Question 20
A 20 kg trolley is pushed up a 10º incline using a force of 160 N.
Determine the coefficient of friction between the trolley and the incline if the trolley is travelling at a uniform velocity. Answer to two decimal places.
At a uniform velocity the push exactly cancels friction and the slope
The trolley climbs at a steady speed the whole time. The forces on it are drawn to scale. Watch the two pulls down the slope add up to the push.
No acceleration means no net force. Everything up the slope equals everything down it.
Build the balance and see which coefficient it gives
Pick the balance equation, then change the angle and the push. The forces are drawn to scale. The question is 10° and 160 N with the first balance.
This is the question. Friction and $F_\parallel$ both point down the slope, so the push has to match both: $F_f=160-34.04=125.96$ N, and $\mu=125.96\div193.02=0.65$.
The same balance with new numbers. A bigger push at the same speed means the surface must be rougher. A steeper slope takes more of the push for $F_\parallel$ and leaves less for friction.
This leaves out $mg\sin\theta$ and treats the slope as flat. It gives 0.83 for the question, and you lose the mark for identifying three forces parallel to the incline.
This puts $F_\parallel$ on the push’s side, as if gravity helped the trolley up the slope. It gives 1.01 for the question. A coefficient above 1 on a gentle slope is a sign your balance is backwards.
The push is now smaller than $F_\parallel$, so no friction could make the trolley climb at a steady speed. A negative coefficient means the situation is impossible.
Work through the solution one mark at a time
Try each step yourself before you reveal it. Two of the five marks are for words and set-up, before any number appears.
Say what “uniform velocity” means before you calculate. The guide’s sample sentence is short.
$$a=0\quad\Rightarrow\quad\sum F_{\text{up}}=\sum F_{\text{down}}$$
The guide accepts either an acceleration of zero because the velocity is uniform, or an equation setting the force up the incline equal to the force down it.
Marker: indicates that the system is in equilibrium
Name the three forces along the slope. The trolley moves up, so friction acts down the slope beside the weight component.
$$160\text{ N}=F_f+mg\sin10^\circ$$
All three are needed: the push (160 N), friction ($\mu_kF_N$) and the weight component down the incline ($mg\sin10^\circ$). The stem does not say the push is parallel to the slope, and the guide treats it as parallel.
Marker: identifies three forces parallel to the incline. “This mark may be implied by subsequent working.”
Write friction as $\mu_k F_N$ with $F_N=mg\cos10^\circ$, then substitute $mg=20\times9.8=196$ N. The trolley is moving, so the coefficient is kinetic.
$$\begin{aligned}&160=\mu_kF_N+mg\sin10^\circ\\&\mu_k\,mg\cos10^\circ\\&\quad=160-mg\sin10^\circ\\&\mu_k(196\cos10^\circ)\\&\quad=160-196\sin10^\circ\end{aligned}$$
Marker: provides correct formula and substituted values
Divide both sides by the normal force to isolate $\mu_k$.
$$\begin{aligned}\mu_k&=\frac{160-196\sin10^\circ}{196\cos10^\circ}\\&=\frac{160-34.04}{193.02}\end{aligned}$$
The guide rounds the denominator to 193. It makes no difference at two decimal places.
Marker: correctly manipulates equation to isolate coefficient of friction
Evaluate and round as the question asks.
$$\begin{aligned}\mu_k&=\frac{125.96}{193.02}\\&=0.6526\end{aligned}$$
A coefficient has no unit. Two decimal places is the instruction here, so the nearest whole unit rule does not apply.
Marker: determines the answer to two decimal places. “Allow FT error/s for incorrect forces parallel to the incline when determining the answer.”
Each phrase of the question gave you something
The 2023 subject report used this question as a sample of effective practice. Strong responses stated that the system was in equilibrium, identified the three forces on the trolley parallel to the incline and gave the coefficient to two decimal places.
QCAA did not publish a band for this question. It reads as simple familiar because it is one balanced slope with one unknown, the Game plan 1.1 routine run backwards to find $\mu$ instead of a force.
If I had to guess where you might lose marks, it would be the first two lines. Jump straight to the number and you can drop the equilibrium mark even with 0.65 at the bottom. The other trap is the slope itself. Treat it as flat and divide 160 by the normal force, and you get 0.83.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording, any figures and marking-guide steps are from the 2023 QCAA Engineering external assessment, © State of Queensland (QCAA) 2023, 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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