Question

Figure shows the top view of a horizontal surface. Two blocks each of mass $m$ are placed on the surface and connected with a string. The friction coefficient is $\mu$ for each block. A horizontal force $F$ is applied on one of the blocks as shown in the figure. $F$ has the maximum value such that there is no sliding at any contact. Then:

• A. If $\theta ={30}^{\circ },F=\frac{2\mu mg}{\sqrt{3}}$ & $T<\mu mg$
• B. If $\theta ={45}^{\circ },F=\sqrt{2}\mu mg$ & $T=\mu mg$
• C. If $\theta ={60}^{\circ },F=2\mu mg$ & $T<\mu mg$
• D. If $\theta ={60}^{\circ },F=\sqrt{3}\mu mg$ & $T=\mu mg$

Answer 1

Answer: (A), (B), (D)

If $\theta ={60}^{\circ },F=\sqrt{3}\mu mg$ & $T=\mu mg$

FBD of the block on which $F$ is acting:


$T\cos \theta =f\sin \alpha$
$\Rightarrow T\cos \theta =\mu mg\sin \alpha$
[$\because f=\mu mg$ in limiting case]

& $F=T\sin \theta +f\cos \alpha$
$=\mu mg\frac{\sin \alpha }{\cos \theta }\sin \theta +\mu mg\cos \alpha$
$=\mu mg\frac{\cos (\alpha -\theta )}{\cos \theta }$

For $F_{max},\alpha =\theta <{45}^{\circ }$
$\Rightarrow F_{max}=\mu mg\sec \theta$

But $T\le \mu mg$ [for no slipping of 2nd block]
$T=\frac{\mu mg\sin \theta }{\cos \theta }<\mu mg$
$\Rightarrow \theta$ should be less than ${45}^{\circ }$

Now, if $\theta >{45}^{\circ },T=\mu mg$
$\Rightarrow \sin \alpha =\cos \theta$ or $\alpha =90°–\theta$
$F=\mu mg\sin \theta +\mu mg\cos \alpha =2\mu mg\sin \theta$

Hence:
If $\theta <{45}^{\circ }$, then $F_{max}=\mu mg\sec \theta$
and if $\theta >{45}^{\circ }$, then $F_{max}=2\mu mg\sin \theta$

Substituting the given angles,
For $\theta ={30}^{\circ },T<\mu mg$ & $F_{max}=\sqrt{2}\mu mg$
For $\theta ={45}^{\circ },T=\mu mg$ & $F_{max}=\sqrt{2}\mu mg$
For $\theta ={60}^{\circ },T=\mu mg$ & $F_{max}=\sqrt{3}\mu mg$

Hence, options (a), (b) and (d) are correct.