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

Answer 1

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


From FBD of block B,
$T=\mu N_1=\mu mg\dots \left(i\right)$

From FBD of block A,
$F\sin \theta +N_2=mg\dots \left(ii\right)$
$F\cos \theta =T+\mu N_2\dots \left(iii\right)$

From equations $\left(i\right)$, $\left(ii\right)$ and $\left(iii\right)$
$F\cos \theta =\mu mg+\mu (mg-F\sin \theta )$
$F(\cos \theta +\mu \sin \theta )=2\mu mg$
$F=\frac{2\mu mg}{\cos \theta +\mu \sin \theta }$
Hence,$F_{min}=\frac{2\mu mg}{\sqrt{1+{\mu }^2}}$

Final Answer: Option(b)