Question

A block of mass m lies on a rough surface of co-efficient of friction force F is applied on it at angle $\theta$ to the horizontal as shown. and the block is rest. The frictional force acting on the block will be

• A. $Fcos\theta$
• B. $\mu (mg+Fsin\theta )$
• C. $\mu (mg-Fsin\theta )$
• D. $\mu mg$

Answer 1

Answer: (A), (C)

The correct options are
A $Fcos\theta$
C $\mu (mg-Fsin\theta )$

$Given$ :- Mass of block = $m$

Coefficient of friction = $\mu$

Angle of applied force $\left(F\right)$ = $\theta$

$ToFind$ :- frictional force ( $f$ ) acting on block

$Solution$ :- Since , the block is at rest i.e, it is balanced by equal and opp. forces

$\therefore$ $N+Fsin\theta =mg$ (according to diagram)

$N=mg-Fsin\theta$

And, $\underset{–}{f_{friction}=Fcos\theta }$

we know, $f_{max}=\mu N$

$\underset{–}{f_{max}=\mu (mg-Fsin\theta )}$

Hence , $\underset{–}{BothOptionA\left(Fcos\theta \right)andC\left(\mu (mg-Fsin\theta )\right)arecorrect.}$