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Masses M1,M...

Question

Masses $M_{1}, M_{2}$ and $M_{3}$ are connected by strings of negligible mass which pass over massless and frictionless pulleys $P_{1}$ and $P_{2}$ as shown in the figure. The $P_{1}$ and $P_{2}$ is parallel to the inclined plane and the portion of the string between $P_{2}$ and $M_{3}$ is horizontal. The masses $M_{2}$ and $M_{3}$ are $4.0 k g$ each and the coefficient of kinetic friction between the masses and the surfaces is $0.25$ . The inclined plane makes an angle of $37^{o}$ with the horizontal. If the mass $M_{1}$ moves downwards with a uniform velocity find
(i) the mass of $M_{1}$
(ii) the tension in the horizontal portion of the string. ( $g = 9.8 m / s^{2}$ and $sin 37^{o} \approx 3 / 5$ )

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Solution

Since, the masses are moving with constant velocity, there is no acceleration and hence no external force is acting on them.
Force balance:

For mass $m_{1}$ :

$T_{1} = m_{1} g - - - - (i)$

$F o r m a s s m_{3}$

$T_{2} = μ m_{3} g$

$T_{2} = 0.25 \times 4 \times 10$

$T_{2} = 10 N$

$F o r m a s s m_{2} :$

$T_{1} = m_{2} μ g sin θ + m_{2} g cos θ + T_{2}$

$T_{1} = 4 \times 0.25 \times 10 \times \frac{3}{5} + 4 \times 10 \times \frac{4}{5} + T_{2}$

$T_{1} = 6 + 32 + 10$

$T_{1} = 48 N$

$P u t t i n g i n e q n (i)$

$48 = m_{1} \times 10$

$m_{1} = 4.8 k g$

Therefore,
The mass $m_{1} = 4.8 k g$ and the tension in the horizontal spring $T_{2} = 10 N$

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