Question

Given, $m_A=30\text{kg},m_B=10\text{kg}$ and $m_C=20\text{kg}$. The co-efficient of friction between $\text{A}$ and $\text{B}$ is ${\mu }_1=0.3$, between $\text{B}$ and $\text{C}$ is ${\mu }_2=0.2$ and between $\text{C}$ and ground is ${\mu }_3=0.1$. The least horizontal force $F$ required to start motion of any part of the system of three blocks resting upon the another as shown below is-

• A. $90\text{N}$
• B. $80\text{N}$
• C. $60\text{N}$
• D. $150\text{N}$

Answer 1

The correct option is C $60\text{N}$

Limiting friction between block $\text{A}$ and $\text{B}$ is,

$f_1={\mu }_1{m}_{A}g=90\text{N}$

Limiting friction between block $\text{B}$ and $\text{C}$ is,

$f_2={\mu }_2(m_A+m_B)g=80\text{N}$

Limiting friction between block $\text{C}$ and ground is,

$f_3={\mu }_3(m_A+m_B+m_C)g=60\text{N}$

As $F$ is gradually increased, the force of friction between $\text{A}$ and $\text{B}$ will increase. When $F=60\text{N}$, block $A$ will exert, force of $60\text{N}$ on $C$ horizontally. Hence, $C$ will be on the point of motion.

Therefore, the least value of $F$ is $60\text{N}$.

Hence, option (c) is the correct answer.