Question

In the given figure, the coefficient of friction between the two blocks is $\mu$, and all other surfaces are smooth. Find the minimum value of $F$ which will prevent slipping.

• A. $\frac{(m_1+m_2)m_{1}g}{\mu m_2}$
• B. $\frac{(m_1+m_2)g}{\mu m_2}$
• C. $\frac{(m_1+m_2)g}{\mu m_1}$
• D. $\frac{(m_1+m_2)m_{1}g}{m_2}$

Answer 1

The correct option is A $\frac{(m_1+m_2)m_{1}g}{\mu m_2}$

Suppose both blocks are moving with same acceleration $a$.

So, from Newton's second law,

$F=(m_1+m_2)a$

$\Rightarrow a=\frac{F}{(m_1+m_2)}..........\left(1\right)$


From FBD of block $1$,

$F-R=m_{1}a$

putting the value of $a$ from eq.(1)

$\Rightarrow R=\frac{F{m}_2}{(m_1+m_2)}..........\left(2\right)$

Net vertical force should be zero on block $1$ to prevent slipping,

So, $f_{\text{max}}\ge m_{1}g$

$\Rightarrow \mu R\ge m_{1}g$

$\Rightarrow \mu \frac{F{m}_2}{(m_1+m_2)}\ge m_{1}g\text{(from eq. (2))}$

$\Rightarrow F\ge \frac{(m_1+m_2)m_{1}g}{m_2\mu }$

for minimum value of $F$,

$\Rightarrow F=\frac{(m_1+m_2)m_{1}g}{m_2\mu }$

Hence, option $\left(a\right)$ is the correct answer.