JEE Main · PhysicsHard

$A block of mass $m$ is placed on a wedge of mass $M$ and angle $\theta$. The wedge is placed on a smooth horizontal surface. If the coefficient of friction between the block and the wedge is $\mu$, find the horizontal force $F$ that must be applied to the wedge so that the block does not slide relative to the wedge.$

A. $(M+m)g\frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}$
B. $(M+m)g\frac{\sin\theta - \mu\cos\theta}{\cos\theta + \mu\sin\theta}$
C. $mg\frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}$
D. $(M+m)g\tan\theta$

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Correct answer: A — $(M+m)g\frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}$

Solution

To prevent sliding, the pseudo force

ma

acting on the block in the non-inertial frame of the wedge must be balanced by the normal and frictional forces. By resolving forces parallel and perpendicular to the inclined plane, we solve for the acceleration

a = g \frac{s i n θ + μ c o s θ}{c o s θ - μ s i n θ}

and then the total force

F = (M + m) a

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