A block of mass m is kept on a wedge of mass | vedclass

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(C) Let the acceleration of the system be $a$. The force $P$ applied to the system is given by $P = (M+m) a$.To keep the block of mass $m$ stationary

Vedclass · Accepted Answer

$(M+m) g 	an \beta$

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(C) Let the acceleration of the system be $a$. The force $P$ applied to the system is given by $P = (M+m) a$.To keep the block of mass $m$ stationary with respect to the wedge,we analyze the forces in the non-inertial reference frame of the wedge. The pseudo-force acting on the block is $ma$ in the direction opposite to the acceleration.Resolving the forces acting on the block $m$ along the inclined plane:$1$. The component of the gravitational force down the incline is $mg \sin \beta$.$2$. The component of the pseudo-force $ma$ up the incline is $ma \cos \beta$.For the block to remain stationary with respect to the wedge,these two components must balance each other:$ma \cos \beta = mg \sin \beta$Dividing both sides by $m \cos \beta$,we get:$a = g 	an \beta$Substituting this value of $a$ into the expression for $P$:$P = (M+m) g 	an \beta$

$A$ block of mass $m$ is kept on a wedge of mass $M$,as shown in the figure,such that mass $m$ remains stationary with respect to the wedge. The magnitude of force $P$ is

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