’s gravitational pull upon it. We know that its acceleration is the free-fall acceleration due to the gravity g. Moreover, we know what an unbalanced force is on the object. It is the pull of gravity on it and we call this force the weight of the object. Therefore, the resultant force on the object is W the object’s weight. Let us now place these values in Eq (3.1). We find:

(3.2)

This is a very important equation. It tells use the relation between mass and weight of an object.

E. Standard Units of Measurement

The system of units based on meter, kilogram, and second is called the mks system. In addition to the units of length, mass, and time, we need other units. One of the most important is the unit of force. The fundamental unit of force is defined in terms of Newton’s second law, F =ma. This definition is easily made in the mks system. We call the unit in this system the Newton (N).

A Newton is the unbalanced force which will give a 1 kg mass an acceleration of 1 m/sec².

Writing this in terms of F = ma, we have

so the Newton is equivalent to the 1 kg • m/sec².

A second system uses centimeter, gram, and seconds as fundamental units. In this, the cgs system, the unit of force is called the dyne.

A dyne is the unbalanced force which will give a 1 gram mass an acceleration of 1 cm/sec².

You can show by converting 1 kg • m/sec² to gram-centimeters per second squared that

1 N = 100,000 dyne

The unit of force in the British system is the pound (lb). It is defined in terms of the Newton as

1 lb = 4.45 N

At present, the pound is fast becoming obsolete except in the United States.

We need yet to discuss the British unit for mass. It is defined in such a way that F = ma is fundamental. The unit of mass in the British system is called the slug.

A pound is the unbalanced force which will give a 1 slug mass an acceleration of 1 ft/sec².

You can also show by converting units that

1 slug = 14.6 kg

In most of the examples and problems in this book we will use the British or mks systems of units.

F. Applications of Newton’s Second Law

In this section we will see how Newton’s second law is used in practical applications.

1. Find the force required to accelerate a 50 kg mass at 20 m/sec².

2. Convert 1000 N to pounds

3. A force if 10 lb applied to the block below produces an acceleration of 2 ft/sec². Find the mass and weight of the block.

4. A hockey puck is accelerated from rest with a force of 50 newtons applied for 0.1 seconds. If the puck has a mass of 200 grams (0.2 kg) what is the final velocity of the puck?

At first glance this seems to be a simple problem of motion in one dimension like the problems we discussed in Chapter 1. First list the information given about the motion of the puck.

Not enough information is given to find v_F since we only know two variables of motion, but we can use other information in the problem to determine the acceleration. Drawing a picture always helps

Now list the information given and determine about the motion of the puck.

We can now use Eq. (1.4) to determine the final velocity.

G. The Normal Force

Examine an object of mass m at rest on a level plane. We see that the object is not accelerating, a = 0. As a result, the second law tells us that the resultant (or net) force on the object is zero, F_net = 0. Although there is no net force on the object we do know that there are forces acting on the object. There is obviously the weight of the object, which is the force of gravity acting down. But the second law tells us that since there is no acceleration the net force is zero. As a result, there must be a second force equal in magnitude and opposite in direction to the weight of the object. As a result, the net force will be zero. This force is called the normal force, R.

The normal force is the perpendicular force with which the supporting surface pushes up on an object.

It is a direct result of Newton’s third law, and is the component of the object’s weight perpendicular to the surface.

H. Motion on a Frictionless Incline Plane

If we a mass m slides down a frictionless inclined plane as shown below, we observe the following:

1. The mass accelerates.

2. The acceleration is less than gravity.

3. The larger the angle q (i.e., the steeper the incline) the greater the acceleration.

4. The mass remains on the incline.

From the first observation and Newton’s second law we can deduce that there is a net force acting on the mass. The second and third observations tell us that this acceleration depends on the angle of inclination.

We can explain the motion on the inclined plane by examining the forces acting on the mass. The two forces acting on the mass are the normal force R and the force of gravity in it (its weight). The normal force acts perpendicular to the surface on the incline, while the weight of the mass acts straight down. The weight can be broken down into two components: one parallel to the incline (Wsinq) and the one perpendicular (Wcosq).

The fourth observation tells us that the net force perpendicular to the surface must be zero. Thus

R = Wcosq = mgcosq

We also recognize that the component of the weight parallel to the incline is unbalanced. Thus there is a force down the incline.

