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a platform of mass M and the counter weight of mass M + M they are connected by the light cord which passes over

Question

A platform of mass M and the counter weight of mass M + M

They are connected by the light cord which passes over...

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Solution

It seems that you are describing a system involving a platform and a counterweight, likely in the context of a physics problem concerning mechanics, pulleys, or equilibrium. Below, I'll outline a comprehensive answer based on the information provided.

Problem Analysis

  1. Understanding the System: The platform of mass M M represents an object that will likely move in response to the force exerted by the counterweight, which has a mass of M+M=2M M + M = 2M . This setup suggests a pulley system where forces will balance or initiate movement.

  2. Weights and Forces:

    • The weight of the platform (downward force) is given by Wplatform=Mg W_{platform} = Mg , where g g is the acceleration due to gravity.
    • The weight of the counterweight is Wcounterweight=(2M)g W_{counterweight} = (2M)g .

Equations of Motion

  1. Setting Up the Forces: When the system is released, the heavier counterweight will accelerate downward, causing the platform to accelerate upward. The net force on each mass will determine their accelerations.

    • For the counterweight, the equation can be set up as: Fnet,counterweight=WcounterweightT=(2M)a F_{net, counterweight} = W_{counterweight} - T = (2M)a
    • For the platform, the upward tension must balance its weight: TWplatform=Ma T - W_{platform} = Ma
  2. Solving the System:

    • First, from the first equation: (2M)gT=2Ma (2M)g - T = 2Ma T=2Mg2Ma T = 2Mg - 2Ma .
    • Substituting T T into the second equation gives us: 2Mg2MaMg=Ma 2Mg - 2Ma - Mg = Ma Simplifying will yield the expression for a a .

Final Calculation

After simplifying, we can find the acceleration a a in terms of g g : a=(2M)gMg3M=g3 a = \frac{(2M)g - Mg}{3M} = \frac{g}{3}

Summary

The system consists of a platform and a counterweight, where the acceleration of the platform upwards, when the counterweight falls, can be calculated to be g3 \frac{g}{3} . The net effect of the counterweight being heavier leads to a downward acceleration of the heavier mass and corresponding upward acceleration of the lighter mass (platform).

Final Answer

The acceleration of the platform is given by: a=g3 a = \frac{g}{3}

This problem has been solved

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