Laws Of Motion Ques 22
- A small block of mass of $0.1 \mathrm{~kg}$ lies on a fixed inclined plane $P Q$ which makes an angle $\theta$ with the horizontal. A horizontal force of $1 \mathrm{~N}$ acts on the block through its centre of mass as shown in the figure. The block remains stationary if (take $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )
(2012)
(a) $\theta=45^{\circ}$.
(b) $\theta>45^{\circ}$ and a frictional force acts on the block towards $P$
(c) $\theta>45^{\circ}$ and a frictional force acts on the block towards $Q$
(d) $\theta<45^{\circ}$ and a frictional force acts on the block towards $Q$
Show Answer
Answer:
Correct Answer: 22.(a, c)
Solution:
Formula:
- $w=m g=0.1 \times 10=1 \mathrm{~N}$
$F_{1}=$ component of weight $=1 \cdot \sin \theta=\sin \theta$ $F_{2}=$ component of applied force $=1 \cdot \cos \theta=\cos \theta$
Now, at $\theta=45^{\circ}: F_{1}=F_{2}$ and block remains stationary without the help of friction.
For $\theta>45^{\circ}, F_{1}>F_{2}$, so friction will act towards $Q$.
For $\theta<45^{\circ}, F_{2}>F_{1}$ and friction will act towards $P$.
BETA
