Figure 6.1. Static Friction.

## 6.1. Static and Kinetic Friction

**Suppose that a horizontal force F is applied to a block resting on a rough surface (see Figure 6.1). As long as the applied force F is less than a certain maximum force (F**

_{max}), the block will not move. This means that the net force on the block in the horizontal direction is zero. Therefore, besides the applied force F, there must be a second force f acting on the block. The force f must have a strength equal to F, and it must be pointing in the opposite direction. This force f is called the friction force, and because the block does not move, we are dealing with

**static friction**. Experiments have shown that the force of static friction is largely independent of the area of contact and proportional to the normal force N acting between the block and the surface. The static friction force is

f <= u

where u_{s}N_{s}is the

**coefficient of static friction**(which is dimensionless). The coefficient of static friction is approximately constant (independent of N). The maximum force that can be applied without moving the block is

F

Once the block has been set in motion, the force F needed to keep it
in motion with a constant velocity is usually less than the critical force
needed to get the motion started. In this situation we are dealing with
_{max}= u_{s}N**kinetic friction**and the friction force f

_{k}is given by

f

where u_{k}= u_{k}N_{k}is the

**coefficient of kinetic friction**. The kinetic friction force is independent of the applied force, but always points in the opposite direction. The equation for f

_{k}is

**not a vector equation**since f

_{k}and N do not point in the same direction.

**Note:**The friction between car tires and the road is static friction. This friction is crucial when you try to stop a car. Since the maximum static friction force is larger than the kinetic friction force, a car can be stopped fastest if we prevent the wheels from locking up. When the wheels lock up, the friction force is changed to kinetic friction (the tires and the ground are moving with respect to each other) thereby reducing the acceleration and increasing the time and length required to bring the car to a halt.

**Sample Problem 6-1**

Figure 6.2 shows a mass m on an inclined slope. At a certain angle [theta] the mass begins to slide down the slope. Calculate the coefficient of static friction.

Figure 6.2. Coordinate System used in Sample Problem 1.

Figure 2 shows the coordinate system used in this problem. Note that with
this choice of coordinate system, the normal force N and the friction force f
have each only one component; N is directed along the y-axis and f is directed
along the x-axis. Since this is the maximum angle at which the object will
remain at rest, the friction force has reached it maximum value:_{ }

_{ }

_{ }

_{ }

_{ }

**Note**The friction force between car tires and the road is reduced when the car travels uphill or downhill. It is harder to drive uphill or downhill when the roads are slick than it is to drive on leveled surface.

Figure 6.3. Free-Body Diagram for Sled.

**Sample Problem 6-3**

A woman pulls a loaded sled (mass m) along a horizontal surface at constant speed. The coefficient of kinetic friction between the runners and the snow is u

_{k}and the angle between the rope and the horizontal axis is [phi] (see Figure 6.3). What is the tension in the rope ?

Since the sled is moving with a constant velocity, the net force on the sled must be zero. Decomposing the net force into its components along the x-axis and the y-axis, we obtain the following equations of force:

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

Figure 6.4. Eraser on the Black Board.

The second equation tells us that the static friction force
f_{s}must be equal to W. This implies the following for the normal force N:

_{ }

_{ }

Figure 6.5. Problem 25P.

**Problem 25P**

In Figure 6.5,

*A*, and

*B*are blocks with weights of 44 N and 22N, respectively. (a) Determine the minimum weight of block

*C*that must be placed on

*A*to keep it from sliding if u

_{s}between

*A*and the table is 0.20. (b) Block

*C*is suddenly lifted off

*A*. What is the acceleration of block

*A*, if u

_{k}between

*A*and the table is 0.15 ?

A weight (block C) is placed on top of block A and prevents it from sliding. The acceleration of the system is therefore 0 m/s

^{2}. Consequently, the net force on each block is equal to 0 N. In order to determine the minimum weight of block C required to accomplish this, we start evaluating the net force on each block. The forces acting on block B, the weight W

_{B}and the tension T, are schematically indicated in Figure 6.6. The net force acting on block B is directed along the y-axis and has a magnitude equal to

Figure 6.6. Forces acting on block B.

_{ }

T = W

The forces acting on block A and block C are indicated in Figure 6.7.
The net force acting in the y-direction is zero and thus_{B}
N = W

Since the system remains at rest, the net force acting on block A and
C along the x-direction must also be zero. This means that the static friction
force f_{A}+ W_{C}_{s}must be equal to the tension T. Experiments show that f

_{s}has a maximum value which is determined by the normal force N and the static friction coefficient u

_{s}

f

The minimum weight of block C that will prevent the system from
slipping can be found by requiring that_{s}<= u_{s}N = u_{s}(W_{A}+ W_{C})
u

and thus_{s}(W_{A}+ W_{C}) >= f_{s}= T = W_{B}_{ }

Figure 6.7. Forces acting on block A and C.

