Question

A block placed on a horizontal surface is being pushed by a force $F$ making an angle $\theta$ with the vertical as shown in figure. 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 equibrium of the block,
$\begin{array}{}N=F\cos \theta +mg& \text{(1)}\end{array}$
While for horizontal motion in equilibrium
$F\sin \theta -\mu N=0$or $\begin{array}{}\frac{F\sin \theta }{\mu }=N& \text{(2)}\end{array}$
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).