Q3.8 Tensions (S)

Three equal masses (m1=m2=m3=m) are hung over a pulley as shown. Find the acceleration of the system and the values of the tensions T1 and T2.
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Hint

It would be a good idea to look at the version of this problem having two masses first: see your notes on S2.7
 

Solution

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Solution

Let a represent the downward acceleration of the left hand rope (and therefore also the upward acceleration of the right hand section).

Consider first the system comprising the mass m1. This mass experiences two forces:

  1. a force due to the rope, of magnitude T2, and directed upwards;
  2. its weight, m1g=mg directed downwards

Newton’s 2nd law tells us that

(3.1)
mg-T2=ma

The left hand side gives the net force acting (on this system, ie the mass m1) vertically downwards; the rhs gives the mass times the downward acceleration.

Now consider the mass m2. This mass experiences three forces:

  1. a force due to the rope below, of magnitude T2, and directed downwards
  2. a force due to the rope above, of magnitude T1, and directed upwards;
  3. its weight, m2g=mg directed downwards

Now Newton’s 2nd law tells us that

(3.2)
mg+T2-T1=ma

Now let’s choose our system to be the mass m3. This mass experiences two forces:

  1. a force due to the rope, of magnitude T1, and directed upwards. [The rope is assumed to be light so the tension is the same throughout].
  2. its weight, m3g=mg directed downwards

Applying Newton’s 2nd law for one final time tells us that

(3.3)
T1-mg=ma

The left hand side gives the net force in the upward direction; the rhs gives the mass times the acceleration of this mass in the upward direction.

We now have three equations with three unknowns (T1, T2 and a). Subtracting Eq 2 from Eq 1 we get

(3.4)
T1-2T2=0

Subtracting Eq 3 from Eq 1 we get

(3.5)
2mg-T1-T2=0

Adding these two equations we identify

\[ T_2=2mg/3 \hspace{0.5cm}\mbox{\rm and thence}\hspace{0.5cm} T_1=4mg/3 \]

Finally, the acceleration follows from Eq 1 as

\[ a=g-T_2/m =\frac{g}{3} \]