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CAF123

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## Homework Statement

A light smooth pulley is attached to a support a fixed height above the ground. An inextensible string passes over the pulley and carries a mass 4m on one side. The other end of the string supports a similar mobile pulley; over this passes a second string, carrying masses of 3m and m on its two ends.

Deduce the acceleration of the three masses

## Homework Equations

N2, Lagrange Equations

## The Attempt at a Solution

I solved this already and got the correct answer using Lagrange's Equations. Now I try again with Newton's Eqns. The final set of three equations I obtain from Newtonian analysis are satisfied by the accelerations I got when I solved via Lagrange's Eqns. However, I cannot actually solve the three equations to obtain a unique solution (they are a linearly dependent set). What I need is another equation which I cannot get.

No picture was given, so I created my own and I will describe what I did.

Forces on 4m mass give eqn: T - 4mg = 4ma1, a1 the acceleration of mass 4m. This is also the tension on the other side of the same pulley. Since this pulley is the fixed pulley, it has an upward force equal to 8ma1 + 8mg

The string then passes down to the mobile pulley. Upward force on this is 4ma1 + 4mg. Consider the 3m mass: Net force is 3ma2 + 3mg and that of the m mass is ma3 + mg.

Now I used these to construct suitable equations. The net upward force on the mobile pulley is equal to the sum of the tensions 3ma2+3mg and ma3+mg. So 4ma1 + 4mg = 3ma2+3mg+mg+ma3 giving 4a1 - 3a2 -a3 = 0. (1st EQN). The tension on either side of the mobile pulley must be the same => 3ma2 + 3mg = mg + ma3 so 3a2 - a3 = -2g. (2nd EQN)

Finally, the net upward force on the fixed pulley is equal to twice the tensions on either side of the mobile pulley => 8ma1 + 8mg = 2(3ma2 + 3mg + 3ma2 + 3mg) giving 8a1 - 12a2 = 4g. (3rd EQN)

This set is linearly dependent ,so I have a free parameter

Is there another eqn I can get somewhere?

Also, now that I think about it, why is my 1st EQN valid? This pulley is mobile so its upward force need not equal the sum f two tensions below it?

Many thanks.

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