A block of mass 3 ,kg is pressed against a ve | vedclass

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Question

(A) The forces acting on the block are as follows:$1$. The normal reaction $N$ exerted by the wall is $N = F \cos 30^{\circ} = F \frac{\sqrt{3}}{2}$.$

Vedclass · Accepted Answer

$30 \,N$

Vedclass · Answer

(A) The forces acting on the block are as follows:$1$. The normal reaction $N$ exerted by the wall is $N = F \cos 30^{\circ} = F \frac{\sqrt{3}}{2}$.$2$. The maximum static friction force is $f_{max} = \mu N = \sqrt{3} 	imes (F \frac{\sqrt{3}}{2}) = \frac{3}{2} F$.$3$. The vertical forces acting on the block are the weight $mg$ acting downwards and the vertical component of the applied force $F \sin 30^{\circ}$ acting downwards.$4$. For the block to be in equilibrium and prevented from falling, the upward friction force must balance the total downward force:$f = mg + F \sin 30^{\circ}$$\frac{3}{2} F = (3 	imes 10) + F 	imes \frac{1}{2}$$\frac{3}{2} F - \frac{1}{2} F = 30$$F = 30 \,N$.

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