Force down the incline = Wsinq

This unbalanced force results in an acceleration down the incline. By the second law F = ma. So

F = ma = Wsinq = mg Wsinq

a = g Wsinq

Thus the acceleration down a frictionless inclined plane is given by gsinq.

Example: A bobsled slides down a slope with a 30° incline. Find the acceleration of the bobsled.

Question: Why is it easier to push a mass up an incline than lifting it straight up?

Example: The Flying Kilometer is a ski hill in Italy that skiers use to try to break the record for speed on skis. The hill is one kilometer long and is inclined at 30 degrees. Assume a skier starts at rest. Find his velocity at the bottom of the hill.

This appears to be a simple one dimensional kinematics problem. First list the information given about the motion if the skier including the acceleration due to the incline.

We can now use the Eq.1.7 to determine the final velocity.

Observation: Although we have done the problem correctly, the result is not believable. It does not seem possible that a person could travel at this speed on a pair of skis. Clearly, something is missing in the stated problem. It is friction--a force that always opposes motion. It is interesting to note that the world records for speed skiing are greater than 140 mph. In 1997 the women’s record of 142 mph was set by Karine Dubouchet and the men’s record was 150 mph set by Philippe Billy. At these speeds serious burns may result during a fall. How can you burn yourself on snow?

I. Frictional Forces

Let us now investigate the frictional force that retards the motion of an object as it slides along a surface. Begin by applying a force to a block of wood that is resting on a rough surface. Start with a very small force and slowly increase it. You will find that the block does not move until this applied force reaches a certain critical value.

In other words, frictional forces exist even where there is no motion. We see from this that there is a maximum friction force that resists the onset of motion. Since there is no acceleration of the block until the critical force is attained, the net force on the block is zero during this time. As a result, this frictional force is equal in magnitude to the applied force and the maximum frictional force is equal in magnitude to the critical force necessary to begin motion. The frictional force acting between surfaces at rest is called a force of static friction. We define

f_s = force of static friction

Now consider what happens after the block begins to slide.

You will notice that a force significantly less than f_s is needed to keep it moving at a constant velocity. Since the velocity is constant the acceleration is zero and the net force is then zero. So there must be a force equal in magnitude and opposite in direction to the applied force that resists the motion of the block. The frictional force acting between surfaces moving with respect to one another is called a force of kinetic friction. We define

f_k = force of kinetic friction

And we note that

f_k £ f_s

The reason why the static force is greater than the kinetic force is easily understood by thinking of the two surfaces in contact with one another. When the block is at rest the rough surfaces penetrate each other and resist sliding. However, when motion begins the surfaces do not have time to settle into each other completely.

There is a rather simple experimental law which allows us to estimate the frictional force. The magnitude of the frictional force should depend on the nature of the two surfaces involved and the force of contact between the two surfaces. This force of contact is simply the normal force R, which measures the force with which the supporting surface pushes on the sliding object. The nature of the two surfaces is accounted for by introducing a coefficient of friction m (Greek letter “mu”) which depends on the two surfaces. The result is that the frictional force (static or kinetic) resisting the motion if a sliding object is given by

Force of static friction = f_k = m_kR and Force of kinetic friction = f_k = m_kR

(3.3)

Where the coefficient (static or kinetic) of friction is μ. It is found that the coefficient of friction μ varies widely from surface to surface. A few typical values are given below.

Surfaces	Coefficient of static friction = m_k	Coefficient of kinetic friction = m_k
Rubber and concrete (dry)	1.0	0.8
Rubber and concrete (wet)	0.5	0.4
Steel on steel (dry)	0.7	0.6
Steel on steel (lubricated)	0.15	0.07
Wood on snow	0.14	0.10
Steel on ice	1.0	0.8
Synovial joints	0.01	0.003

The joints between your bones are covered with a material called the synovial sheath. This sheath secretes a lubricant which greatly reduces the friction between the joints

Examples:

1. A 5N force is just able to start a 2 kg mass in motion. Find R, f_s, and m_k. We assume in this problem that the surface is horizontal so

R = W = mg

R = (2 kg) (10 m/sec²) = 20 N

Since the 5N force just starts motion it is the critical force equal to f_s. Hence,

We may now use Eq. 3.3 to find μ_s

2. A 200 lb refrigerator rests on a level surface. If the coefficient of static friction is 0.5, find the force required to start the refrigerator sliding.

