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In Figure, a body A of mass m slides on plane inclined at angle $\theta _{1}$ to the horizontal and $\mu _{1}$ is the coefficient of friction between A and the plane. A is connected by a light string passing over a frictionless pulley to another body B, also of mass m, sliding on a frictionless plane inclined at angle $\theta _{2}$ to the horizontal. Which of the following statements are true?

(a) A will never move up the plane.
(b) A will just start moving up the plane when $\mu = \frac{\sin \theta _{2} - \sin \theta _{1}}{\cos \theta _{1}}$
(c) For A to move up the plane, $\theta _{2}$ must always be greater than $\theta _{1}$ .
(d) B will always slide down with constant speed.

Answers (1)

The correct answer is the option:

(b) A will just start moving up the plane when $\mu = \frac{\sin \theta _{2} - \sin \theta _{1}}{\cos \theta _{1}}$

(c) For A to move up the plane, $\theta _{2}$ must always be greater than $\theta _{1}$ .

Explanation:

In the fig. below, $\mu$ = coefficient of friction of A whereas B is on a frictionless surface.

(i) when A is about to start

$f =\mu N_{1} = \mu mg \cos \theta _{1}$

$mg sin\theta _{1} +f=mg \sin \theta _{2}$

When A just start moving upward

$mg sin\theta _{1} +\mu mg cos\theta _{1}=mg \sin \theta _{2}$

$\mu=\frac{sin\theta_2-sin\theta_1}{cos\theta_1}$

If on the plane, the body A moves upward, and B moves downward

$mg \sin \theta _{2} - mg \sin \theta _{1 }>0$

$\sin \theta _{2} - \sin \theta _{1 }>0$

$\theta _{2} - \theta _{1 }>0$

Hence, opt (c) is verified and opt (a) is rejected.

(ii) If the body B moves upward and A moves downward then,

$mg\sin \theta_{ 2} -f - mg\sin \theta_{ 2} > 0$

$mg\sin \theta _{1} - \mu mg \cos \theta_{1} - mg\sin \theta _{2} > 0$

$\sin \theta _{1} - \sin \theta_{1} > \mu \cos\theta_{ 1}$

Since $\theta _{1}$ and $\theta _{2}$ are acute angles, $\theta _{1}$ > $\theta _{2}$ , thus, $\sin\theta > \mu \cos \theta_{1}$ may be true

Since it is now clear that body B can also move upward, opt (d) is also rejected.

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