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2022 · Section 3 · Calculation

Question 22

[9 marks] Complex unfamiliar
Topic guidesPremiumTopic guide 6 · Resolving the weight on a slopeTopic guide 3 · The two conversions every question needs
Game plansPremium1.3 Two stages chained2.3 Turns, distance and the drive between them
The question

An airport luggage handling system consists of three connected sections.

  • Section 1: Luggage slides away from the aircraft down a 25-metre-long steel ramp set at 45° to the horizontal.
  • Section 2: Luggage travels along a 45-metre-long motorised horizontal conveyor.
  • Section 3: Luggage travels along a 30-metre-long motorised variable-speed horizontal conveyor before reaching the luggage carousel.
Section 3 conveyor, not to scale. A belt runs over a row of rollers. The conveyor is marked 30 m long and 250 mm high, and the roller at the right-hand end is labelled motor-driven roller.

Determine the revolutions per minute (rpm) of the variable-speed conveyor motor in Section 3 if the coefficient of kinetic friction (µk) between luggage and steel is 0.42 and the luggage

  • takes 20 seconds to move from the aircraft to the carousel
  • spends half as long in Section 2 as Section 3.
Answer to the nearest whole unit. State any assumptions made.[9 marks]
Watch the situation first

The ramp takes 3.53 s, and two thirds of what is left belongs to Section 3

The clock runs in real time, so the bag leaves the aircraft at 0 s and reaches the carousel at 20 s. The strip at the top is to one scale. Below it is QCAA’s Section 3 conveyor, which starts moving when the bag arrives.

Section 1: from rest down the ramp at 4.02 m/s² Section 2: 45 m in half the Section 3 time Section 3: 30 m in 10.98 s is 2.73 m/s, so 209 rpm aircraft Section 1 · 25 m at 45° Section 2 · 45 m Section 3 · 30 m carousel mg sin 45° = 6.93 N Ff = 2.91 N per kg: 6.93 − 2.91 = 4.02 N down the ramp Section 3 conveyor 30 m 250 mm Not to scale Motor-driven roller π × 0.25 = 0.785 m per turn2.73 ÷ 0.785 × 60 = 209 rpm ramp 3.53 s Section 2: 5.49 s Section 3: 10.98 s 0 3.53 9.02 20 s

Friction only matters on the ramp. After that the question is about time: 20 s in total, 3.53 s on the ramp, and the rest split 1 to 2.

Now for the engineering

Change the ramp, the time split and the roller

Each change ripples down the chain to the rpm. The question is 45°, $\mu_k=0.42$, 20 s in total, Section 2 taking half the Section 3 time, and a 250 mm roller.

Section 2 time compared with Section 3

QCAA marking guide · 9 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it.

Section 1 · Step 11 mark

Resolve the weight along the ramp. The component $mg\sin45^\circ$ pulls the bag down the ramp, and kinetic friction $F_f=\mu_k F_N=0.42\,mg\cos45^\circ$ acts up the ramp, against the slide.

45° mg mg sin 45° FN = mg cos 45° Ff = 0.42 mg cos 45°

The guide’s sample response draws this diagram with the $\mu_k=0.42$ label. Name both forces along the ramp in your working.

Marker: correctly identifies the forces acting up and down the ramp in Section 1

Section 1 · Step 22 marks

The net force down the ramp is the difference. The mass cancels when you divide by $m$.

$$F_{\text{net}}=ma$$
$$=mg(\sin45^\circ-0.42\cos45^\circ)$$
$$a=9.8\times(0.707-0.297)$$
$$=4.02\text{ m/s}^2$$

The guide also accepts working with a 1 kg bag: $6.93-2.91=4.02$ N, so $a=4.02\div1=4.02$ m/s². No mass is given, and either route earns all three of the first marks.

Marker: correctly determines the net force parallel to the ramp in Section 1. Correctly determines the acceleration down the ramp in Section 1. “Accept acceleration down the ramp range from 4.019194944 to 4.02 m/s2.” “Allow FT error for forces acting up and down the ramp when determining acceleration.”

Assumptions · Step 31 mark

The stem says “State any assumptions made”, and one valid assumption is a mark. Write it in a sentence before you use it.

  • The bag starts from rest at the top of the ramp, so $u=0$.
  • The time to transfer from one section to the next is negligible.
  • The bag moves at belt speed as soon as it reaches each conveyor, and the roller diameter is the 250 mm shown.

The third one is worth knowing. The bag reaches the foot of the ramp at about 14 m/s, faster than either belt, so the model only works if each change of speed is instant.

