Question

A block placed on a horizontal surface is being pushed by a force $F$ making an angle $\theta$ with the vertical. The coefficient of friction between block and surface is $\mu$. The force required to slide the block with uniform velocity on the floor is

• A. $\frac{\mu mg}{(\sin \theta -\mu \cos \theta )}$
• B. $\frac{(\sin \theta -\mu \cos \theta )}{\mu mg}$
• C. $\mu mg$
• D. None of these

Answer 1

The correct option is A $\frac{\mu mg}{(\sin \theta -\mu \cos \theta )}$

For vertical equilibrium of the block,
$N=F\cos \theta +mg$ ... (1)
While for horizontal motion
$F\sin \theta -\mu N=ma$ ... (2)
or $F\sin \theta =\mu N$
From Eqs. (1) and (2),
$F\sin \theta =\mu (F\cos \theta +mg)$
i.e., $F=\frac{\mu mg}{(\sin \theta -\mu \cos \theta )}$
Hence, the correct answer is option (a).