When block C is removed the static friction force is changed (since
the normal force is changed). The maximum static friction force is now
u_{s}W

_{A}= 8.8 N which is less than the weight of block B. Obviously block A will slip and both blocks will accelerate. At this point the friction force acting on block A is the kinetic friction force f

_{k}whose magnitude is equal to

f

The net force acting on block A is given by_{k}= u_{k}N = u_{k}W_{A}_{ }

_{ }

_{ }

Figure 6.8. Problem 36P.

**Problem 36P**

Two masses,

*m*

_{1}= 1.65 kg and

*m*

_{2}= 3.30 kg, attached by a massless rod parallel to the inclined plane on which they both slide (see Figure 6.8), travel along the plane with

*m*

_{1}trailing

*m*

_{2}. The angle of incline is 30deg.. The coefficient of kinetic friction between

*m*

_{1}and the incline is u

_{1}= 0.226; that between

*m*

_{2}and the incline is u

_{2}= 0.113. Compute (a) the tension in the rod and (b) the common acceleration of the two masses. (c) How would the answers to (a) and (b) change if

*m*

_{2}trailed

*m*

_{1}?

The forces acting on mass m

_{1}are schematically shown in Figure 6.9. The x and y-components of the net force acting on m

_{1}are given by

Figure 6.9. Forces acting on m

_{1}._{ }

_{ }

_{1}must therefore be equal to m

_{1}g cos([theta]). This fixes the kinetic friction force

f

Mass m_{1k}= u_{1k}N_{1}= u_{1k}m_{1}g cos([theta])_{1}will accelerate down hill with an acceleration a. The acceleration a is related to the x-component of the net force acting on mass m

_{1}

_{ }

_{2}are schematically shown in Figure 6.10. The friction force f

_{2k}acting on mass m

_{2}can be determined easily (see calculation of f

_{1k}):

f

The x-component of the net force acting on mass m_{2k}= u_{2k}N_{2}= u_{2k}m_{2}g cos([theta])_{2}is given by

_{ }

_{2}

_{ }

Figure 6.10. Forces acting on m

We now have two equations with two unknown (a and T). Eliminating the
tension T from these two equations we obtain the following expression for a_{2}._{ }

^{2}. The tension T in the rod can now be determined easily

_{ }

_{1}and mass m

_{2}are reversed, we will still obtain the same acceleration, but the tension in the rod will be negative (which means that the rod is being compressed).

## 6.2. Drag Force

**The friction force we have discussed so far acts when two surfaces touch. The force that tends to reduce the velocity of objects moving through air is very similar to the friction force; this force is called the drag force. The drag force D acting on an object moving through air is given by**

_{ }

Figure 6.11. Drag Force.

Because of the drag force, a falling body will eventually fall with a
constant velocity, the so called terminal velocity v_{t}. When the object is moving with its terminal velocity v

_{t}the net force on it must be zero (no change in velocity means no acceleration). This occurs when D = mg, and the terminal velocity v

_{t}has to satisfy the following relation:

_{ }

_{t}is calculated to be

_{ }

_{t}shows that the terminal velocity of an object increases with a decreasing effective area.

The terminal velocity of an object is the final velocity it obtains during free fall. The object will obtain this velocity independent of whether its initial velocity is larger or smaller than the terminal velocity (see Figure 6.11).

## 6.3. Uniform Circular Motion

**In chapter 4 we have seen that when a particle moves in a circle, it experiences an acceleration a, directed towards the center of the circle, with a magnitude equal to**

_{ }

**centripetal acceleration**. To account for the centripetal acceleration, a

**centripetal force**must be acting on this object. This force must be directed towards the center of the circle, and can be calculated from Newton's second law:

_{ }

_{m}, can be calculated:

_{ }

_{ }

_{m}is the mass of the moon. The centripetal force is supplied by the gravitational attraction between the earth and the moon. In Chapter 15 we will see that the strength of the gravitational interaction can be calculated as follows:

_{ }

_{e}is the mass of the earth. For a constant circular motion, the gravitational force must provide the required centripetal force:

_{ }

_{ }

^{-11}m

^{3}/(s kg) and the mass of the earth is known to be m

_{e}= 5.98 x 10

^{24}kg. The measured period of the moon is 27.3 days (2.3 x 10

^{6}s). The distance between the moon and the earth can therefore be calculated:

r = 3.82 x 10

which agrees nicely with the distance obtained using other techniques
(for example the measurement of the time it takes for light to travel from the
earth to the moon and back).^{8}m
Figure 6.12. Forces acting on a car while rounding an unbanked curve.

