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A rectangular block of mass M and height a is resting on a smooth level surface. A force F is applied to one corner as shown in figure. At what point should a parallel force 3F be applied in order that the block shall undergo pure translational motion? Assume normal contact force between the block and surface passes through the centre of gravity of the block.

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$\frac{a}{3}$ vertically above centre of gravity

$\frac{a}{6}$ vertically above centre of gravity

No such point exists

it is not possible

answer is B.

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Detailed Solution

The free-body diagram of the block can be drawn as shown. As the body has to move in pure translational motion, the torque about the centre of gravity must be zero.

$3 F \times x = F \times \frac{a}{2} \Rightarrow x = \frac{a}{6}$

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