Question 7
A 20 kg box sits just on the point of sliding on an incline plane. If the coefficient of static friction is 0.27, what is the angle of repose?
Not quite. 5° comes from $20\times0.27=5.4$, a mass times a coefficient. That product is not an angle, and the mass cancels out of the angle of repose anyway.
Not quite. 13° comes from dividing the 196 N weight by $\tan^{-1}(0.27)$. That mixes a force with an angle. The angle of repose depends on the coefficient alone.
That is it. Step through the hints below to see why.
Not quite. 16° is $\sin^{-1}(0.27)=15.7^\circ$, the wrong inverse function. On the point of sliding $\mu_s=\tan\theta$, so you need inverse tan.
Arrows drawn to scale. The dashed arrow is the component of the weight down the slope.
“Just on the point of sliding” means static friction is at its maximum and exactly matches the pull of the weight down the slope.
$$\mu_s F_N=F_\parallel$$
$$\mu_s\,mg\cos\theta=mg\sin\theta$$
Divide both sides by $mg\cos\theta$. The mass and $g$ cancel.
$$\mu_s=\frac{\sin\theta}{\cos\theta}=\tan\theta$$
The 20 kg is a distractor. Any answer that changes when you change the mass is wrong. That rules out (A) and (B).
Undo the tangent with inverse tan, with your calculator in degrees.
$$\theta=\tan^{-1}(0.27)=15.11^\circ$$
Inverse sine gives $15.66^\circ$, which rounds to (D). At that angle the box is already sliding.
$$\begin{aligned}\theta&=\tan^{-1}(\mu_s)=\tan^{-1}(0.27)\\&=15.11^\circ\approx15^\circ\end{aligned}$$
On the point of sliding, $\mu_s=\tan\theta$. The formula book gives you this line, and the mass plays no part.
The 2020 subject report built (D) from inverse sine instead of inverse tan, and (A) and (B) from the 20 kg. If you used the mass, stop and ask whether the angle of repose should change with it.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and any figures are from the 2020 QCAA Engineering external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and have been adapted, except where a third-party source is named. Tangent Tuition is not affiliated with the QCAA.
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