(C) To find the minimum force required to topple the block,we consider the torque about the edge of the block where it is about to rotate (the pivot p

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(C) To find the minimum force required to topple the block,we consider the torque about the edge of the block where it is about to rotate (the pivot point $P$).At the critical condition for toppling,the normal reaction $N$ acts through the pivot point $P$.The torque due to the gravitational force $mg$ about point $P$ is $\tau_{mg} = mg \times (L/2)$.The torque due to the applied horizontal force $F$ about point $P$ is $\tau_{F} = F \times L$.For the block to topple,the torque due to the applied force must be greater than the torque due to gravity:$\tau_{F} > \tau_{mg}$$F \times L > mg \times (L/2)$$F > mg/2$Thus,the minimum force required to topple the block is $F = mg/2$.

Class 11 Physics System of Particles and Rotational Motion Toppling and Instantaneous Axis of Rotation

Difficult

$A$ cubical block of side $L$ rests on a rough horizontal surface with coefficient of friction $\mu$. $A$ horizontal force $F$ is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling,the minimum force required to topple the block is

IIT 2000 All IIT

A
$Infinitesimal$
B
$mg/4$
C
$mg/2$
D
$mg(1 - \mu)$

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Class 11

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System of Particles and Rotational Motion

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Toppling and Instantaneous Axis of Rotation

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The instantaneous velocity of point $B$ of a rod of length $0.5 \ m$ is $3 \ m/s$ at an angle of $30^{\circ}$ with the rod as shown. Find the angular velocity of the rod such that the velocity of end $A$ is minimum.

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