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Question: Two identical weights of mass $M$ are connected by a thread wrapped around a fixed frictionless pull...

Two identical weights of mass MM are connected by a thread wrapped around a fixed frictionless pulley. A small weight of mass 'm' is placed on one of the weights and the system is allowed to move. What is reaction force between mm and MM?

A

2Mg(M+m)2M+m\frac{2Mg(M + m)}{2M + m}

B

Mg(M+m)2M+m\frac{Mg(M + m)}{2M + m}

C

mg(M+m)2M+m\frac{mg(M + m)}{2M + m}

D

2mMg2M+m\frac{2mMg}{2M + m}

Answer

(d) 2mMg2M+m\frac{2mMg}{2M + m}

Explanation

Solution

To determine the reaction force between mass 'm' and mass 'M', we need to analyze the forces acting on each part of the system and apply Newton's second law.

Let's denote:

  • MM = mass of each identical weight
  • mm = mass of the small weight
  • TT = tension in the thread
  • aa = acceleration of the system
  • NN = reaction force between mm and MM
  1. Determine the direction of motion:

The total mass on the left side is MM.
The total mass on the right side is M+mM + m.
Since (M+m)>M(M + m) > M, the right side (mass M+mM + m) will move downwards, and the left side (mass MM) will move upwards.

  1. Free Body Diagram and Equations of Motion for the System:

  • For the mass MM on the left side (moving upwards):
    The forces acting on MM are tension TT upwards and its weight MgMg downwards.
    Applying Newton's second law: T−Mg=MaT - Mg = Ma (Equation 1)

  • For the combined mass (M+m)(M + m) on the right side (moving downwards):
    The forces acting on the combined mass (M+m)(M + m) are tension TT upwards and its total weight (M+m)g(M + m)g downwards.
    Applying Newton's second law: (M+m)g−T=(M+m)a(M + m)g - T = (M + m)a (Equation 2)

  1. Calculate the acceleration aa of the system:

Add Equation 1 and Equation 2 to eliminate TT: (T−Mg)+((M+m)g−T)=Ma+(M+m)a(T - Mg) + ((M + m)g - T) = Ma + (M + m)a
T−Mg+Mg+mg−T=(M+M+m)aT - Mg + Mg + mg - T = (M + M + m)a
mg=(2M+m)amg = (2M + m)a
a=mg2M+ma = \frac{mg}{2M + m}

  1. Calculate the reaction force NN:

Now, consider the free body diagram of the small mass mm placed on top of MM (on the right side).
The forces acting on mm are:

  • Its weight mgmg acting downwards.
  • The normal reaction force NN from mass MM acting upwards.

Since mass mm is accelerating downwards with acceleration aa:
Applying Newton's second law to mass mm: mg−N=mamg - N = ma
Now, substitute the value of aa we found:
N=mg−maN = mg - ma
N=mg−m(mg2M+m)N = mg - m \left( \frac{mg}{2M + m} \right)
N=mg(1−m2M+m)N = mg \left( 1 - \frac{m}{2M + m} \right)
To simplify, find a common denominator:
N=mg((2M+m)−m2M+m)N = mg \left( \frac{(2M + m) - m}{2M + m} \right)
N=mg(2M2M+m)N = mg \left( \frac{2M}{2M + m} \right)
N=2Mmg2M+mN = \frac{2Mmg}{2M + m}

This reaction force NN is the force exerted by MM on mm, and by Newton's third law, it is also the force exerted by mm on MM.