(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

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

(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: $0.2 \times (10 + m_C) = 5$.$2 + 0.2 m_C = 5$.$0.2 m_C = 3$.$m_C = \frac{3}{0.2} = 15\, kg$.

Class 11 Physics Newton's Laws of Motion and Friction Static and Limiting Friction and Minimum Force Required to Move

Medium

Two masses $A$ and $B$ of $10\, kg$ and $5\, kg$ respectively are connected with a string passing over a frictionless pulley fixed at the corner of a table as shown. The coefficient of static friction of $A$ with the table is $0.2$. The minimum mass of $C$ that may be placed on $A$ to prevent it from moving is ........ $kg$.

A
$15$
B
$10$
C
$5$
D
$12$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Newton's Laws of Motion and Friction

Subtopic

Static and Limiting Friction and Minimum Force Required to Move

The maximum value of the applied force $F$ such that the block as shown in the arrangement does not move is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $\,N$)

EasyAP EAMCET 2019

View Solution

$A$ person initially at rest starts walking towards the east without slipping or skidding. What is the type of friction acting on the person from the ground, and in which direction does the frictional force act?

EasyTS EAMCET 2023

View Solution

$A$ box of mass $15 \text{ kg}$ is kept on the floor of a stationary trolley. The coefficient of static friction between the box and the trolley is $0.12$. Keeping the box in a stationary state over the trolley, the maximum acceleration with which the trolley can be moved horizontally in $\text{m s}^{-2}$ is: $(g = 10 \text{m s}^{-2})$

MediumNEET 2026

View Solution

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

Easy

View Solution

$A$ block of mass $M$ is being pulled along a rough horizontal surface. The coefficient of friction between the block and the surface is $\mu$. If another block of mass $M/2$ is placed on the block and it is again pulled on the surface,the coefficient of friction between the block and the surface will be

Easy

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

The maximum value of the applied force $F$ such that the block as shown in the arrangement does not move is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $\,N$)

$A$ person initially at rest starts walking towards the east without slipping or skidding. What is the type of friction acting on the person from the ground, and in which direction does the frictional force act?

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module