A block of mass m rests on a rough horizontal | vedclass

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(D) The forces acting on the block are its weight $mg$ downwards,the normal reaction $R$ upwards,and the applied force $F = mg$ at an angle $	heta$ w

Vedclass · Accepted Answer

$\cot \frac{	heta}{2} \geq \mu$

Vedclass · Answer

(D) The forces acting on the block are its weight $mg$ downwards,the normal reaction $R$ upwards,and the applied force $F = mg$ at an angle $	heta$ with the vertical.Resolving the force $F$ into components:Horizontal component $= F \sin 	heta = mg \sin 	heta$Vertical component $= F \cos 	heta = mg \cos 	heta$For vertical equilibrium:$R + F \cos 	heta = mg$$R = mg - mg \cos 	heta = mg(1 - \cos 	heta)$The frictional force $f_r$ is given by:$f_r = \mu R = \mu mg(1 - \cos 	heta)$The block will be pulled if the horizontal component of the force is greater than or equal to the frictional force:$mg \sin 	heta \geq \mu mg(1 - \cos 	heta)$$\sin 	heta \geq \mu(1 - \cos 	heta)$Using trigonometric identities $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ and $1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$:$2 \sin \frac{	heta}{2} \cos \frac{	heta}{2} \geq \mu (2 \sin^2 \frac{	heta}{2})$$\cos \frac{	heta}{2} \geq \mu \sin \frac{	heta}{2}$$\cot \frac{	heta}{2} \geq \mu$

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