Discussion Forum

A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle $\theta$ with the horizontal. A horizontal force of 1 N acts on the block through its centre of mass as shown in the figure. The block remains  stationary if (take g=10m/s²)

Answer:

Explanation:

W=mg=0.1 x10=1N

 $F_{1}=$ component of weight

 =1$.\sin \theta= \sin \theta$

 $F_{2}=$ component of applied force

 =$1. \cos \theta= \cos \theta$

 now,  at $\theta= 45^{0}$ :$F_{1}=F_{2}$  and biock remains stationary without the help of friction.

 For  $\theta > 45^{0}$ ; $F_{2}>F_{1}$,  so friction will act

towards Q.

 For $\theta <45^{0}$ , $F_{2} >F_{1}$ ,

and friction will act towards P.