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(B) The forces acting on the block in the frame of the plank are the pseudo force $ma$ (acting backwards at the centre of mass),the weight $mg$ (actin

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$(-l/20, 0)$

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(B) The forces acting on the block in the frame of the plank are the pseudo force $ma$ (acting backwards at the centre of mass),the weight $mg$ (acting downwards at the centre of mass),the static friction $f$ (acting at the base),and the normal reaction $N$ (acting at the base at some distance $x$ from the centre line).For the block to be in rotational equilibrium,the net torque about the centre of mass must be zero.The torque due to the pseudo force $ma$ and the weight $mg$ is zero because their lines of action pass through the centre of mass.The torque due to friction $f$ is $	au_f = f \cdot (l/2)$.The torque due to the normal reaction $N$ is $	au_N = N \cdot x$.Setting $	au_N = 	au_f$,we get $N \cdot x = f \cdot (l/2)$.Since $N = mg$ and $f = ma$,we have $mg \cdot x = ma \cdot (l/2)$.Thus,$x = (a/g) \cdot (l/2)$.Given $a = g/10$,we get $x = (1/10) \cdot (l/2) = l/20$.Since the pseudo force acts in the positive $x$-direction (relative to the block's frame,it acts backwards),the normal reaction must shift in the negative $x$-direction to balance the torque. Therefore,the point of application is $(-l/20, 0)$.

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