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. $\frac{F_{2}sin\theta }{(mg+F_{2}cos\theta )}$
• B. $\frac{(F_{1}cos\theta +F_2)}{mg-F_{2}sin\theta }$
• C. $\frac{(F_1+F_{2}sin\theta )}{(mg+F_{2}cos\theta )}$
• D. $\frac{(F_{1}sin\theta -F_2)}{(mg-F_{2}cos\theta )}$

Answer 1

Answer: (C)

We know $f=\mu N$ and $N=mg+F_{2}cos\theta$
Now, since wedge is in equilibrium,
$(F_1+F_{2}sin\theta )=\mu (mg+F_{2}cos\theta )$
$\mu =\frac{F_1+F_{2}sin\theta }{mg+F_{2}cos\theta }$