A plank is moving in a horizontal direction w | vedclass

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Question

$(b)$ Given situation is Forces acting on the block are

In above diagram, ma=pseudo force due to acceleration of plank,

$f=$ force of stati

Vedclass · Accepted Answer

$(-l / 20,0)$

Vedclass · Answer

$(b)$ 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.

$\Rightarrow \quad 	au_{\substack{	ext { Normal } \ 	ext { reaction }}}=	au_{	ext {Friction }}$

$m g$ and $m a$ does not produces any torque about centre of mass as their line of action passes through centre of mass.

$\Rightarrow N \cdot x=f \cdot l / 2$

$\Rightarrow m g x=m a \frac{l}{2}$

${[\because N=m g 	ext { and } f=m a]}$

$\Rightarrow x=\left(\frac{a}{g}ight) \frac{l}{2}$

$	ext { or } x=\frac{1}{10} \cdot \frac{l}{2} \quad\left[\because a=\frac{g}{10}ight]$

$	ext { or } x=\frac{l}{20}$

So, coordinates of point from which reaction passes are $\left(-\frac{l}{20}, 0ight)$.

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