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Question

One vertical face is subjected to a gradually increasing horizontal force. If the force is applied near the bottom of the face, the block will slide b

Vedclass · Accepted Answer

$\frac{{mg}}{2}$

Vedclass · Answer

One vertical face is subjected to a gradually increasing horizontal force. If the force is applied near the bottom of the face, the block will slide before it topples as the force increases. If the force is applied towards the top, the block will topple over before sliding.

Step $1:$ To find the minimum force required to topple the block.

Then,

$N$ and the weight of the block $W = mg$.

Since no sliding, $F = f$ and also, $N = mg$.

As the block starts to topple the normal force shifts to the edge $O$ of the block and so the perpendicular distance from the centre of gravity becomes $\frac{1}{2}$ as shown in figure $b$.

Then the torque acting in the clockwise direction will be $	au_{	ext {clock }}=F \frac{L}{2}+f \frac{L}{2}$

The torque acting in the anticlockwise direction will be $	au_{	ext {enticlock }}= N \frac{L}{2} \ldots$. $(2)$. We have to find the minimum applied force for which the block topples.

Step $2$: Express the torque balance equation of the block.

The torque balance equation can be expressed as $	au_{	ext {clock }}=	au_{2 	ext { nticlock }}-(3)$

Substituting equations $(1)$ and $(2)$ in equation $(3)$ we get, $F \frac{L}{2}+f_{\frac{L}{2}}= N \frac{1}{2}$ $\Rightarrow F + f = N$

Substituting for $F=f$ and $N=m g$ in the above expression we get, $F+F=m g$ $\Rightarrow F =\frac{ m z}{2}$

$	herefore$ the minimum force required to topple the block will be $F=\frac{m s}{2}$.

Therefore the correct option is ' $C$ ' is $\frac{m g}{2}$.

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