A block of mass M is placed on a horizontal s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For the block $m$ to remain stationary,the tension $T$ in the string must be equal to the weight of the block $m$,so $T = mg$.For the block $M$ (w

Vedclass · Accepted Answer

$\frac{m}{\mu} - M$

Vedclass · Answer

(A) For the block $m$ to remain stationary,the tension $T$ in the string must be equal to the weight of the block $m$,so $T = mg$.For the block $M$ (with $m_0$ on top) to remain stationary,the tension $T$ must be balanced by the maximum static friction force $f_{max}$ acting between block $M$ and the horizontal surface.The normal force $N$ acting on the surface is $N = (M + m_0)g$.The maximum static friction force is $f_{max} = \mu N = \mu(M + m_0)g$.For equilibrium,$T = f_{max}$,which implies $mg = \mu(M + m_0)g$.Dividing both sides by $g$,we get $m = \mu(M + m_0)$.Rearranging for $m_0$,we have $\frac{m}{\mu} = M + m_0$,which gives $m_0 = \frac{m}{\mu} - M$.

Similar Questions

What can be inferred from the relation $f_S \leqslant \mu_S N$?

Which of the following forces does not depend on the velocity: the frictional force between the contact surfaces of two solids or the viscous force?

How does static friction oppose motion? On which factors does the value of the coefficient of friction depend?

Vedclass Test Series

Exam Paper Generator

Online Exam Module