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(C) For vertical equilibrium,the normal force $N$ is given by:$N + F \sin 	heta = W \Rightarrow N = W - F \sin 	heta$The block will move when the ho

Vedclass · Accepted Answer

$\frac{W \sin \alpha}{\cos (	heta - \alpha)}$

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(C) For vertical equilibrium,the normal force $N$ is given by:$N + F \sin 	heta = W \Rightarrow N = W - F \sin 	heta$The block will move when the horizontal component of the applied force $F$ is greater than or equal to the limiting friction $f_{\max}$:$F \cos 	heta \geq f_{\max}$Since $f_{\max} = \mu N$ and $\mu = 	an \alpha$,we have:$F \cos 	heta \geq 	an \alpha (W - F \sin 	heta)$Substituting $	an \alpha = \frac{\sin \alpha}{\cos \alpha}$:$F \cos 	heta \geq \frac{\sin \alpha}{\cos \alpha} (W - F \sin 	heta)$$F \cos 	heta \cos \alpha \geq W \sin \alpha - F \sin 	heta \sin \alpha$$F (\cos 	heta \cos \alpha + \sin 	heta \sin \alpha) \geq W \sin \alpha$Using the trigonometric identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$:$F \cos (	heta - \alpha) \geq W \sin \alpha$Therefore,the minimum force required is:$F = \frac{W \sin \alpha}{\cos (	heta - \alpha)}$

$A$ pulling force $F$ making an angle $\theta$ with the horizontal is applied on a block of weight $W$ placed on a horizontal table. If the angle of friction is $\alpha$,then the magnitude of the force required to move the body is equal to

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