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(A) The weight of the block,$W = Mg$,acts vertically downwards.We can resolve this weight into two rectangular components:$1$. $A$ component perpendic

Vedclass · Accepted Answer

$M g \cos 	heta$

Vedclass · Answer

(A) The weight of the block,$W = Mg$,acts vertically downwards.We can resolve this weight into two rectangular components:$1$. $A$ component perpendicular to the inclined plane: $Mg \cos 	heta$.$2$. $A$ component parallel to the inclined plane: $Mg \sin 	heta$.Since the block is at rest on the inclined plane,the net force perpendicular to the plane must be zero.The plane exerts a normal reaction force $N$ on the block,which balances the perpendicular component of the weight.Therefore,$N = Mg \cos 	heta$.The total force exerted by the plane on the block is the normal reaction force $N$ (assuming a frictionless surface or considering only the normal component as implied by the options).Thus,the magnitude of the force exerted by the plane on the block is $Mg \cos 	heta$.

$A$ block of mass $M$ is at rest on a plane surface inclined at an angle $\theta$ to the horizontal. The magnitude of the force exerted by the plane on the block is

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