Question 20
A parcel slides down an inclined ramp, which is at 20° to the horizontal plane, at a constant velocity. The parcel experiences a normal force of 98.2 N. A coefficient of kinetic friction of 0.37 and a coefficient of static friction of 0.39 exist between the parcel and the ramp.
Constant velocity means the forces balance, so the normal force tells you the weight
The parcel keeps the same speed the whole way down. Watch which force cancels the normal force once the weight is split along and into the ramp. The arrows are drawn to scale.
Into the ramp, only $F_N$ and $mg\cos\theta$ act, and the question gives you $F_N$. One balance and one division give the mass.
Pick a relationship and see whether the mass balances the parcel
Each button is an equation a student could write for this parcel. The figure keeps the real forces from the stem, $F_N=98.2$ N and kinetic friction $F_f=0.37\times98.2=36.33$ N, and draws the weight from the mass that equation gives. The question is Method 1.
This is the question, Method 1. The parcel is not accelerating, so the forces into the ramp balance and $F_N=mg\cos\theta$. It uses the one force the stem gives you and no coefficient at all. The dashed $mg\cos20^\circ$ is exactly as long as $F_N$.
This is Method 2. Along the ramp, the pull down the slope balances kinetic friction. It lands on 10.84 kg, not 10.66 kg, because the stem’s 0.37 is slightly more than $\tan20^\circ=0.364$. No single mass balances both directions exactly, so QCAA accepts 10.66 to 11 kg.
Static friction applies to something at rest or on the point of sliding. This parcel is already sliding, so 0.39 is a distractor. The mass of 11.43 kg is outside the range, and its weight would push 105.22 N into the ramp against a 98.2 N normal force.
Sine and cosine are swapped. The component into the ramp is the side next to the angle, so it takes the cosine. With 29.30 kg the weight would press 269.80 N into the ramp, nearly three times the normal force. Check with $\theta=0^\circ$: a flat surface must give $F_N=mg$, and only cosine does that.
$F_N=mg$ only holds on a horizontal surface. On a 20° ramp the surface carries just the part of the weight pressing into it, so the parcel must weigh more than 98.2 N. The mass of 10.02 kg is too small and outside the range.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The guide gives two methods worth the same four marks. Each card shows Method 1 first, then the Method 2 version of the same mark.
Say why you are allowed to balance the forces before you write any equation. Constant velocity means zero acceleration, so the net force is zero in every direction. Perpendicular to the ramp:
$$F_\perp=F_N$$
This sentence earns a mark of its own, separate from the calculation. Method 2 makes the same move along the ramp instead, $F_\parallel=F_f$.
Marker: recognises that forces are balanced, therefore $F_\perp=F_N$. Method 2: recognises that forces are balanced, therefore $F_\parallel=F_f$
Resolve the weight. The component into the ramp is the side next to the 20° angle, so it takes the cosine.
$$F_N=mg\cos\theta$$
Method 2 needs two relationships for this mark, $F_\parallel=mg\sin\theta$ and $F_f=\mu_kF_N$, so $mg\sin\theta=\mu_kF_N$. Use $\mu_k$ because the parcel is already sliding. The static coefficient never enters this question.
Marker: recognises $F_N=mg\cos\theta$. Method 2: recognises $F_\parallel=mg\sin\theta$ and $F_f=\mu_kF_N$. “This mark may be implied by subsequent working”
Rearrange for $m$ and substitute. Write the substitution out in full, because it is the mark.
$$m=\frac{F_N}{g\cos\theta}=\frac{98.2}{9.8\cos20^\circ}$$
Method 2 substitutes into its own rearrangement:
$$m=\frac{\mu_kF_N}{g\sin\theta}=\frac{0.37\times98.2}{9.8\sin20^\circ}$$
The guide’s Method 1 prints 9.82 on top of the fraction. That is a typo for 98.2, since 9.82 would give 1.07 kg and the guide’s own answer is 10.66 kg.
Marker: correctly substitutes values
Evaluate, with the calculator in degrees.
Method 2 gives $m=10.84$ kg. The two methods disagree because the stem is slightly inconsistent. A parcel sliding at constant velocity needs $\mu_k=\tan20^\circ=0.364$, and the stem says 0.37. The guide covers both answers with one range, and 11 kg to the nearest whole unit is inside it too.
Marker: determines the mass of the parcel. “Accept 10.66-11 kg for mass inclusive”
Each phrase of the question gave you something
The 2024 subject report says effective responses understood the forces acting on the parcel and the components of its weight, which the report calls horizontal and vertical but which are really along and perpendicular to the ramp. It printed excerpts using both valid methods, so either route can earn all four marks.
QCAA did not publish a band for this question. It reads as simple familiar because it is the balanced-forces shape you have practised, with one balance equation and one rearrangement, and the stem hands you the normal force directly.
If I had to guess where you might lose marks, it would be the first one. Jumping straight to $98.2\div(9.8\cos20^\circ)$ gets the right mass, but the guide pays a separate mark for stating that the forces balance, so $F_\perp=F_N$. Write that sentence down. The next most likely loss is reaching for the 0.39 because it is the last number in the stem. The parcel is sliding, so the static coefficient does not apply, and 11.43 kg falls outside the range.
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 2024 QCAA Engineering external assessment, © State of Queensland (QCAA) 2024, 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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