Question

A wedge of mass $m$ , lying on a rough horizontal plane, is acted upon by a horizontal force $F_{1}$ and another force $F_{2}$ , inclined at an angle $\theta $ to the vertical. The block is in equilibrium, then the minimum coefficient of friction between it and the surface is

• A. $\frac{F_{2} sin\theta }{\left(\right. mg + F_{2} cos\theta \left.\right)}$
• B. $\frac{\left(\right. F_{1} cos\theta + F_{2} \left.\right)}{mg - F_{2} sin\theta }$
• C. $\frac{\left(\right. F_{1} + F_{2} sin\theta \left.\right)}{\left(\right. mg + F_{2} cos\theta \left.\right)}$
• D. $\frac{\left(\right. F_{1} sin\theta - F_{2} \left.\right)}{\left(\right. mg - F_{2} cos\theta \left.\right)}$

Answer 1

Answer: (C)

We know $f=μN$ and $N=mg+F_{2}cos\theta $
Now, since wedge is in equilibrium,
$\left(\right.F_{1}+F_{2}sin\theta \left.\right) \, = \, \mu \left(\right.mg+F_{2}cos\theta \left.\right)$
$\mu =\frac{F_{1} + F_{2} s i n \theta }{m g + F_{2} c o s \theta }$