Question 8
A chain block is used to lift an engine with a weight of 1200 N. The effort force required is 250 N and the chain block is 80% efficient.
What is the velocity ratio of the chain block?
Not quite. You used η as 80 instead of 0.8: $\dfrac{4.8}{80} = 0.06$.
Not quite. You multiplied by the efficiency instead of dividing: $4.8 \times 0.8 = 3.84$.
Not quite. You stopped at MA and gave 4.8 as the VR.
That is it. Step through the hints below to see why.
Pull the hand chain and the engine rises by the chain pulled divided by the VR of the option you are testing. Black bar: the MA from the forces. Orange bar: that option's VR. The efficiency bar must stop at the dashed 80% line, and can never pass 100%.
A VR of 0.06 gives an efficiency of 8000%, which is impossible. The engine would rise 100 m for every 6 m of chain pulled. You needed 0.8, not 80.
A VR of 3.84 is smaller than the MA of 4.8, so the efficiency would be 125%. No machine puts out more work than goes in.
A VR of 4.8 equals the MA, so the efficiency would be 100%. That is a chain block with no friction, but this one is 80% efficient.
A VR of 6 gives 4.8 ÷ 6 × 100% = 80%, the efficiency the question gives. Pull 6 m of chain and the engine rises 1 m.
Start with the two forces you are given: the load is the engine’s weight and the effort is the pull on the chain.
$$MA=\frac{F_L}{F_E}=\frac{1200}{250}=4.8$$
Put the MA and the efficiency into the efficiency formula and solve for VR.
$$\begin{aligned}\eta&=\frac{MA}{VR}\times100\%\\80&=\frac{4.8}{VR}\times100\\VR&=\frac{4.8\times100}{80}=6\end{aligned}$$
This is option (D).
Friction makes MA smaller than VR, so VR is MA divided by 0.8.
$MA=\dfrac{1200}{250}=4.8$, and with an efficiency of 80%, $VR=\dfrac{4.8}{0.8}=6$.
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.
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