Question 7
A 10 kg wheelbarrow sits stationary but just on the point of sliding on an inclined ramp. When the ramp is raised by 5° to make an angle of 20° to the horizontal, the wheelbarrow slowly starts to slide.
The minimum amount of additional force required to stop the wheelbarrow from sliding is
Not quite. 59 N adds friction to the pull down the ramp, as if friction pushed the wheelbarrow down. Friction is on your side here. It holds the wheelbarrow back, so it comes off the pull. In 2022, 7.9% of Queensland students chose (A).
Not quite. 34 N is $mg\sin20^\circ$, the whole pull down the ramp. Static friction still supplies most of the holding force, so you only have to make up the shortfall. In 2022, 31.5% of Queensland students chose (B).
Not quite. 25 N is the friction force itself, about $\mu_s mg\cos20^\circ$. The question asks for the extra force you add on top of friction. In 2022, 21.1% of Queensland students chose (C).
That is it. Step through the hints below to see why. In 2022, 38.8% of Queensland students chose (D).
“Just on the point of sliding” at 15° tells you that 15° is the angle of repose. That gives you the coefficient of static friction.
$$\mu_s=\tan15^\circ=0.26795$$
The coefficient belongs to the two surfaces, so it stays the same when the ramp is raised.
At 20° the pull down the ramp grows and the maximum friction shrinks, because the normal force drops.
$$\begin{aligned}F_\parallel&=10\times9.8\times\sin20^\circ=33.52\text{ N}\\F_f&=\mu_s mg\cos20^\circ=0.26795\times92.09\\&=24.68\text{ N}\end{aligned}$$
Either of these on its own is a distractor: (B) is the pull and (C) is the friction.
Friction and your extra force both act up the ramp. Together they have to match the pull down it.
$$\begin{aligned}F_{extra}+F_f&=F_\parallel\\F_{extra}&=33.52-24.68\\&=8.84\text{ N}\end{aligned}$$
The only option close to this is (D) 8 N. Adding friction instead of taking it away gives (A).
$$\begin{aligned}F_{extra}&=mg\sin20^\circ-\tan15^\circ\times mg\cos20^\circ\\&=33.52-24.68\\&=8.84\text{ N}\end{aligned}$$
The 15° angle of repose gives $\mu_s$. At 20° the extra force only has to make up what static friction cannot supply.
8.84 N rounds to 9 N, which is not an option. QCAA’s 8 N matches $mg\sin20^\circ-mg\sin15^\circ=8.15$ N, which keeps friction at its 15° value of 25.36 N and ignores the drop in the normal force. Its 59 N fits the same shortcut, 33.52 N plus 25.36 N. Either way, (D) is the only option near the answer.
The most common wrong answer was (B), 34 N, from finding the pull down the ramp and forgetting that friction still holds back most of it. Only 38.8% of students chose (D), the lowest of the five friction multiple choice questions with published results.
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 2022 QCAA Engineering external assessment, © State of Queensland (QCAA) 2022, 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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