A force F = mg is acting on a block of mass m | vedclass

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(C) The normal force $N$ acting on the block is given by balancing the vertical forces: $N = mg + F \cos 	heta$. Since $F = mg$,we have $N = mg + mg

Vedclass · Accepted Answer

$	an \frac{	heta}{2} \ge \mu$

Vedclass · Answer

(C) The normal force $N$ acting on the block is given by balancing the vertical forces: $N = mg + F \cos 	heta$. Since $F = mg$,we have $N = mg + mg \cos 	heta = mg(1 + \cos 	heta)$.Using the trigonometric identity $1 + \cos 	heta = 2 \cos^2 \frac{	heta}{2}$,we get $N = 2 mg \cos^2 \frac{	heta}{2}$.The horizontal component of the force is $F_x = F \sin 	heta = mg \sin 	heta$.Using the double angle identity $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$,we get $F_x = 2 mg \sin \frac{	heta}{2} \cos \frac{	heta}{2}$.For the block to move,the horizontal force must be greater than or equal to the limiting friction: $F_x \ge \mu N$.Substituting the expressions: $2 mg \sin \frac{	heta}{2} \cos \frac{	heta}{2} \ge \mu (2 mg \cos^2 \frac{	heta}{2})$.Dividing both sides by $2 mg \cos \frac{	heta}{2}$ (assuming $\cos \frac{	heta}{2} 
eq 0$),we get $\sin \frac{	heta}{2} \ge \mu \cos \frac{	heta}{2}$.Therefore,$	an \frac{	heta}{2} \ge \mu$.

$A$ force $F = mg$ is acting on a block of mass $m$ as shown in the figure. The pushing of the block along the surface is possible if:

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