Question

Two blocks each of mass $1kg$ are placed as shown. They are connected by a string which passes over a smooth (massless) pulley. There is no friction between $m_2$ and the ground. The coefficient of friction between $m_1$ and $m_2$ is $0.2$. A force $F$ is applied to $m_1$. Which of the following statements is/are correct?

• A. The system will be in equilibrium if $F\le 4N$
• B. If $F>4N$, tension in the string will be $4N$
• C. If $F>4N$, the frictional force between the blocks will be $2N$
• D. If $F=6N$, tension in the string will be $3N$

Answer 1

Answer: (A), (C), (D)

The correct options are
A The system will be in equilibrium if $F\le 4N$
C If $F>4N$, the frictional force between the blocks will be $2N$
D If $F=6N$, tension in the string will be $3N$

Given:

• $m_1=m_2=1kg$
• $\mu =0.2$ (between $m_1$ and $m_2$)

Frictional force $f=\mu N$ where the normal force $N=1kg\times g=g$

$\Rightarrow f=0.2g$

Let tension in the string be $T$.

Note: Frictional force is experienced by both $m_1$ and $m_2$ due to the relative motion between them. $m_2$ experiences $f_{m_2}$ to the left and $m_1$ experiences $f_{m_1}$ to the right. $f_{m_2}=f_{m_1}=f$.

Forming the equations for the system to be in rest,

$F=T+f_{m_1}⟶$(1)

$T=f_{m_2}⟶$(2)

From (1) and (2) $F=2f$

$\Rightarrow F=0.4g=4N$ ($\because g\approx 10m/s^2$)

Hence system will remain in equilibrium for $F\le 4N$.

Note: Frictional force value never exceeds the opposing force value. It always tries to get even with the opposing force (if f’s upper limit > opposing force)

If $F>2N$ (opposing force greater than the f's upper limit) then $f=0.2g=2N$.

If $F>4N$ there is an acceleration in $m_1$ and $m_2$ of the system. Acceleration of $m_1$ and $m_2$ is the same.

$\therefore \frac{T-f_{m_2}}{m_2}=\frac{F-(T+f_{m_1})}{m_1}$

$\Rightarrow T-f=F-T-f$

$\Rightarrow T=\frac{F}{2}$

If $F=6N\Rightarrow T=3N$.

Hence options A, C and D are correct.