A block of mass m is lying on a wedge having inclination of α=tan−1(15). Wedge is moving with a constant acceleration of 2m/s2. Find the minimum value of coefficient of friction so that m remains stationary with respect to wedge. Take g=10m/s2.
A
29
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B
512
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C
15
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D
25
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Solution
The correct option is B512 In the FBD the pseudo force will be acting on the block towards left as the wedge is accelerating towards right,
Balancing the forces acting perpendicular to the plane: N+masinα=mgcosα ⇒N=mgcosα−masinα Along the plane mgsinα+macosα=f=μN mgsinα+macosα=μ(mgcosα−masinα) μ=gtanα+ag−atanα, μ=g(15)+ag−a(15) putting the values we get, μ=512