6-1   6-2   6-3 - 6-5   next        

Chapter Review

---

This chapter is a continuation of chapter five. Here we will discuss further details in the application of Newton's laws by adding several specific forces. These forces are friction, tension, and Hooke's law force. We also have the concepts of centripetal force and the associated centripetal acceleration that relate to circular motion.

6-1 Frictional Force

When two surfaces are in direct contact and one surface either slides or attempts to slide across the other, a force, called friction, that opposes the motion (or attempted motion) is generated between the surfaces. When the surfaces are sliding across each other we call it static friction. For example, if you apply a small horizontal force to a heavy object, in an attempt to slide it, the object may not move at all because of the friction between it and the floor; this is a case where there is definitely friction but no motion - static friction.

The orgin of friction is based, complicated ways, on the microscopic structure of the surfaces involved. Often in physics we handle these types of situations by identifying those factors on which the dependence is fairly simple and representing the rest of the complicated physics by one measured quantity. For kinetic friction we note that the frictional force, f_k, is directly proportional to the normal force, N, between the surfaces. The measured proportionality factor, , is called the coefficient of kinetic friction; so we have

.

Note that this equation relates only the magnitudes of f_k and N because they are not in the same direction; they are perpendicular to each other.

For static friction the force, f_s, takes on a range of values depending on the strength of the force it opposes. Static friction will cancel out any force trying to slide two surfaces across each other up to some maximum value beyond which static friction is overcome. This maximum force of static friction is also directly proportional to the normal force. The measured proportionality factor, , is called the coefficient of static friction; so we have

which, for the more general case yields

.

Typically, for a given pair of surfaces we have .

Physlet Illustration: Static Friction
	Interactive Help on      off
F = Newtons m = kg
A block is being pulled across a rough floor by a constant, horizontal force, as shown. It starts from rest, and its velocity is shown in m/s. Adjust the mass (1 kg < m < 500 kg) and/or the pulling force (0 N < F < 10000 N), and determine the coefficient of static friction between the block and the floor. Start
Hints If you pull with a small force, why doesn't the box move? What minimum force is needed to just start a particular mass moving? How does this minimum moving force relate to the force of static friction? How does the coefficient of friction relate to the force of friction?
Reference See Walker, Section 6-1

Exercise 6.1 Getting it Going: A 105 lb crate sits on a floor. If the minimum horizontal force required to start it sliding is 33 lb, what is the coefficient of static friction between the crate and the floor?

Solution: The free-body diagram appears below. The following information is given in the problem:

Given: W = 105 lb, F_min = 33 lb; Find:

The fact that we seek the minimum force needed to get the crate moving means that we want the force that is just barely large enough to overcome static friction. The force right at this critical condition equals the maximum force of static friction

.

To determine N we apply Newton's second law in the y-direction (recognizing that the vertical acceleration is zero): N - W = 0. This tells us that N = W. Hence, we have

Physlet Illustration: Accelerating a Train
	Interactive Help on      off
F = Newtons m1= kg  m2= kg m3= kg
A train, consisting of three blocks connected by rods, is pulled by a constant force (shown as a black arrow) across a frictionless floor. The velocity is given in m/s, and the times shown are in seconds. Adjust the masses (20 kg < m < 100 kg) and/or the pulling force (0  N < F < 100 N), and measure the train's acceleration. Does the train obey Newton's Second Law?  What are the tensions in each of the connecting rods? Start
Hints Measure acceleration by finding the change in velocity over a specific time interval. How do the accelerations of the individual blocks compare? Draw a free-diagram for each block. Equate the sum the forces on each block to its acceleration.
Reference See Walker, Section 6-2

Practice Quiz

A person pushes an object across a room, on a horizontal floor, by applying a horizontal force. If instead, he pushed it with a force that makes an angle below the horizontal, the object would have experienced greater friction less friction the same frictional force zero friction [none of the above]

A person pulls an object across a room, on a horizontal floor, by applying a horizontal force. If instead, she pulled it with a force that makes an angle above the horizontal, the object would have experienced greater friction less friction the same frictional force zero friction [none of the above]

The coefficient of kinetic friction represents. the force that one surface applies to another when they slide across each other. the force that one surface applies to another when they do not slide across each other. the force that one surface applies to another that is perpendicular to the interface between them. the ratio of the normal force to the force of friction between two sliding surfaces. [none of the above]

6-1   6-2   6-3 - 6-5   next