A wooden wedge of mass M and inclination angl | vedclass

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Question

(A) Let the acceleration of the wedge be $a$ towards the left. Since the block is stationary with respect to the wedge,the block also moves with the s

Vedclass · Accepted Answer

$(M+m) g 	an \alpha$

Vedclass · Answer

(A) Let the acceleration of the wedge be $a$ towards the left. Since the block is stationary with respect to the wedge,the block also moves with the same acceleration $a$ towards the left.For the entire system (wedge + block),the total force is $F = (M+m) a \dots (i)$.Now,consider the free body diagram of the block of mass $m$ in the non-inertial frame of the wedge. The forces acting on the block are:$1$. Gravitational force $mg$ downwards.$2$. Normal force $N$ perpendicular to the inclined surface.$3$. Pseudo force $ma$ acting horizontally towards the right.For the block to remain stationary on the wedge,the component of the pseudo force along the incline must balance the component of the gravitational force along the incline:$ma \cos \alpha = mg \sin \alpha$$a = g \frac{\sin \alpha}{\cos \alpha} = g 	an \alpha \dots (ii)$Substituting equation $(ii)$ into equation $(i)$,we get:$F = (M+m) g 	an \alpha$.

$A$ wooden wedge of mass $M$ and inclination angle $\alpha$ rests on a smooth floor. $A$ block of mass $m$ is kept on the wedge. $A$ force $F$ is applied on the wedge as shown in the figure such that the block remains stationary with respect to the wedge. The magnitude of the force $F$ is:

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