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. |
Hint
Reveal
Solution
Reveal

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:
- a force due to the rope, of magnitude T2, and directed upwards;
- its weight, m1g=mg directed downwards
Newton’s 2nd law tells us that
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:
- a force due to the rope below, of magnitude T2, and directed downwards
- a force due to the rope above, of magnitude T1, and directed upwards;
- its weight, m2g=mg directed downwards
Now Newton’s 2nd law tells us that
Now let’s choose our system to be the mass m3. This mass experiences two forces:
- 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].
- its weight, m3g=mg directed downwards
Applying Newton’s 2nd law for one final time tells us that
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
Subtracting Eq 3 from Eq 1 we get
Adding these two equations we identify
![\[ T_2=2mg/3 \hspace{0.5cm}\mbox{\rm and thence}\hspace{0.5cm} T_1=4mg/3 \]](mastermathpng-0.png)
Finally, the acceleration follows from Eq 1 as
![\[ a=g-T_2/m =\frac{g}{3} \]](mastermathpng-1.png)