In the figure shown,there is friction between | vedclass

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(B) The block $P$ is carried along with block $Q$ due to the static friction force acting between them. The maximum static friction force available is

Vedclass · Accepted Answer

$x=+A$

Vedclass · Answer

(B) The block $P$ is carried along with block $Q$ due to the static friction force acting between them. The maximum static friction force available is $f_{max} = \mu N = \mu m_P g$,where $m_P$ is the mass of block $P$.The acceleration of the system is given by $a = -\omega^2 x$. The force required to keep block $P$ moving with block $Q$ is $F = m_P a = -m_P \omega^2 x$.Block $P$ will slip when the required force exceeds the maximum static friction,i.e.,$|m_P \omega^2 x| > \mu m_P g$,or $|x| > \frac{\mu g}{\omega^2}$.Since the acceleration magnitude $|a| = \omega^2 |x|$ is maximum at the extreme positions $(x = \pm A)$,the required force is maximum at these points. Therefore,block $P$ is most likely to slip at the extreme positions,$x = +A$ or $x = -A$.

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