A plank is moving in a horizontal direction with a constant accel… - Vidyadip

MCQ

A plank is moving in a horizontal direction with a constant acceleration $\alpha \hat{ i }$. A uniform rough cubical block of side $l$ rests on the plank and is at rest relative to the plank. Let the centre of mass of the block be at $(0, l / 2)$ at a given instant. If $\alpha =g / 10$, then the normal reaction exerted by the plank on the block at that instant acts at

A
$(0,0)$
✓
$(-l / 20,0)$
C
$(-l / 10,0)$
D
$(l / 10,0)$

✓

Answer

Correct option: B.

$(-l / 20,0)$

b
$(b)$ Given situation is Forces acting on the block are

In above diagram, ma=pseudo force due to acceleration of plank,

$f=$ force of static friction, $m g=$ weight of block and $N=$ normal reaction.

Here, we must note that due to motion of plank, normal reaction force does not passes through centre of mass.

If line of action of normal force at a distance $x$ from centre, then for equilibrium of block, net torque about centre of mass must be zero.

$\Rightarrow \quad \tau_{\substack{\text { Normal } \\ \text { reaction }}}=\tau_{\text {Friction }}$

$m g$ and $m a$ does not produces any torque about centre of mass as their line of action passes through centre of mass.

$\Rightarrow N \cdot x=f \cdot l / 2$

$\Rightarrow m g x=m a \frac{l}{2}$

${[\because N=m g \text { and } f=m a]}$

$\Rightarrow x=\left(\frac{a}{g}\right) \frac{l}{2}$

$\text { or } x=\frac{1}{10} \cdot \frac{l}{2} \quad\left[\because a=\frac{g}{10}\right]$

$\text { or } x=\frac{l}{20}$

So, coordinates of point from which reaction passes are $\left(-\frac{l}{20}, 0\right)$.

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