Q.
A block of mass $m$ on a rough horizontal surface is acted upon by two forces as shown in figure. For equilibrium of block, the coefficient of friction between block and surface is
A) $\frac{F_{1} + F_{2}\sin\theta}{mg + F_{2}\cos\theta}$
B) $\frac{F_{1}\sin\theta + F_{2}}{mg + F_{2}\sin\theta}$
C) $\frac{F_{1} + F_{2}\cos\theta}{mg + F_{2}\sin\theta}$
D) $\frac{F_{1}\sin\theta - F_{2}}{mg - F_{2}\cos\theta}$
Explanation
(a) $R = mg + F_{2}\cos{\theta,f = \mu R}$
$f = \mu(mg + F_{2}\cos\theta)$
[image:5_image269.emf]
Also, $f = F_{1} + F_{2}\sin\theta$
Equating, $\mu(mg + F_{2}\cos\theta)$
$= F_{1} + F_{2}\sin\theta$
or $\mu = \frac{F_{1} + F_{2}\sin\theta}{mg + F_{2}\cos\theta}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.