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Mass $m_{1}$ moves on a slope making an angle $\theta$ with the horizontal and is attached to mass $m_{2}$ by a string passing over a frictionless pulley as shown in Figure. The co-efficient of friction between $m_{1}$ and the sloping surface is $\mu$ .

Which of the following statements are true?
(a) If $m_{2} > m_{1} \sin \theta$ , the body will move up the plane.
(b) If $m_{2} > m_{1} \sin \theta +\mu \cos \theta$ , the body will move up the plane.
(c) If $m_{2} < m_{1} \sin \theta +\mu \cos \theta$ , the body will move up the plane.
(d) If $m_{2} < m_{1} \sin \theta -\mu \cos \theta$ , the body will move down the plane.

Answers (1)

The correct answer is: -

(b) If $m_{2} > m_{1} \sin \theta +\mu \cos \theta$ , the body will move up the plane.

(d) If $m_{2} < m_{1} \sin \theta -\mu \cos \theta$ , the body will move down the plane.

Explanation:

Case I-

Considering the first case, normal reaction (N) = $m_{1}g\cos \theta$

Now, $f = \mu N = \mu m_{1}g\cos \theta$ , which becomes,

$T - m_{1}g\ sin \theta - \mu m_{1}g\cos \theta$

$= m_{1}a$

m₁ will be up, and m₂ will be down when $m_{2}g - (m_{1}g\cos \theta + f) > 0$

$m_{2}g - m_{1}g \sin \theta - g \mu m_{1}g \cos g \theta > 0$

$m_{2}g > m_{1}g (\sin\theta + \mu \cos \theta )$

or $m_{2} > m_{1} (\sin\theta + \mu \cos \theta )$

hereby, opt (b) is verified and opt (a) is rejected.

Case II-

Now, considering that if the body m1 moves down and the body m2 up then, direct of f becomes upward (i.e., opposite to the motion).

$-f + m_{1}g\sin\theta > m_{2}g$

$-\mu m _{1}g \cos \theta + m_{2}g \sin \theta > m _{2}g$

$m_{1} (-\mu \cos \theta + \sin\theta ) > m_2$

$m_{2} < m_{1} (\sin \theta - \mu \cos \theta )$

Hence, option (d) is verified here, as well as opt (c) is rejected.

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