Discussion
A force of 400 N is applied to the brake drum of 0.5 m diameter in a band-brake system as shown in the figure, where the wrapping angle is 180°. If the coefficient of friction between the drum and the band is 0.25, the braking torque applied, in N.m is
Answer: B
The key's answer is 54.4 N.m.
The band wraps 180° (\(\pi\) rad) with \(\mu = 0.25\), so the tension ratio is:
\[ \frac{T_{1}}{T_{2}} = e^{\mu\theta} = e^{0.25\pi} = 2.193 \]
The key takes the 400 N pull as the tight-side tension T1 and the anchored end as the slack side:
\[ \begin{aligned} T_{2} &= 400/2.193 = 182.4\ \text{N} \\ T &= (T_{1}-T_{2})r = (217.6)(0.25) \\ &= 54.4\ \text{N.m} \end{aligned} \]
Read strictly, the drum's rotation as drawn drags the band towards the anchored end, which would make that end the tight side: its tension would be 877 N and the torque 119.3 N.m. That value is not among the options; the key's 54.4 N.m is the torque with the 400 N side tight. 100.6, 22.1 and 15.7 N.m do not follow from either reading.
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