Laws of Motion 3 Question 14
15. A small block of mass of $0.1 kg$ lies on a fixed inclined plane $P Q$ 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 $g=10 m / s^{2}$ )
(a) $\theta=45^{\circ}$.
(2012)
(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 $Q$
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Answer:
Correct Answer: 15. (a, c)
Solution:
- $w=m g=0.1 \times 10=1 N$
$F _1=$ 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 $P$.
