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(B) For the box to move,the applied force $F$ must be equal to the limiting friction force:$F = \mu mg \dots (1)$For the box not to topple,the torque

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(B) For the box to move,the applied force $F$ must be equal to the limiting friction force:$F = \mu mg \dots (1)$For the box not to topple,the torque about the front edge must be zero or balanced. The force $F$ is applied at a height $h = \frac{a}{2} + b$ from the base. The normal force $N$ shifts to the front edge to prevent toppling. Taking torque about the front edge:$F \left( \frac{a}{2} + b ight) = mg \left( \frac{a}{2} ight) \dots (2)$Substituting $F = \mu mg$ from equation $(1)$ into equation $(2)$:$\mu mg \left( \frac{a}{2} + b ight) = mg \left( \frac{a}{2} ight)$$\mu \left( \frac{a}{2} + b ight) = \frac{a}{2}$Given $\mu = 0.4 = \frac{2}{5}$,we have:$\frac{2}{5} \left( \frac{a}{2} + b ight) = \frac{a}{2}$Multiply by $5$:$2 \left( \frac{a}{2} + b ight) = 2.5a$$a + 2b = 2.5a$$2b = 1.5a$$\frac{b}{a} = \frac{1.5}{2} = 0.75$Therefore,$100 	imes \frac{b}{a} = 100 	imes 0.75 = 75$.

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