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2020 · Section 3 · Calculation

Question 20

[9 marks] Complex unfamiliar

Topic guidesPremiumTopic guide 3 · Worm drives, gear trains and compound machines Topic guide 3 · Worked example: 2020 Question 20

Game plansPremium2.2 Gear trains and compound machines 3.4 Running an efficiency

The question

A double-threaded worm and 40-tooth worm wheel are used to drive a mechanical winch. The efficiency of the machine is 45%. The pitch of a double-threaded worm is halved.

a)Determine the mass of the load lifted by the winch using an effort of 50 N on the circumference of the worm wheel. Answer to the nearest whole unit.[5 marks]

b)An effort wheel is added to the worm axle, decreasing the efficiency of the compound machine to 36%. Determine the velocity ratio of the added component. Answer to two decimal places.[4 marks]

Watch the situation first

Two starts move the wheel two teeth per turn, so the VR is 40 ÷ 2 = 20

The worm is drawn from the side with its two threads in two colours. Watch the orange tooth on the wheel while the worm turns three times, and watch the drum on the same axle wind up the load.

The VR counts teeth and starts. The pitch never enters it, so “the pitch is halved” changes nothing in this question.

Now for the engineering

Change the starts, the teeth and the efficiencies

Set the worm and wheel, then run the chain to the mass. Halve the pitch and watch which numbers move. The question is 2 starts, 40 teeth, 45%, 50 N and 36% for the compound machine.

Starts on the worm

Worm pitch

Teeth on the worm wheel:

This is the question. 40 teeth over 2 starts gives $\text{VR}=20$. At 45% the MA is 9, so 50 N lifts 450 N, a mass of 46 kg. The added wheel must be 80% efficient to bring the compound machine down to 36%.

QCAA marking guide · 9 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it.

Part a) · Step 11 mark

Double-threaded means two starts. Each turn of the worm moves the wheel on by two teeth.

$$\text{VR}=\frac{\text{worm wheel teeth}}{\text{worm starts}}$$

$$=\frac{40}{2}=20$$

The halved pitch does not appear. Pitch sets the size of each tooth, and the VR only counts teeth and starts. The guide writes “worm teeth” on the bottom of this fraction, meaning the starts.

Marker: provides correct working to give correct gear ratio (VR)

Part a) · Step 21 mark

You know the VR and want the MA. That is the forwards direction, so multiply by the efficiency as a decimal.

$$\eta=\frac{\text{MA}}{\text{VR}}\times100\%$$

$$0.45=\frac{\text{MA}}{20}$$

$$\text{MA}=0.45\times20=9$$

Marker: provides correct formula and substituted values for mechanical advantage. “Formula is considered correct if substituted and constant values are correctly applied.”

Part a) · Step 31 mark

The 50 N is the effort going into the machine. Rearrange the formula book MA for the load.

$$\text{MA}=\frac{F_L}{F_E}$$

$$F_L=\text{MA}\times F_E$$

$$=9\times50=450\text{ N}$$

The guide writes L and E. The 2026 formula book writes $F_L$ and $F_E$, and either is fine.

Marker: provides correct working to give correct L

Part a) · Step 42 marks

The question asks for a mass, so 450 N is not the answer yet. The load force is the weight, $F=mg$.

$$m=\frac{F}{g}=\frac{450}{9.8}=45.92\text{ kg}$$

$m\approx46\text{ kg}$

Round only here, and give the unit. A wrong MA from step 2 still earns these marks if you carry it through correctly.

Marker: provides correct formula and substituted values for mass. Provides correct working to give correct answer to the nearest whole unit with correct unit provided. “Allow FT error for incorrect MA in determining the answer.”

Part b) · Step 52 marks

The effort wheel drives the worm, so the two machines are in series. Their efficiencies multiply.

$$\eta_{\text{total}}=\eta_1\times\eta_2$$

$$0.36=0.45\times\eta_2$$

$$\eta_2=\frac{0.36}{0.45}=0.80$$

This rule is not in the formula book. The added wheel is 80% efficient on its own, and the compound machine is less efficient than either stage.

Marker: provides correct formula and substituted values for effort wheel efficiency. Provides correct working to give correct effort wheel efficiency.

Part b) · Step 62 marks

Now run the efficiency backwards. The VR is the ideal, bigger number, so you divide by the efficiency.

$$\eta=\frac{\text{MA}}{\text{VR}}$$

$$\text{VR}=\frac{\text{MA}}{\eta}=\frac{9}{0.80}$$

$\text{VR}=11.25$

This is QCAA’s published working and it earned the marks in 2020. It pairs the worm’s MA of 9 with the effort wheel’s efficiency, because the question gives no MA for the added wheel on its own. If the compound machine still lifts 450 N with 50 N, its overall VR would be $9\div0.36=25$ and the wheel would add $25\div20=1.25$. Learn the guide’s route for this question, and expect a 2026 question to give you the forces you need.

Marker: provides correct formula and substituted values for velocity ratio. Provides correct working to give correct answer to two decimal places. “Allow FT error for incorrect η₂ in determining the answer.”

Putting it all together

Each phrase of the question gave you something

“A double-threaded worm and 40-tooth worm wheel”

The 2 starts go on the bottom of the VR fraction and the 40 teeth on top, so $\text{VR}=20$.

“The efficiency of the machine is 45%”

$\eta_1=0.45$. Forwards from the VR, so $\text{MA}=0.45\times20=9$.

“The pitch of a double-threaded worm is halved”

You do not use it. The VR counts teeth and starts, and no mark in the guide touches the pitch.

“an effort of 50 N on the circumference of the worm wheel”

$F_E=50\text{ N}$, the force going into the machine. The guide multiplies it by the MA of 9.

“Determine the mass of the load”

Finish with $m=F\div g$. Stopping at 450 N leaves two marks behind.

“decreasing the efficiency of the compound machine to 36%”

This is $\eta_{\text{total}}$, the product of the two stages. Divide it by 0.45 to get the added wheel’s 0.80.

“Answer to two decimal places”

The guide’s answer is 11.25, so the two decimal places come out exactly.

What makes this complex unfamiliar

There is no diagram, the stem carries a sentence you must ignore, and part b) bolts a second machine onto the first and gives you only the combined efficiency. The 2020 subject report lists two complex unfamiliar calculation questions worth 18 marks in Section 3, and Questions 20 and 22 are the only 9-mark questions there, so the report places this one as complex unfamiliar.

If I had to guess where you might lose marks, it would be the pitch sentence. Use it to halve or double the VR and you lose the VR mark, then lean on follow-through for everything after it. The other trap is the efficiency direction. Part a) multiplies by 0.45 and part b) divides by 0.80. The 2023 subject report calls the shortcut of multiplying by 1.2 for an 80% machine incorrect practice, so divide by the decimal every time you go from output back to input.

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 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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