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. $\text{None of these}$

Answer 1

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

The given condition can be reframed as shown below in the figure

For vertical equilibrium of the block
$N=F\cos \theta +mg$......(1)
As block will move with uniform velocity, the acceleration $a$ will be zero.
Hence, for horizontal motion
$F\sin \theta -\mu N=0$
$\Rightarrow F\sin \theta =\mu N$....(2)

From Eqs. (1) and (2) we get
$F\sin \theta =\mu (F\cos \theta +mg)$
i.e., $F=\frac{\mu mg}{(\sin \theta -\mu \cos \theta )}$