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Constrained Motion : Intuitive Approach

Masses M1, M2...

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 fig. The masses move such that the portion of the string between $P_{1}$ 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 surface is $0.25.$ The inclined plane makes an angle of $37^{\circ}$ with the horizontal. If the mass $M_{1}$ moves downwards with a uniform velocity. find the mass of $M_{1}$ in $kg$ .

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Solution

For $M_{1}$ to move with uniform velocity net acceleration must be zero. So, $M_{1} g = μ M_{2} g c o s θ + μ M_{3} g + M_{2} g s i n θ$
$\Rightarrow M_{1} = μ (M_{2} c o s θ + M_{3}) + M_{2} s i n θ$
$= 0.25 ((4) \frac{4}{5} + 4) + (4) \frac{3}{5} = \frac{21}{5} = 4.2 k g$

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For M1 to move with uniform velocity net acceleration must be zero. So,M1g=μM2gcosθ+μM3g+M2gsinθ ⇒M1=μ(M2cosθ+M3)+M2sinθ =0.25((4)45+4)+(4)35=215=4.2kg

For $M_{1}$ to move with uniform velocity net acceleration must be zero. So, $M_{1} g = μ M_{2} g c o s θ + μ M_{3} g + M_{2} g s i n θ$
$\Rightarrow M_{1} = μ (M_{2} c o s θ + M_{3}) + M_{2} s i n θ$
$= 0.25 ((4) \frac{4}{5} + 4) + (4) \frac{3}{5} = \frac{21}{5} = 4.2 k g$