Question

Two blocks each of mass $m$ are placed on a rough horizontal surface and connected by a massless inelastic string as shown in the figure. The coefficient of friction between each block and horizontal surface is $\mu$. The string connecting both the blocks initially has zero tension. The minimum force to be applied on block A to just move the two block system horizontally (without the string getting slack, given acceleration due to gravity $=g$) is:

• A. $2\mu mg$
• B. $\frac{2\mu mg}{\sqrt{{\mu }^2+1}}$
• C. $\frac{2\mu mg}{\sqrt{4{\mu }^2+1}}$
• D. $\frac{\mu mg}{\sqrt{{\mu }^2+1}}$

Answer 1

The correct option is B $\frac{2\mu mg}{\sqrt{{\mu }^2+1}}$

From FBD of B:

$T=\mu N_1=\mu mg....\left(i\right)$

From FBD of A:
$Fsin\theta +N_2=mg.....\left(ii\right)$
$Fcos\theta =T+\mu N_2.....\left(iii\right)$

From equations (i), (ii) and (iii)
$F\cos \theta =\mu mg+\mu (mg-Fsin\theta )$
$F(cos\theta +\mu sin\theta )=2\mu mg$

$F=\frac{2\mu mg}{cos\theta +\mu sin\theta }$....(i)

as this force is minimum differentiating w.r.t. $\theta$

$\frac{dF}{d\theta }=-\frac{2\mu mg}{(cos\theta +\mu sin\theta {)}^2}(-sin\theta +\mu cos\theta )=0$

$\Rightarrow tan\theta =\mu$....(ii)

substituting $sin\theta$ and $cos\theta$ in equation (i) from equation (ii)

Hence, $F_{min}=\frac{2\mu mg}{\sqrt{1+{\mu }^2}}$