Question 9
A drum winch is used to vertically raise a 50 kg motor as shown. It has a velocity ratio of 3 and is 90% efficient.
What is the minimum effort force required to raise the motor using the drum winch?
Not quite. $50\div2.7=18.52$ divides the mass, not the weight. The load force is $mg=490\text{ N}$, so you need the 9.8. In 2024, 8.4% of Queensland students chose (A).
Not quite. $490\div3=163.33\text{ N}$ is the effort for a perfect winch. At 90% the MA is $0.9\times3=2.7$, so the effort is larger. In 2024, 9.3% of Queensland students chose (B).
That is it. Step through the hints below to see why. In 2024, 74.5% of Queensland students chose (C).
Not quite. $490\times2.7=1323\text{ N}$ multiplies by the MA. A winch reduces the force you need, so the effort has to come out smaller than the 490 N load. In 2024, 7.3% of Queensland students chose (D).
The winch lifts the motor “vertically”, so the load is the full weight.
$$F_L=mg=50\times9.8=490\text{ N}$$
The 50 kg is a mass. Leaving out $g$ is how you land on (A).
Rearrange the formula book’s $\eta=\frac{\text{MA}}{\text{VR}}\times100\%$, with $\eta$ as a decimal.
$$\text{MA}=\eta\times\text{VR}=0.90\times3=2.7$$
Using the VR of 3 in place of the MA gives the perfect-winch effort, (B).
The “minimum effort” is just enough to raise the motor steadily. Rearrange $\text{MA}=\frac{F_L}{F_E}$.
$$\begin{aligned}F_E&=\frac{F_L}{\text{MA}}=\frac{490}{2.7}\\[10pt]&=181.48\text{ N}\end{aligned}$$
A machine with an MA above 1 makes the effort smaller than the load. That rules out (D).
$$\begin{aligned}F_E&=\frac{mg}{\eta\times\text{VR}}\\[10pt]&=\frac{50\times9.8}{0.90\times3}\\[10pt]&=181.48\text{ N}\end{aligned}$$
Weight first, then the MA from the efficiency and the VR, then the load over the MA.
The most common wrong answer was (B), the effort for a perfect winch. A real winch wastes some of the effort to friction, so it always needs more than the load divided by the VR.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and any figures 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.
Spot a mistake on this page? Report it