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Principle of Superposition

Masses M 1 ...

Question

Masses $M_{1}, M_{2}$ and $M_{3}$ are connected by string of negligible mass which pass over massless and frictionless pulley $P_{1}$ and $P_{2}$ as shown in the figure. The masses move such that the portion of the inclined plane and the portion of the string between $P_{1}$ and $P_{2}$ is parallel to the incline plane and the portion of the sting 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 surface is $0.25$ . The inclined plane makes an angle of $37^{o}$ with the horizontal. If the mass $M_{1}$ moves downward with a uniform velocity find the
$(a)$ mass of $M_{1}$
$(b)$ tension in the horizontal portion of the string. $(9 = 9.8 m / s^{2}$ and $sin 37^{o} \approx 3 / 5)$

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Solution

Let $T_{1}$ be the tension in the string connecting $m_{1}$ and $m_{2}$ and $T_{2}$ be the tension in the string connecting $m_{2}$ and $m_{3}$ . Form the figure.
$M_{1} g = T_{1}$
$T_{2} = μ m_{3} g = (0.5) 4 g$
Or, $T_{2} = g = 9.8 N$
Also
$T_{1} = T_{2} + (0.25) \times 4 g cos 37^{o} + 4 g sin 37^{o}$
$= g (1 + \frac{4}{5} + \frac{4 \times 3}{5})$
$T_{1} = \frac{21}{5} g$
Or, $M_{1} g = \frac{21}{5} g$
$∴ M_{1} = \frac{21}{5} = 4.2 k g$

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P_{1}

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