Question

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. $\frac{F_1\cos \theta +F_2}{mg+F_2\sin \theta }$
• C. $\frac{F_1+F_2\cos \theta }{mg+F_2\sin \theta }$
• D. $\frac{F_1+F_2\cos \theta }{mg+F_2\cos \theta }$

Answer 1

Answer: (A)

$\frac{F_1+F_2\sin \theta }{mg+F_2\cos \theta }$

Normal force acting on the block $R=mg+F_2\cos \theta ,$
Friction force $f=\mu R$
$f=\mu [mg+F_2\cos \theta ]$
Also force is balanced in horizontal direction, $f=F_1+F_2\sin \theta$
Equating,
$\mu [mg+F_2\cos \theta ]$ $=F_1+F_2\sin \theta$
We get $\mu =\frac{F_1+F_2\sin \theta }{mg+F_2\cos \theta }$.