Forces and Equilibrium: Question 9

Worked solution

Step 1: Set up the equilibrium equations

Let the block have mass $m$. Resolving perpendicular to the plane, the normal reaction balances the perpendicular component of the weight: $R = mg\cos\theta$

Resolving along the plane, at the maximum angle $\theta$ the block is on the point of sliding down, so friction acts up the plane at its limiting value $F = \mu R$, exactly balancing the weight’s component down the plane: $\mu R = mg\sin\theta$

Step 2: Combine the equations

Substituting $R = mg\cos\theta$ into the second equation: $\mu \, mg\cos\theta = mg\sin\theta$

The mass $m$ and $g$ cancel: $\mu\cos\theta = \sin\theta$ $\tan\theta = \mu$

Step 3: Solve for $\theta$

$\tan\theta = 0.36$ $\theta = \arctan(0.36) \approx 19.8^\circ$

Why the other options are wrong

• B ($21.1^\circ$): this comes from solving $\sin\theta = \mu$ (i.e. $\arcsin(0.36)$), incorrectly comparing friction directly to the weight instead of to the reduced normal reaction $mg\cos\theta$.
• C ($36.0^\circ$): this simply reads the coefficient of friction $0.36$ as if it were ”$36$ degrees”, with no trigonometry applied at all.
• D ($70.2^\circ$): this comes from computing $\arctan(1/\mu)$ instead of $\arctan(\mu)$, the reciprocal relationship, not the correct one.

Final answer

• The maximum angle is $\theta \approx \boxed{19.8}^\circ$, option A.