Sum
Find the mass M of the hanging block in the following figure that will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulleys are light.
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Solution
The free-body diagram of the system is shown below:
Block ‘m’ will have the same acceleration as that of M', as it does not slip over M'.
From the free body diagrams,
T + Ma – Mg = 0 ...(i)
T – M'a – Rsinθ = 0 ...(ii)
Rsinθ – ma = 0
Rcosθ – mg = 0
Eliminating T, R and a from the above equations, we get:
\[M = \frac{M' + m}{\cot \theta - 1}\]
Concept: Newton’s Second Law of Motion
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