GATE 2022 ME (ME2) – Question 30
A rope with two mass-less platforms at its two ends passes over a fixed pulley as shown in the figure. Discs with narrow slots and having equal weight of 20 N each can be placed on the platforms. The number of discs placed on the left side platform is $n$ and that on the right side platform is $m$. It is found that for $n=5$ and $m=0$, a force $F=200$ N is just sufficient to initiate upward motion of the left side platform. If the force $F$ is removed then the minimum value of $m$ required to prevent downward motion of the left side platform is ______ (in integer).
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Correct answer: 3
Explanation
Use the capstan (rope-friction) relation, $T_1/T_2=e^{\mu\theta}$, together with the equilibrium of each platform.
**Step 1: limit of motion with $F=200$ N.** The left platform carries $n=5$ discs, i.e. $5\times20=100$ N. Pulling the right side with 200 N is just enough to start the left side moving **up**. So the larger tension (200 N, on the pulling side) and the smaller tension (100 N, the load) are in the ratio
$$e^{\mu\theta}=\frac{200}{100}=2.$$
**Step 2: with $F$ removed.** The right platform carries $m$ discs of 20 N each. The left platform (100 N) tends to move **down**, so friction opposes it. At the verge of motion
$$\frac{T_{left}}{T_{right}}=e^{\mu\theta}=2\;\Rightarrow\;T_{right}=\frac{100}{2}=50\text{ N}.$$
**Step 3: number of discs.** The right side must hold at least 50 N:
$$20m\geq50\;\Rightarrow\;m\geq2.5.$$
Since $m$ is a whole number, the minimum is $\mathbf{m=3}$.