Figure 6.1. Static 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_{s} N

where u_{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_{max} = u_{s} N

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
**kinetic friction** and the friction force f_{k} is given by

f_{k} = u_{k} N

where u_{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:

_{
}

Since the object is at rest, the net force on the object equals zero:

_{
}

In terms of the components of the net force along the x-axis and the y-axis:

_{
}

_{
}

The coefficient of static friction can be easily obtained from these two equations:

_{
}

**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:

_{
}

_{
}

The second equation can be used to eliminate N:

_{
}

Substituting this expression in the first equation we obtain:

_{
}

The tension T can now be calculated:

_{
}

The normal force N is always perpendicular to the surface. In the previous two sample problems, the normal force N was proportional to the weight of the object. However, this is not always true. For example, suppose I am pressing an eraser against the blackboard. I ask myself, what is the minimum force that I need to apply in order to prevent the eraser from slipping ? This situation is shown schematically in Figure 6.4. Since the eraser is at rest, the net force acting on it must be zero (and therefore, the components of the net force in both the x and the y-direction must be equal to zero):

_{
}

_{
}

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:

_{
}

However, the normal force N is equal to the applied force F. In order to prevent the eraser from slipping, the force F will need to exceed a minimum threshold:

_{
}

This relation shows that if the mass of the eraser is increased, the applied force needed to prevent the eraser from slipping will increase (the minimum force is proportional to the mass). This example also illustrates a situation in which the normal force is not related to the mass of the object.

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.

_{
}

Since the net force acting on block B must be zero we conclude that

T = W_{B}

_{}

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

N = W_{A} + W_{C}

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_{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_{s} <= u_{s} N = u_{s} (W_{A}
+ W_{C})

The minimum weight of block C that will prevent the system from slipping can be found by requiring that

u_{s} (W_{A} + W_{C}) >= f_{s} =
T = W_{B}

and thus

_{
}

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_{k} = u_{k} N = u_{k} W_{A}

The net force acting on block A is given by

_{
}

The net force acting on block B is given by

_{
}

Eliminating the tension T from these last two equations we obtain for the acceleration 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}.

_{
}

_{
}

In the coordinate system chosen, there is no acceleration along the
y-axis. The normal force N_{1} must therefore be equal to
m_{1} g cos([theta]). This fixes the kinetic friction force

f_{1k} = u_{1k} N_{1} = u_{1k}
m_{1} g cos([theta])

Mass m_{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}

_{
}

The forces acting on mass m_{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_{2k} = u_{2k} N_{2} = u_{2k}
m_{2} g cos([theta])

The x-component of the net force acting on mass m_{2} is given
by

_{
}

and is related to the acceleration a of mass m_{2}

_{
}

Figure 6.10. Forces acting on m_{2}.

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

_{
}

Substituting the values of the parameters given we find that a = 3.62
m/s^{2}. The tension T in the rod can now be determined easily

_{
}

which is equal to 1.06 N. If mass m_{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).

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

_{
}

where A is the effective cross-sectional area of the body, [rho] is the density of air and v is the speed of the object. C is a dimensionless drag coefficient that depends on the shape of the object and whose value generally lies in the range between 0.5 and 1.0. The direction of the drag force is opposite to the direction of the velocity.

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:

_{
}

and v_{t} is calculated to be

_{
}

The equation for v_{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).

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

_{
}

where v is the velocity of the particle, and r is the radius of the
circle. The acceleration a is called the **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:

_{
}

An example of uniform circular motion is the motion
of the moon around the earth. Suppose the period of this motion is T. What
does this tell us about the distance r between the earth and the moon ? During
one period, the moon covers a total distance equal to 2[pi]r. The velocity of
the moon, v_{m}, can be calculated:

_{
}

The corresponding centripetal force is

_{
}

Here we assumed that m_{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:

_{
}

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

_{
}

The distance between the earth and the moon can therefore be calculated:

_{
}

The constant of gravity is known to be G = 6.67 x 10^{-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^{8} m

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).

Figure 6.12. Forces acting on a car while rounding an unbanked 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

_{
}

In order for the car to carry out this circular motion there must exist a radial force with a strength equal to

_{
}

This force can only be supplied by the static friction force and therefore we require that

_{
}

The static friction force f_{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:

_{
}

We conclude that the car will be able to make a turn with radius R and velocity v if the coefficient of static friction between the tires and the road is

_{
}

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

_{
}

and fixes the normal force N

_{
}

The radial component of the normal force is given by

_{
}

This component of the normal force can produce the radial acceleration required to allow the car to round the curve, even in the absence of friction. If the car has to round a curve with radius R and with velocity v, we require that

_{
}

or

_{
}

This last equation shows that the banking angle of a highway curve is designed for a specific velocity and radius of curvature.

**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:

_{
}

This expression allows us to calculate the tension T:

_{
}

The net force in the radial direction can now be determined:

_{
}

Figure 6.14. Sample Problem 6-9.

The centripetal acceleration a can now be calculated:

_{
}

From the radius R of the trajectory and the centripetal acceleration a, the velocity of the object can be calculated:

_{
}

The period T can be calculated from the known velocity v and radius R:

_{
}

For L = 1.7 m and [theta] = 37deg. the period T = 2.3 s.

**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:

_{
}

The centripetal force F_{c} is given by:

_{
}

In this situation, the centripetal force is provided by the static friction force. If no slipping occurs, the maximum static friction force must exceed the required centripetal force:

_{
}

The minimum coefficient of static friction can be obtained from this equation:

_{
}

If the velocity of the car is 72 km/hr (20 m/s) and the radius of curvature R = 190 m, the minimum value of the coefficient of static friction is 0.21. Note that the mass of the car does not enter in the calculation, and the friction coefficient is therefore the same for all objects moving with the same velocity. The minimum coefficient of static friction scales with the square of the velocity; a reduction of the velocity by a factor if two, will reduce the minimum friction coefficient by a factor of four.

**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_{r} = W - N

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.

_{
}

The maximum velocity of the car can now be found easily

_{
}

Suppose the car is driving with a velocity less than 178 km/h. The
normal force N can now be calculated and will be a function of the velocity v.
If the car carries out a uniform circular motion than we know that a net radial
force must be acting on it and that its magnitude is equal to mv^{2}/R.
The net radial force acting on the car is equal to W - N. We conclude that

_{
}

or

_{
}

**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:

_{
}

In order for the object to carry out such a uniform circular motion it must provide a radial force with magnitude equal to

_{
}

The only radial force acting on the object is the static friction
force. The friction force f_{s} has a maximum value given by

_{
}

If the object remains on the table, the static friction coefficient needs to satisfy the following relation:

_{
}

In this problem the distance to the rotation axis is 0.1 m. The block remains on the table when the table rotates at 33 1/3 rev/min. This corresponds to 1 revolution per 1.80 s, and a linear velocity of 0.35 m/s. This implies that the coefficient of static friction must be at least 0.12. When the table rotates at 45 rev/min the block leaves the table. This implies the coefficient of static friction is less than 0.22

Send comments, questions and/or suggestions via email to **wolfs@pas.rochester.edu** and/or visit the home page of Frank Wolfs.