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(B) $1$. Consider the free body diagram of the block of mass $m = 2.5 	ext{ kg}$ placed on the wedge.$2$. The normal force $N$ exerted by the wedge o

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$12$

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(B) $1$. Consider the free body diagram of the block of mass $m = 2.5 	ext{ kg}$ placed on the wedge.$2$. The normal force $N$ exerted by the wedge on the block balances the component of gravity perpendicular to the inclined surface: $N = mg \cos 37^{\circ}$.$3$. Given $m = 2.5 	ext{ kg}$, $g = 10 	ext{ m/s}^2$, and $\cos 37^{\circ} = 0.8$, we have $N = 2.5 	imes 10 	imes 0.8 = 20 	ext{ N}$.$4$. Now, consider the wedge. The block exerts a normal force $N$ on the wedge in the direction perpendicular to the incline. The horizontal component of this force is $N \sin 37^{\circ}$ (or $N \cos 53^{\circ}$).$5$. Since the wedge is in equilibrium, the spring force $F_s$ must balance this horizontal component: $F_s = N \sin 37^{\circ}$.$6$. Substituting the values: $F_s = 20 	imes 0.6 = 12 	ext{ N}$.$7$. Thus, the reading of the spring balance is $12 	ext{ N}$.

Find the reading of the spring balance as shown in the figure. Assume that mass $M$ is in equilibrium and the block of mass $2.5 \text{ kg}$ is also in equilibrium. All surfaces are smooth. (Take $g = 10 \text{ m/s}^2$) (in $\text{ N}$)

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