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. $\frac{F_1\sin \theta -F_2}{mg-F_2\cos \theta }$

Answer 1

The correct option is A $\frac{F_1+F_2\sin \theta }{mg+F_2\cos \theta }$

By the free diagram of the block,
and resolving the force $F_2$ along the horizontal and the vertical components.
Here the reaction force $R$ will be ,
$R=mg+F_2\cos \theta ,f=\mu R,$ hence the frictional force we get as $f=\mu [mg+F_2\cos \theta ]$

Also, for the equilibrium condition $f=F_1+F_2\sin \theta$

Equating the above equations we get,

$\mu [mg+F_2\cos \theta ]=F_1+F_2\sin \theta$ or $\mu =\frac{F_1+F_2\sin \theta }{mg+F_2\cos \theta }$

Hence option A is correct answer.