## 6.4. Rounding a Curve

**Friction is critical if we want to round a curve while driving a car or bicycle. This can be easily understood if we consider the forces that act on a car while it is making a turn. Suppose that the car in question make a turn with radius R and velocity v. Figure 6.12 shows the forces acting on the car. There is no motion in the vertical direction and the net force in this direction must therefore be equal to zero. This requires that the normal force N is equal to the weight of the car:**

N = m g

When the car rounds the curve it carries out uniform circular motion.
The corresponding centripetal acceleration of this motion is given by_{ }

_{ }

_{ }

_{s}has a maximum value equal to u

_{s}N and this limits the velocity and the radius of curvature of the curve that the car can take:

_{ }

_{ }

Figure 6.13. Forces acting on a car while rounding an banked curve.

If the road is frictionless (u_{s}= 0) because of a cover of ice, the car will not be able to round any curve at all. In order to avoid problems like this, curves on highways are usually banked. The effect of banking the curves can be easily understood. Figure 6.13 shows the forces acting on an automobile when it is rounding a curve on a banked highway. We assume that there is no friction between the tires and the road. The normal force N has components both along the radial and the vertical axes. Since there is no motion along the vertical direction, the net force along the vertical axis must be zero. This requires that

_{ }

_{ }

_{ }

_{ }

_{ }

**Sample Problem 6-9**

A conical pendulum whirls around in a horizontal circle at constant speed v at the end of a cord whose length is L. The cord makes an angle [theta] with the vertical. What is the period of the pendulum ?

The pendulum is shown schematically in figure 6.14. Since the pendulum is carrying out a uniform circular motion, the acceleration of the pendulum has to point toward the center of the circle (direction along the position vector r) and the magnitude of the acceleration equals v

^{2}/r, where v is the velocity of the pendulum and r is the radius of the circle. The net force in the horizontal plane should therefore be always directed towards the center of the circle and have strength determined by Newton's second law.

The coordinate system chosen is such that the origin coincides with the center of the circle describing the motion of the pendulum. Since the horizontal component of the force is always directed towards the center we will be using an r-axis (rather than an x-axis). The y-axis coincides with the vertical direction (see Figure 6.14). Since the y-coordinate of the bob is constant, the acceleration in y-direction must be zero. The net force in this direction must therefore be zero:

_{ }

_{ }

_{ }

Figure 6.14. Sample Problem 6-9.

The centripetal acceleration a can now be calculated:_{ }

_{ }

_{ }

**Sample Problem 6-10**

A Cadillac with mass m moves at a constant speed v on a curved (unbanked) roadway whose radius of curvature is R. What is the minimum coefficient of static friction u

_{s}between the tires and the roadway ?

Figure 6.15. Sample Problem 6-10.

The situation is schematically shown in Figure 6.15. Since there is
no acceleration in the y-direction, the net force in this direction must be
zero:_{ }

_{c}is given by:

_{ }

_{ }

_{ }

**Problem 58E**

A stunt man drives a car over the top of a hill, the cross section of which can be approximated by a circle of radius 250 m. What is the greatest speed at which he can drive without the car leaving the road at the top of the hill ?

The car will not leave the road at the top of the hill if the net radial force acting on it can supply the required centripetal acceleration. The only radial forces acting on the car are the gravitational force and the normal force (see Figure 6.16). The net radial force F

_{r}acting on the car is equal to

Figure 6.16. Forces acting on the car.

F

Since the normal force N is always directed along the positive y-axis,
the radial force will never exceed the weight W of the car. This therefore
also limits the centripetal force and therefore the speed of the car._{r}= W - N_{ }

_{ }

^{2}/R. The net radial force acting on the car is equal to W - N. We conclude that

_{ }

_{ }

**Problem 60P**

A small object is placed 10 cm from the center of a phonograph turntable. It is observed to remain on the table when it rotates at 33 1/3 revolutions per minute but slides off when it rotates at 45 revolutions per minute. Between what limits must the coefficient of static friction between the object and the surface of the turntable lie ?

The object is located a distance R away from the rotation axis. During one revolution the object covers a distance 2[pi]R. If one revolution is completed during a time T, the linear velocity of the object can be obtained using the following equation:

_{ }

_{ }

_{s}has a maximum value given by

_{ }

_{ }

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