Two masses A and B of 10, kg and 5, kg respec | vedclass

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

Question

(A) For the system to be in equilibrium,the tension $T$ in the string must be equal to the weight of block $B$,so $T = m_B g = 5g$.For block $A$ (with

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) For the system to be in equilibrium,the tension $T$ in the string must be equal to the weight of block $B$,so $T = m_B g = 5g$.For block $A$ (with block $C$ on top) to remain stationary,the limiting friction $f_L$ must be at least equal to the tension $T$.The normal force $N$ on block $A$ is $(m_A + m_C)g$.The limiting friction is $f_L = \mu N = \mu (m_A + m_C)g$.Setting $f_L = T$,we get $\mu (m_A + m_C)g = m_B g$.Substituting the given values: $0.2 	imes (10 + m_C) = 5$.$2 + 0.2 m_C = 5$.$0.2 m_C = 3$.$m_C = \frac{3}{0.2} = 15\, kg$.

Similar Questions

$A$ body of mass $m$ is kept on a rough horizontal surface (coefficient of friction $= \mu$). $A$ horizontal force is applied on the body,but it does not move. The resultant of the normal reaction and the frictional force acting on the object is given by $F$,where $F$ is:

$A$ body of mass $4\, kg$ is at rest on a horizontal table. The coefficient of friction between the body and the table is $0.2$. $A$ horizontal force of $6\, N$ is applied on the body. The acceleration of the body is .......... $m/s^2$.

$A$ man walks over a rough surface; the angle between the force of friction and the instantaneous velocity of the person is

If the mass of an object is doubled,what will be the effect on the coefficient of friction?

Which of the following is self-adjusting in nature?

Vedclass Test Series

Exam Paper Generator

Online Exam Module