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(A) The block is in equilibrium. We resolve the forces acting on it.Given: $	an 	heta = \frac{3}{4}$, so $\sin 	heta = \frac{3}{5}$ and $\cos 	het

Vedclass · Accepted Answer

$N_1 = 37.2 \ N$ and $N_2 = 9.6 \ N$

Vedclass · Answer

(A) The block is in equilibrium. We resolve the forces acting on it.Given: $	an 	heta = \frac{3}{4}$, so $\sin 	heta = \frac{3}{5}$ and $\cos 	heta = \frac{4}{5}$.The applied force of $12 \ N$ has components:Horizontal component $F_x = 12 \cos 	heta = 12 	imes \frac{4}{5} = 9.6 \ N$.Vertical component $F_y = 12 \sin 	heta = 12 	imes \frac{3}{5} = 7.2 \ N$.For horizontal equilibrium, the normal reaction from the wall $N_2$ must balance the horizontal component of the applied force:$N_2 = F_x = 9.6 \ N$.For vertical equilibrium, the upward normal reaction from the ground $N_1$ must balance the downward forces (weight of the block, the $10 \ N$ downward force, and the vertical component of the $12 \ N$ force):Weight of block $W = mg = 2 	imes 10 = 20 \ N$.$N_1 = W + 10 + F_y = 20 + 10 + 7.2 = 37.2 \ N$.Thus, $N_1 = 37.2 \ N$ and $N_2 = 9.6 \ N$.

$A$ block is placed between two surfaces as shown in the figure. Find the normal reaction at both surfaces. [Assume $g = 10 \ m/s^2$]

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