GRE Physics Companion: Friction

Timed Friction problems are usually won or lost before the algebra begins. The key questions are: Is there relative sliding? What is the Normal force? Which way would the surfaces tend to slip? If the contact is static, what friction force is actually required?

1 Fast friction triage

For static contact,

(1)

Do not automatically set f_s = μ_sN. Solve for the friction required by the no-slip assumption, then check whether the required value is available.

For sliding contact,

(2)

The friction direction is opposite relative sliding at the contact.

2 The graph to recognize

For a horizontal block subject to an increasing horizontal force, static friction tracks the applied force while the block remains at rest. At impending slip,

(3)

After sliding begins, the elementary model uses

(4)

3 Useful incline results

For a simple block on an incline,

(5)

The threshold for impending downhill slip is

(6)

so

(7)

If the block slides downhill,

(8)

4 Worked GRE example 1: is the block actually moving?

A 10.0 kg block rests on a horizontal floor with μ_s = 0.30 and μ_k = 0.20. A horizontal force of 25.0 N is applied. Find the friction force and acceleration.

The maximum static friction is

(9)

Only 25.0 N is required to maintain equilibrium, so

(10)

The kinetic coefficient is irrelevant because the block never starts sliding.

5 Worked GRE example 2: angle of repose shortcut

A block just begins to slide when a board is tilted to 30.0^∘. Estimate the coefficient of static friction.

At impending slip,

(11)

Therefore

(12)

No mass information is needed.

6 GRE speed questions

1. A block is at rest on a horizontal surface. A 12 N horizontal force is applied, and the block remains at rest. The static friction magnitude is (A) 0 (B) less than 12 N (C) 12 N (D) μ_sN regardless of μ_s.
2. A 4.00 kg block slides on a horizontal surface with μ_k = 0.25. The kinetic friction magnitude is closest to (A) 1.0 N (B) 4.0 N (C) 9.8 N (D) 39 N.
3. A horizontal block is pulled upward at an angle while maintaining contact with the floor. Compared with a purely horizontal pull of the same magnitude, the normal force is (A) larger (B) smaller (C) unchanged (D) zero.
4. A block just begins to slide down an incline at angle 𝜃. In the elementary dry friction model, μ_s equals (A) sin 𝜃 (B) cos 𝜃 (C) tan 𝜃 (D) 1∕ tan 𝜃.
5. Two blocks of masses m and 2m slide down the same incline with the same μ_k. Their accelerations are (A) equal (B) in the ratio 1 : 2 (C) in the ratio 2 : 1 (D) impossible to compare.
6. A box initially at rest is placed on a conveyor belt moving right. While the belt slips under the box, friction on the box points (A) left (B) right (C) upward (D) zero.
7. A block on an incline is held at rest by a strong force directed uphill. Static friction (A) must always point uphill (B) can point downhill (C) must be zero (D) must equal μ_sN.

7 Answers and rationales

1. C. Static friction supplies exactly the 12 N needed for equilibrium, provided this is below its maximum value.
2. C. f_k = μ_kmg = (0.25)(4.00)(9.81) = 9.81 N.
3. B. The upward component of the pull reduces the normal force.
4. C. At impending downhill slip, mg sin 𝜃 = μ_smg cos 𝜃.
5. A. The mass cancels from a = g(sin 𝜃 − μ_k cos 𝜃).
6. B. The belt slips right relative to the box, so friction on the box acts right and accelerates it.
7. B. Static friction opposes the tendency to slip, which can be uphill if the external force is sufficiently strong.

References