Question 21
An electric winch lifts pallets of roof tiles vertically onto a roof at a slow, constant speed. The winch motor is connected to a 240 V supply and draws a current of 10 A. The motor has an electrical efficiency of 85%. The motor drives the winch drum through a gearbox, and the gearbox and drum have a combined mechanical efficiency of 70%.
Each pallet and its lifting cradle have a combined mass of 600 kg and are lifted 12 m from the ground to the roof. Allow 2 minutes after each lift for the cradle to be unloaded and lowered back to the ground.
Determine the maximum number of pallets that the winch can lift to the roof in one hour.
Two efficiencies shrink the power, then the hour fills with cycles
The width of the orange band is the power, drawn to scale. The building is drawn to scale for the 12 m lift, and the hour bar is drawn to scale in time. Solid orange blocks are lifts, grey blocks are the 120 s unload and return.
Each efficiency is applied to the power that reaches that stage, so the band narrows twice. The leftover quarter of a cycle never gets a pallet to the roof.
Change the winch, then try each slip
Move the sliders and watch the power band narrow and the hour fill with cycles. The pallet count is always rounded down. Then try each slip to see the wrong count it gives. The question is 240 V, 10 A, 85%, 70%, 600 kg, 12 m and 120 s.
These are the settings in the question. 1428 W lifts each pallet in 49.41 s, a cycle is 169.41 s and 21.25 cycles fit in the hour. Only 21 of them finish, so the answer is 21 pallets.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Start with the electrical power the motor takes from the supply. Multiply the supply voltage by the current.
$$\begin{aligned}P&=VI\\P_{in}&=240\times10\\&=2400\text{ W}\end{aligned}$$
The supply voltage and current set the power going into the motor.
Marker: identifies that the electrical input power to the motor is calculated using $P=VI$
Marker: correctly calculates the electrical input power
Write the efficiency formula, substitute 85% and the 2400 W input, and solve for the output.
$$\begin{aligned}\eta&=\frac{P_{out}}{P_{in}}\times100\%\\85&=\frac{P_{out}}{2400}\times100\\P_{out}&=0.85\times2400\\&=2040\text{ W}\end{aligned}$$
The motor converts only 85% of its input to mechanical output.
Allow FT marks for errors in prior working.
Marker: provides correct formula and substituted values for the motor output power using the electrical efficiency
Use the same formula again, with 70% and the 2040 W that comes out of the motor as the input.
$$\begin{aligned}\eta&=\frac{P_{out}}{P_{in}}\times100\%\\70&=\frac{P_{out}}{2040}\times100\\P_{out}&=0.70\times2040\\&=1428\text{ W}\end{aligned}$$
The motor's output is the gearbox's input, and the gearbox and drum pass on 70% of it.
Accept the overall efficiency, $\eta=0.85\times0.70\times100\%=59.5\%$, being used to determine the useful power output in one step ($0.595\times2400=1428$ W). This earns the third and fourth bullets.
Marker: determines the useful power output using the mechanical efficiency of the gearbox and drum
Say what the useful work is, then calculate it for one pallet and its cradle.
$$\begin{aligned}W&=\text{PE}=mgh\\\text{PE}&=600\times9.8\times12\\&=70\,560\text{ J}\end{aligned}$$
The pallet rises at constant speed, so all the useful work goes into potential energy.
Marker: identifies that the useful work done on each pallet is equal to its gain in potential energy
Marker: correctly calculates the potential energy gained by each pallet
Rearrange the power formula for the time, and use the useful power from Step 3, not the 2400 W input.
$$\begin{aligned}P&=\frac{W}{t}\\t&=\frac{W}{P}=\frac{70\,560}{1428}\\&=49.41\text{ s}\end{aligned}$$
Power is the rate of doing work, so the lift time is the work divided by the useful power.
Accept lift time range from 49.41–49.42 s inclusive.
Marker: determines the time taken to lift one pallet
Add the 2 minutes for unloading and lowering, converted to seconds.
$$\begin{aligned}t_{cycle}&=49.41+(2\times60)\\&=169.41\text{ s}\end{aligned}$$
Each pallet ties up the winch for its lift plus 2 min (120 s) to unload and lower the cradle.
Accept cycle time range from 169.41–169.42 s inclusive.
Marker: determines the time for one complete cycle, including the 2 minutes (120 s) to unload and lower the cradle
Divide one hour, in seconds, by the cycle time.
$$\begin{aligned}\text{cycles}&=\frac{3600}{169.41}\\&=21.25\end{aligned}$$
One hour is 3600 s.
Accept number of cycles 21.25.
Marker: determines the number of cycles in one hour
Round down. The 22nd lift would still be under way when the hour ends.
$$21.25\ \Rightarrow\ 21\text{ pallets}$$
A part-cycle does not deliver a pallet, so round down.
Accept 21 pallets whether or not the final unload and return is counted: 21 lifts take at most 3558 s, and 22 lifts need at least 3607 s.
The final mark requires 21 (not 22): a rounded-up answer does not earn it. Allow FT for a whole number rounded down from the response's own value.
Marker: determines that the maximum number of pallets is 21, recognising that only complete cycles count
Each phrase of the question gave you something
If I had to guess where you might lose marks…
- Ignoring both efficiencies. Using 2400 W as the lifting power gives $t=29.40$ s, 24.10 cycles, so 24 pallets. Marks 3 and 4 are lost. The rest can earn FT.
- Applying only the 85% motor efficiency. 1428 W is never found, so $t=34.59$ s, 23.29 cycles, so 23 pallets.
- Leaving out the 120 s unload-and-return time. $3600\div49.41=72.86$, giving 72 pallets. Mark 8 is lost. The rest can earn FT.
- Rounding 21.25 up to 22 pallets. The final mark needs 21.
The past QCAA questions this one was modelled on
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it