Marker: provides one or more valid assumption/s. “Valid assumptions may also include that the time for luggage to transfer from section to section is negligible and that the initial velocity on the steel ramp is 0.”

Section 1 · Step 41 mark

You know $s$, $u$ and $a$ and want $t$. Choose the motion equation without $v$.

$$s=ut+\tfrac{1}{2}at^2$$
$$25=0\times t+\tfrac{1}{2}\times4.02\,t^2$$

Marker: provides correct formula and substituted values for displacement

Section 1 · Step 51 mark
$$t^2=\frac{50}{4.02}$$
$$t=\sqrt{12.438}=3.53\text{ s}$$

Marker: correctly determines the time required to complete Section 1. “Accept time to complete Section 1 range from 3.526728079 to 3.53 s.” “Allow FT error for incorrect acceleration down the ramp when determining time to complete Section 1.”

Section 3 · Step 61 mark

Take the ramp time off the 20 s. Section 2 takes half as long as Section 3, so the remaining time splits 1 to 2 and Section 3 gets two thirds.

$$20-3.53=16.47\text{ s}$$
$$t_3=\frac{16.47}{3}\times2=10.98\text{ s}$$

Check it: $t_2=5.49$ s is half of 10.98 s, and $3.53+5.49+10.98=20$ s.

Marker: correctly determines the time required for Section 3. “Accept time required for Section 3 range from 10.9 to 11 s.” “Allow FT error for incorrect time to complete Section 1 when determining time to complete Section 3.”

Section 3 · Step 71 mark

The Section 3 belt runs at a steady speed, so this is the average speed formula.

$$v=\frac{s}{t}=\frac{30}{10.98}$$
$$=2.73\text{ m/s}$$

Marker: correctly determines the velocity in Section 3. “Accept velocity in Section 3 range from 2.7 to 2.8 m/s.” “Allow FT error for incorrect time required for Section 3 when determining velocity.”

Section 3 · Step 81 mark

The belt moves one roller circumference for every turn of the motor-driven roller. Turns per second times 60 gives rpm.

$$C=\pi D=\pi\times0.250$$
$$=0.785\text{ m}$$
$$\text{rpm}=\frac{2.73}{0.785}\times60$$
$$=208.66$$
$209\text{ rpm}$

Carrying every value unrounded gives 208.69, which also rounds to 209. The guide accepts 206 to 214 rpm, so early rounding does not cost you here, but the $\times60$ and the unit do.

Marker: correctly determines the answer to the nearest whole unit with correct unit provided. “Allow FT error for incorrect velocity in Section 3 when determining the answer.” “Accept answer range from 206 to 214 rpm”

Putting it all together

Each phrase of the question gave you something

“slides away from the aircraft down a 25-metre-long steel ramp set at 45°”
An incline with kinetic friction, $s=25$ m and $u=0$. Resolve the weight along the slope.
“coefficient of kinetic friction (µk) between luggage and steel is 0.42”
It only applies on the steel ramp. The conveyors are belts and carry the bag at belt speed.
“a 45-metre-long motorised horizontal conveyor”
You never use the 45 m. Section 2 only enters through its time.
“takes 20 seconds to move from the aircraft to the carousel”
The total time for all three sections, $t_1+t_2+t_3=20$ s.
“spends half as long in Section 2 as Section 3”
$t_2=\tfrac{1}{2}t_3$, so $t_3$ is two thirds of what is left after the ramp.
“30-metre-long” and “250 mm” on the Section 3 conveyor
The 30 m gives the belt speed and the 250 mm is the roller diameter for $\pi D$.
“State any assumptions made”
A mark of its own. Write $u=0$ and negligible transfer time in words.
What QCAA said about this question

The 2022 subject report says the strongest responses found the ramp time by naming the forces up and down the incline and the net force before the acceleration, then found the Section 3 time and finished with the rpm to the nearest whole unit. The same report asks teachers to give students more work on the parallel and perpendicular components of the weight on an incline.

What makes this complex unfamiliar

It chains three topics, friction on an incline, motion and a roller conversion, and the link between them is a time split you have to set up yourself. It also hides information you do not need, the 45 m, and asks for assumptions. The 2022 subject report calls it a complex unfamiliar question.

If I had to guess where you might lose marks, it would be the time split. Read “half as long in Section 2 as Section 3” backwards and you give Section 3 one third instead of two thirds, which doubles the answer to 417 rpm. The other easy loss is the assumption mark, because nothing in the working forces you to write one.

Only mark this done when you could do it without help. Reading the solution does not count.

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Question wording, any figures and marking-guide steps 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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