Question

Two blocks each of mass $m$ are placed on a rough horizontal surface and connected by a massless inelastic string as shown. 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) 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 block B,
$T=\mu N_1=\mu mg$ .... (i)
From FBD of block A,
$Fsin\theta +N_2=mg$ .... (ii)
$Fcos\theta =T+\mu N_2$ .... (iii)
From equations (i), (ii) and (iii)
$Fcos\theta =\mu mg+\mu (mg-Fsin\theta )$
$F(cos\theta +\mu sin\theta )=2\mu mg$
$F=\frac{2\mu mg}{cos\theta +\mu sin\theta }$
hence, $F_{min}=\frac{2\mu mg}{\sqrt{1+{\mu }^2}}$