Since the surface is level R = W = 200 lb. By definition the critical force required to start the sliding is f_s.

Thus we must apply a force of 100 lb to start the refrigerator sliding.

3. Suppose a force of 150 lb is applied to the refrigerator on the previous example. If m_k = 0.25 find f_k, the net force, and the acceleration of the refrigerator.

First find Fmk. From Eq 3.3.

Now find the net force by totaling all horizontal forces.

Thus there is a net unbalanced force of 100 lb to the right. By Newton’s second law there is an acceleration to the right. So we just need to use F_net = ma. Now be careful! Although we know the weight is 200 lb we have yet to determine the mass. Recall Eq. 3.2.

Now we can find the acceleration

4. As a final example consider a 10 lb object sliding down a 30° incline at constant velocity. First identify and draw all forces, then find F_net, R, and f_k. Begin by drawing a picture which includes all information we have been given.

We know the weight of the object gives a force vertically down. This force can be broken into a component parallel to the surface, W sinq and a component and a component perpendicular to the surface W cosq. The surface pushes on the sliding object with a normal force R. Finally, the force of kinetic friction resists the motion.

The net force on the object is easily determined in this problem. Since the velocity is constant there is not acceleration. From the second law we must have

F_net = ma =0

The normal acceleration is also easy to determine. Since there is no acceleration perpendicular to the surface, R is equal and opposite to the component of the weight perpendicular to the surface.

R = Wcosq = (10 lb) cos30°

R = (10 lb) (0.866) = 8.66 lb

Finally, finding f_k is a little more difficult. Our usual method is to use Eq 3.3 (f_k = m_kR), but in this problem we are not given m_k. We must look at the forces to find f_k. Examine the force parallel to the incline. There is a component of the weight Wsinq down the incline. The friction force f_k opposes the motion and is directed up the incline. Since the net force is zero these two forces must be equal and opposite.

f_k = Wsinq = (10 lb) sin30°

f_k = (10 lb) (0.5) = 5 lb

Summary

Newton’s Law of Motion Normal Force

Inertia Inclined Plane

Mass Static Friction

Weight Kinetic Friction

Standard Units Coefficient of Friction

Problems

1. What force is necessary to accelerate a 25 kg mass at 8 m/sec²?

2. What is the weight of a 17 kg mass?

3. What is the mass of a 224 lb weight?

4. Convert your weight from pounds to Newtons.

5. Determine your mass in kilograms.

6. What force is necessary to accelerate a 224 lb weight at 5 ft/sec²?

7. A net force of 10 lb produces an acceleration of 2 ft/sec²: What is the mass of the object? What is its weight?

8. A car traveling 20 m/sec (45 mph) is involved in a collision and stops over a distance of 1.8 meters (6 ft). (This is a comparatively clam collision.)

What is the deceleration of the car during this collision?
What force is necessary to hold a 70 kg (154 lb) adult in his seat during the collision?
Assume an adult is wearing his seatbelt and holding a 15 kg (33lb) child on his lap. What force must the adult exert to hold this child during the collision? Could you exert this large of a force?

9. A skier of mass 50 kg skis down a frictionless hill. The hill is sloped at an angle of 37°.

What is her acceleration?
What is her weight?
What is the net force parallel to the hill acting on her?
What is the normal force on her?
What is the net force perpendicular to the hill acting on her?

10. For the three cases sketched below the block has a mass of 20 kg and the coefficient of kinetic friction is 0.25. Each block is moving with an initial velocity of 8 m/sec to the right. For each case determine the net force acting on the block, including the direction of this force, and describe the acceleration.

11. The coefficient of static friction between a sled and a particular surface is 0.4. If the maximum force a person can exert is 200 lb, what is the maximum weight sled that can be started in motion on a level surface?

12. A hockey puck is sliding on the ice and slowing down. Identify all force on the puck.

13. A force of 5N just starts a 2 kg mass moving on a horizontal surface.

What is the force of static friction?
What is the normal force?
What is the coefficient of static friction?

14. Find the force required to hold a 200 lb refrigerator stationary on a 15° incline. Assume the refrigerator is on rollers and friction can be ignored.

15. Find the force required to push a 200 lb refrigerator up a 15 incline at constant velocity if

the refrigerator is on rollers and friction can be ignored.
the coefficient of kinetic friction is 0.25.