SAT Physics Subject Test

Chapter 3 Newton’s Laws


When an object is in contact with a surface, the surface exerts a contact force on the object. The component of the contact force that’s parallel to the surface is called the friction force on the object. Friction, like the normal force, arises from electrical interactions between atoms that make up the object and those that make up the surface.

We’ll look at two main categories of friction: (1) static friction and (2) kinetic (sliding) friction. If you attempt to push a heavy crate across a floor, at first you meet with resistance, but then you push hard enough to get the crate moving. The force that acted on the crate to cancel out your initial pushes was static friction, and the force that acts on the crate as it slides across the floor is kinetic friction. Static friction occurs when there is no relative motion between the object and the surface (no sliding); kinetic friction occurs when there is relative motion (when there’s sliding).

The strength of the friction force depends, in general, on two things: the nature of the surfaces and the strength of the normal force. The nature of the surfaces is represented by the coefficient of friction, denoted by µ (mu). The greater this number is, the stronger the friction force will be. For example, the coefficient of friction between rubber-soled shoes and a wooden floor is 0.7, but between rubber-soled shoes and ice, it’s only 0.1. Also, since kinetic friction is generally weaker than static friction (it’s easier to keep an object sliding once it’s sliding than it is to start the object sliding in the first place), there are two coefficients of friction: one for static friction (µs) and one for kinetic friction (µk). For a given pair of surfaces, it’s virtually always true that µk < µs. The magnitude of these two types of friction forces are given by the following equations:

Fstatic friction, max = µs FN

Fkinetic friction = µk FN

Notice that the equation for the magnitude of the static friction force is for the maximum value. This is because static friction can vary, counteracting weaker forces that are less than the minimum force required to move an object. For example, suppose an object feels a normal force of FN = 100 N, and the coefficient of static friction between it and the surface it’s on is 0.5. Then, the maximum force that static friction can exert is (0.5)(100 N) = 50 N. However, if you push on the object with a force of, say, 20 N, then the static friction force will be 20 N (in the opposite direction),not 50 N: The object won’t move. The net force on a stationary object must be zero. Static friction can take on all values, up to a certain maximum, and you must overcome the maximum static friction force to get the object to slide.

Kinetic vs. Static

For a person walking, the
friction between the
person’s shoes and the
floor is static (no sliding)
and is directed forward
(in the direction the
person is walking). The
person pushes on the
floor in the backward
direction. Static friction
prevents it from moving
backward, and so
therefore must be
forward. For similar
reasons, objects that are
rolling without slipping—
rolling normally, not
skidding—roll because
of static friction.

The direction of Fkinetic friction = Ff (kinetic) is opposite to that of motion (sliding), and the direction of Fstatic friction = Ff (static) is opposite to that of the intended motion.

Questions 15-16

A crate of mass 20 kg is sliding across a wooden floor. The coefficient of kinetic friction between the crate and the floor is 0.3.

15. Determine the magnitude of the friction force acting on the crate.

16. If the crate is being pulled by a force of 90 N (parallel to the floor), find the acceleration of the crate.

Here’s How to Crack It

First draw a free-body diagram.

15. The normal force on the object balances the object’s weight, so FN = mg = (20 kg)(10 m/s2) = 200 N. Therefore, Ff (kinetic) = µk FN = (0.3)(200N) = 60 N.

16. The net horizontal force that acts on the crate is F – Ff = 90 N – 60 N = 30 N, so the acceleration of the crate is a = Fnet/m = (30 N)/(20 kg) = 1.5 m/s2.

17. A crate of mass 100 kg rests on the floor. The coefficient of static friction is 0.4. If a force of 250 N (parallel to the floor) is applied to the crate, what’s the magnitude of the force of static friction on the crate?

Here’s How to Crack It

The normal force on the object balances its weight, so FN = mg = (100 kg)(10 m/s2) = 1,000 N. Therefore, Fstatic friction, max = Ff (static), max = µs FN = (0.4)(1,000 N) = 400 N. This is the maximum force that static friction can exert, but in this case it isn’t the actual value of the static friction force. Since the applied force on the crate is only 250 N, which is less than the Ff (static), max, the force of static friction will be less also: Ff (static) = 250 N, and the crate will not slide.