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If we stick two block masses by a string and pass it over a pully and let the blocks free then mathematically we can obtain that the whole weight on the pully is more if the blocks are kept at rest, than if the heavy block is allowed to fall freely. But what is the physical meaning of it?

Qmechanic
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    "In physics, the reduced mass is the "effective" inertial mass appearing in the two-body problem of Newtonian mechanics. It is a quantity which allows the two-body problem to be solved as if it were a one-body problem." https://en.wikipedia.org/wiki/Reduced_mass . I do not think that you are expressing well the problem. strings tie, do not stick; do you have a link showing the set up? – anna v Apr 27 '16 at 11:34
  • @annav Sorry, I think my mentioning 'reduced mass' was not right. If the strings tie the two blocks and then if we release them, the heavier mass falls down and we can find that in this case the total weight acting on the pully is less than the sum of the mass of the blocks, through mathematical equation, but what is the physical meaning of this ? – Soubhadra Maiti Apr 27 '16 at 12:23
  • Possible duplicates: https://physics.stackexchange.com/q/159271/268448 and links therein. – JAlex Apr 08 '21 at 21:35

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Any time you have an equal and opposite force acting on two bodies, the effect is proportional to $\frac{1}{m}$ for each one.

The combined effect is proportional to $$\text{(effect)} = \left( \frac{1}{m_1} + \frac{1}{m_2} \right) \text{(action)}$$

When inverted to find which action has a desired effect we get the reduced mass

$$ \text{(action)} = \left( \frac{1}{m_1} + \frac{1}{m_2} \right)^{-1} \text{(effect)} = \frac{m_1 m_2}{m_1+m_2} \text{(effect)} $$

This concept applies to any interaction between two things, like for example:

  • Two masses attached to a spring, rod or rope.
  • Two masses under gravity.
  • Two masses under electrostatics.
  • Two objects colliding

and probably many more.

John Alexiou
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When you state that the two masses are 'kept at rest' I assume you mean that the pulley is prevented from turning. The load on the pulley is then $(M+m)g$.

[If the pulley is not prevented from turning, then to keep the masses at rest we would have to lift the heavier mass $M$ until it became, effectively, the same weight as the smaller mass. The total load on the pulley is then $2mg$. Alternatively we could pull down on the lighter mass $m$ to make it as heavy as the larger mass $M$. The total load on the pulley is then $2Mg$.]

If the pulley is not prevented from turning, the tension $T$ in the string is the same on both sides of the pulley. The load supported by the pulley is $2T$.

Write the equations of motion for the two masses $M$ and $m$ : $$Mg - T = Ma$$ and $$T - mg = ma$$ where $a$ is the common acceleration of the two masses. Subtracting we get : $$2T = (M+m)g - (M-m)a$$ which is less than $(M+m)g$ because $M\gt m$ and $a\gt 0$.

So the load on the pulley is less if the pulley can turn freely and the heavier mass is allowed to fall.

lucas
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sammy gerbil
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When the blocks are kept at rest , for the vertical equilibrium the pulley has to exert the force equal to the weight of two blocks. But when the heavier block starts moving down the mechanical advantage of pulley with respect to the heavier block increases and hence the weight on the pulley decreases.

  • "and hence the weight on the pulley decreases." This is actual question. Why? Why it decrease? – Anubhav Goel Apr 27 '16 at 11:55
  • Search on Google for mechanical advantage. As much as the distance between the pulley and rope increases, the more mechanical advantage the pulley has and thus the weight on it decreases. – Vaibhav Singh Apr 27 '16 at 12:21