Question 18
A 900 kg machine is designed to rest without slipping on a slope of up to 30° on the surface of Mars. The acceleration due to gravity on Mars is one-third of that on Earth.
Gravity sets the size of every force, and the slope sets their ratio
The machine sits at the steepest slope it is designed for, so friction is at its limit. Every arrow is drawn to the same scale. Watch what happens to the arrows, and to their ratio, when the machine moves to Earth.
The mass and $g$ scale every force by the same factor. The coefficient is a ratio of two of those forces, so only the angle is left in it: $\mu_s=\tan\theta$.
Change the planet, the mass and the slope
The machine is held at the steepest slope it must survive, so friction is at its limit. Change one thing at a time and watch the last line. The question is Mars, 900 kg and 30°.
Arrows drawn to scale. The weight is always drawn the same length, so read the scale bar to see how big the forces are.
This is the question. On a 30° slope on Mars, static friction has to supply 1470 N against a normal force of 2546 N. Their ratio is $\tan30^\circ=0.58$, the coefficient part a) asks for.
Every force changed size, but the triangle of forces kept its shape. Friction and the normal force grew or shrank together, so the coefficient you need is still $\tan30^\circ=0.58$. That is the whole of part c).
The angle is the only thing that changes the shape of the triangle. A steeper slope asks friction to hold more of the weight while the normal force shrinks, so the coefficient you need rises as $\tan\theta$.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
“Up to 30°” means 30° is the steepest slope, so at 30° the machine is on the point of slipping. Friction is at its maximum and equals the pull down the slope: $\mu_s mg\cos\theta=mg\sin\theta$. The mass and $g$ cancel.
$$\mu_s=\tan\theta=\tan30^\circ=0.577\ldots$$
The formula book gives $\mu_s=\tan\theta$ directly. You do not need the 900 kg or Mars gravity for this part.
Marker: provides correct working to give correct coefficient of static friction to two decimal places. QCAA’s note: “Correct result for μs range: 0.57 or 0.58 only”.
Work out Mars gravity first, then write the friction formula with the normal force on a slope, $F_N=mg\cos\theta$.
$$g_{\text{Mars}}=\frac{9.8}{3}=3.27\text{ m s}^{-2}$$
$$\begin{aligned}F_f&=\mu_s F_N=\mu_s mg\cos\theta\\&=0.58\times900\times3.27\times\cos30^\circ\end{aligned}$$
This is the line where 9.8 costs you the mark. The machine is on Mars, so $g$ is 3.27.
Marker: provides correct formula and substituted values. QCAA’s note: “Formula considered correct if substituted and constant values are correctly applied.”
$$F_f=1478.25\text{ N}$$
The 1478 N comes from the rounded 0.58 and 3.27. Keep full precision and you get exactly 1470 N, because $\tan30^\circ\cos30^\circ=\sin30^\circ$. The friction is just $mg\sin\theta=900\times\tfrac{9.8}{3}\times0.5$. The guide accepted anything from 1451 N to 1478 N, and follow-through applies if your part a) value was wrong.
Marker: provides correct answer to the nearest whole unit with correct unit provided. QCAA’s notes: “Correct result for force of friction range: 1451 to 1478 N” and “Allow FT error for incorrect μs in determining the force of static friction.”
Explain asks for the reason, so give it before the conclusion. QCAA’s sample response opens with this sentence:
The coefficient of static friction depends only on the angle of repose.
Back it with the working from part a). In $\mu_s=\tan\theta$ the mass and $g$ have already cancelled, so changing $g$ cannot change it. Strictly, $\mu_s$ belongs to the two surfaces and the angle of repose follows from it. Either way round, $g$ does not appear.
Marker: explains using wording indicative of μs depends only on the angle of repose
Now state the consequence for this machine.
Therefore μs will be the same quantity on both Mars and Earth.
The friction force does change. On Earth it would be three times bigger, $900\times9.8\times\sin30^\circ=4410\text{ N}$. Say “coefficient” when you mean the coefficient.
Marker: explains using wording indicative of μs is the same on Mars and Earth. QCAA’s note: “Explanation for Part C must be appropriate.”
Each phrase of the question gave you something
Each part is one formula book line used once: $\mu_s=\tan\theta$, then $F_f=\mu_s F_N$, then the observation that $g$ cancelled in the first line. The only twist is the Mars gravity. The 2020 subject report lists two simple familiar calculation questions worth 10 marks in Section 3, and Questions 18 and 21 are the only five-mark questions in that section.
If I had to guess where you might lose marks, it would be part b) with 9.8 in it. The stem mentions Mars gravity once, in its second sentence, and using 9.8 gives 4430 N, three times too big. In part c) the trap is writing that $\mu_s$ goes up because Earth’s gravity is stronger. The friction force triples while the coefficient stays at 0.58.
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 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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