Physics Pulley

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Pulley

Pulley

A pulley is simply a collection of one or more wheels over which you loop a rope to make it easier to lift things.

Pulley systems are used to provide us with a mechanical advantage, where the amount of input effort is multiplied to exert greater forces on a load.
They are typically used for hauling and lifting loads but can also be used to apply tension within a system such as in a Tensioned Line or Tyrolean. This page explains the basic principles of pulley systems and how they work, for information on how to use them in hauling see the hauling systems page.

A pulley is simply a collection of one or more wheels over which you loop a rope to make it easier to lift things.

Pulley systems are used to provide us with a mechanical advantage, where the amount of input effort is multiplied to exert greater forces on a load.
They are typically used for hauling and lifting loads but can also be used to apply tension within a system such as in a Tensioned Line or Tyrolean. This page explains the basic principles of pulley systems and how they work, for information on how to use them in hauling see the hauling systems page.

Constraints [The Virtual Work]

Consider the Atwood machine shown. If one holds both the masses such that the string is slack. There will not be any tension in the string. Once released the string will become taut and it will oppose the motions of one block where as it will support the motion of other one.

Conclusion:
It is transferring the effort of one block to the other one. So, on its own it will not do any work on the system of two blocks. Work done by tension on block having mass ` m_1=vec T-X_(L)`, where `X_L` is the displacement of block `m_1` Similarly work done by tension on block `m_2` will be `T_(x_2)` .Since sum of these two has to be zero; we have
`vec T * x_1+ vec T * x_2=0`

`=> x_1+x_2=0`

`=> x_1=-x_2`......(i)

Differentiating equation (i) we have

`V_1=-V_2` (ii) and `a_1=-a_2` (iii)

Consider the figure shown

Tension on block `m_1 = T`

Tension on block `m_2 = T //2`

Tension on block `m_3 = T //2`

Assuming displacements of masses as shown in the figure.

Now using sum of work done by tension on the entire system `= 0`.

`Tx_1+T/2 x_2+T/2x_3=0`

`=> 2x_1+x_2+x_3=0`

Differentiating it twice we get the relation among the accelerations of three blocks

`2a_1+a_2+a_3=0`

In this system we have only one block on which tension will work on.

`Tx=0`

`=> x=0` and `a=0`

`=> 2a_1+a_2+a_3=0`

We have made a mistake which gives a wrong result. We cancelled `T` against zero that was our mistake. We should always make sure the quantity getting cancelled itself is non zero. Looking at the `F.B.D.` of pulley `(2)` we realize the mistake

Since the pulley is massless

`2T-T=0 * a_(text(pulley))=>T=0`

Hence the mass m is going to fall freely under gravity with which we have over burdended the system. Massless pulleys and strings are hypothetical entities.

Consider the Atwood machine shown. If one holds both the masses such that the string is slack. There will not be any tension in the string. Once released the string will become taut and it will oppose the motions of one block where as it will support the motion of other one.

Conclusion:
It is transferring the effort of one block to the other one. So, on its own it will not do any work on the system of two blocks. Work done by tension on block having mass ` m_1=vec T-X_(L)`, where `X_L` is the displacement of block `m_1` Similarly work done by tension on block `m_2` will be `T_(x_2)` .Since sum of these two has to be zero; we have
`vec T * x_1+ vec T * x_2=0`

`=> x_1+x_2=0`

`=> x_1=-x_2`......(i)

Differentiating equation (i) we have

`V_1=-V_2` (ii) and `a_1=-a_2` (iii)

Consider the figure shown

Tension on block `m_1 = T`

Tension on block `m_2 = T //2`

Tension on block `m_3 = T //2`

Assuming displacements of masses as shown in the figure.

Now using sum of work done by tension on the entire system `= 0`.

`Tx_1+T/2 x_2+T/2x_3=0`

`=> 2x_1+x_2+x_3=0`

Differentiating it twice we get the relation among the accelerations of three blocks

`2a_1+a_2+a_3=0`

In this system we have only one block on which tension will work on.

`Tx=0`

`=> x=0` and `a=0`

`=> 2a_1+a_2+a_3=0`

We have made a mistake which gives a wrong result. We cancelled `T` against zero that was our mistake. We should always make sure the quantity getting cancelled itself is non zero. Looking at the `F.B.D.` of pulley `(2)` we realize the mistake

Since the pulley is massless

`2T-T=0 * a_(text(pulley))=>T=0`

Hence the mass m is going to fall freely under gravity with which we have over burdended the system. Massless pulleys and strings are hypothetical entities.

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