A block of mass m is placed on a smooth wedge | vedclass

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Question

(D) To keep the block stationary relative to the wedge,the net force acting on the block along the inclined plane must be zero.$1$. Resolve the forces

Vedclass · Accepted Answer

$mg/\cos 	heta$

Vedclass · Answer

(D) To keep the block stationary relative to the wedge,the net force acting on the block along the inclined plane must be zero.$1$. Resolve the forces acting on the block in the frame of the wedge:- Gravitational force $mg$ acting downwards.- Pseudo force $ma$ acting horizontally in the direction opposite to the acceleration of the wedge.- Normal reaction $R$ exerted by the wedge perpendicular to the inclined surface.$2$. For the block not to slip,the component of $mg$ down the plane must be balanced by the component of the pseudo force $ma$ up the plane:$mg \sin 	heta = ma \cos 	heta$$a = g 	an 	heta$$3$. The normal force $R$ balances the components of $mg$ and $ma$ perpendicular to the inclined plane:$R = mg \cos 	heta + ma \sin 	heta$Substitute $a = g 	an 	heta = g \frac{\sin 	heta}{\cos 	heta}$:$R = mg \cos 	heta + m(g \frac{\sin 	heta}{\cos 	heta}) \sin 	heta$$R = mg \frac{\cos^2 	heta + \sin^2 	heta}{\cos 	heta}$$R = \frac{mg}{\cos 	heta}$

$A$ block of mass $m$ is placed on a smooth wedge of inclination $\theta$. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block ($g$ is acceleration due to gravity) will be

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