A block of mass m is at rest relative to the stationary wedge of mass M. The coefficient of friction between block and wedge is µ. The wedge is now pulled horizontally with acceleration ‘a’ as shown in figure. Then the minimum magnitude of ‘a’ for the friction between block and wedge to be zero is:
Text Solution
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The free body diagram of the block is as shown in the figure.
N is the normal reaction exerted by wedge on the block.
The wedge moves towards left with acceleration ‘a’, then the component of acceleration of block
normal to the plane is.
Applying Newtons second law to the block normal to plane.
mg cos θ – N = ma sin θ
For N to be zero a = g cot θ .
Hence the friction shall be zero when a = g cot θ .
mg cos θ – N = ma sin θ
N a = g cot θ .
a = g cot θ .
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