Laws of Motion 3 Question 2

2. A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force $2 N$ down the inclined plane. The maximum external force up the inclined plane that

does not move the block is $10 N$.

The coefficient of static friction between the block and the plane is

(2019 Main, 12 Jan I)

(Take, $g=10 m / s^{2}$ )

(a) $\frac{2}{3}$

(b) $\frac{\sqrt{3}}{2}$

(c) $\frac{\sqrt{3}}{4}$

(d) $\frac{1}{2}$

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Answer:

Correct Answer: 2. (b)

Solution:

  1. Block does not move upto a maximum applied force of $2 N$ down the inclined plane.

So, equating forces, we have;

$$ 2+m g \sin \theta=f $$

or $2+m g \sin \theta=\mu m g \cos \theta$

Similarly, block also does not move upto a maximum applied force of $10 N$ up the plane.

Now, equating forces, we have

$$ m g \sin \theta+f=10 N $$

or $m g \sin \theta+\mu m g \cos \theta=10$

Now, solving Eqs. (i) and (ii), we get

$$ m g \sin \theta=4 $$

and $\quad \mu m g \cos \theta=6$

Dividing, Eqs. (iii) and (iv), we get

$$ \begin{array}{rlrl} & \mu \cot \theta & =\frac{3}{2} \\ \Rightarrow & \mu & =\frac{3 \tan \theta}{2}=\frac{3 \tan 30^{\circ}}{2} \\ \Rightarrow \quad \mu & =\frac{\sqrt{3}}{2} \end{array} $$



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Dual Pane