Question 2
A 25 kg block is pulled up an incline using a force P as shown. What is the minimum value of P required to just move the box from rest if the coefficient of static friction is 0.4?
That is it. Step through the hints below to see why.
Not quite. 103.5 N is $mg\sin25^\circ$, the pull of the weight down the slope on its own. Static friction also holds the box back, so P has to beat that as well.
Not quite. 41.4 N is $0.4\times103.5$, the coefficient times the pull down the slope. Friction is $\mu_s$ times the normal force, $mg\cos25^\circ$, and P still has to beat the pull down the slope on top of it.
Not quite. 14.7 N is $103.5-88.8$, which puts friction on the side of P as if it helped the pull. The box is about to move up the slope, so friction points down the slope and adds to what P has to beat.
“Just move the box from rest” means the box is on the point of sliding up the slope. Friction opposes the slide that is about to start, so it points down the slope, on the same side as the weight.
At that threshold static friction is at its maximum, $F_f=\mu_s F_N$. P has to beat the weight component and friction together. That rules out (D).
Tilt your axes onto the slope and split the weight, $mg=245\text{ N}$, into its two components.
$$\begin{aligned}F_N&=mg\cos25^\circ=25\times9.8\times\cos25^\circ\\&=222.05\text{ N}\end{aligned}$$
$$\begin{aligned}F_\parallel&=mg\sin25^\circ=25\times9.8\times\sin25^\circ\\&=103.54\text{ N}\end{aligned}$$
The coefficient multiplies $F_N$, the component pressing into the slope. Multiplying the component along the slope gives (C).
Find the maximum static friction, then balance the forces along the slope.
$$\begin{aligned}F_f&=\mu_s F_N=0.4\times222.05\\&=88.82\text{ N}\end{aligned}$$
$$\begin{aligned}P&=F_\parallel+F_f=103.54+88.82\\&=192.36\text{ N}\end{aligned}$$
Stopping at $F_\parallel$ with friction left out gives (B).
$$\begin{aligned}P&=mg\sin25^\circ+\mu_s mg\cos25^\circ\\&=103.54+88.82\\&=192.36\text{ N}\end{aligned}$$
The box is about to slide up, so static friction at its maximum points down the slope and joins the weight component. P has to balance both.
QCAA printed the answer as 192.3 N. Rounded to one decimal place, 192.36 N is 192.4 N. QCAA’s figure matches adding the two components after rounding each one to one decimal place, $103.5+88.8$. No other option is anywhere near it, so (A) stands.
If I had to guess the most common wrong answer, it would be (B), from resolving the weight correctly and then forgetting friction. On the point of sliding, friction is always in the balance.
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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