THE PROBLEM
A block of mass $m$ rests on a rough inclined plane of angle $\theta$. The coefficient of static friction between the block and the plane is $\mu_s$. The block is at rest, and $\tan\theta < \mu_s$. What is the magnitude of the friction force acting on the block?
(A) $\mu_s mg\cos\theta$
(B) $\mu_s mg\sin\theta$
(C) $mg\sin\theta$
(D) $mg\cos\theta$
(E) $\mu_s mg$
Difficulty: AP Physics 1 / Physics Bowl D1 · ★★☆☆☆
Source: OlympiadVault Original
Topics: Mechanics · Friction · Inclined Planes · Equilibrium
Answer: (C) $mg\sin\theta$
THE KEY IDEA
The block is at rest, so the forces along the incline must balance.
Static friction is not automatically equal to its maximum value. It adjusts to whatever value is needed to maintain equilibrium, up to a maximum limit.
Here, the friction force simply needs to balance the component of gravity pulling the block down the incline.
THE SOLUTION
Resolve the gravitational force $mg$ into two components:
- Along the incline: $mg\sin\theta$, pulling the block down the slope
- Perpendicular to the incline: $mg\cos\theta$, pressing the block into the surface
The perpendicular component is balanced by the normal force:
$$
N = mg\cos\theta
$$
Since the block is at rest, its acceleration along the incline is zero. Therefore, the friction force $f_s$ must exactly balance the downhill component of gravity:
$$
f_s = mg\sin\theta
$$
But static friction has a maximum possible value:
$$
f_{s,\max} = \mu_s N
$$
Substituting $N = mg\cos\theta$:
$$
f_{s,\max} = \mu_s mg\cos\theta
$$
The given condition $\tan\theta < \mu_s$ guarantees that the required friction is within this limit:
$$
mg\sin\theta < \mu_s mg\cos\theta
$$
Dividing by $mg\cos\theta$ gives:
$$
\tan\theta < \mu_s
$$
which is exactly the condition given in the problem. Therefore, the block can remain at rest, and the actual friction force is:
$$
\boxed{f_s = mg\sin\theta}
$$
So the correct answer is (C).
WHY THE OBVIOUS APPROACH FAILS
The tempting answer is (A), $\mu_s mg\cos\theta$.
This comes from memorizing:
$$
f = \mu N
$$
and immediately substituting:
$$
N = mg\cos\theta.
$$
But this confuses actual static friction with maximum static friction.
For static friction, the correct relationship is:
$$
f_s \leq \mu_s N
$$
The friction force takes whatever value is required for equilibrium, provided that value does not exceed $\mu_s N$.
In this problem:
$$
f_s = mg\sin\theta
$$
while:
$$
f_{s,\max} = \mu_s mg\cos\theta.
$$
Because $\tan\theta < \mu_s$, the block is not at the threshold of slipping, so:
$$
f_s < f_{s,\max}.
$$
The mistake is not in knowing $\mu_s N$ — it is in assuming that static friction must always equal $\mu_s N$.
THE INSIGHT
Static friction is a constraint force, not a fixed force.
Its magnitude is determined by the conditions of the problem:
$$
f_s \leq \mu_s N
$$
with equality only when the object is on the verge of slipping.
So whenever a problem says an object is at rest, first ask:
What friction force does equilibrium require?
Only after finding that value should you check whether static friction can provide it. This simple habit prevents one of the most common mistakes in introductory mechanics.
GENERALIZATION
As the angle $\theta$ increases, so does the friction required to maintain equilibrium:
$$
f_s = mg\sin\theta
$$
At the same time, the maximum available static friction decreases:
$$
f_{s,\max} = \mu_s mg\cos\theta
$$
The block remains at rest as long as:
$$
\tan\theta \leq \mu_s
$$
At the critical angle $\theta_c$:
$$
\tan\theta_c = \mu_s
$$
so:
$$
\theta_c = \arctan(\mu_s)
$$
At this angle, the block is on the verge of slipping and static friction reaches its maximum value.
For $\theta > \theta_c$, static friction can no longer maintain equilibrium, and the block begins to slide. This critical angle is called the angle of repose.
CHALLENGE
A block of mass $m$ is placed on a rough inclined plane with coefficient of static friction $\mu_s = 0.5$. The angle is slowly increased from $0^\circ$.
Find the angle $\theta_c$ at which the block begins to slide, and the friction force acting on it at that instant.