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# Problem: Two masses on a pulley

Two masses of 80 kg and 140 kg hang from a rope that runs over a pulley. You can assume that the rope is massless and inextensible, and that the pulley is frictionless. Find the upward acceleration of the smaller mass and the tension in the rope.

## Solving the problem

Let's start by drawing a sketch of what is happening:

We have a massless rope that runs over a frictionless pulley, this means that the two masses are subject to upward tensions equal in magnitude. We will indicate the magnitude of the tensions with T.

In the case of the smaller mass, the tension wins the force of gravity, which means that the smaller mass is accelerating upward.

However, in the case of the larger mass, the force of gravity wins the tension, which means that the larger mass is accelerating downward.

Also, since the rope is inextensible, the two masses move with accelerations that are equal in magnitude. We will indicate the magnitude of the accelerations with a.

With all of that said, let's list all the forces that act on the two masses.

The smaller mass is subject to 2 forces:

• the upward tension, T
• and the force of gravity, mg

The larger mass is also subject to 2 forces:

• the upward tension, T
• and the force of gravity, Mg

Here are the free-body diagrams of the two masses:

Smaller mass:

Larger mass:

We know the smaller mass (80 kg) and the larger mass (140 kg).

We want to find the acceleration of the smaller mass (which, as we saw has the same magnitude as the acceleration of the larger mass), and the tension in the rope.

m = 80 kg
M = 140 kg

### We want to know

a = ?
T = ?

The smaller mass is subject to two forces: T (directed upward) and mg (directed downward), where T is larger in magnitude. Therefore, the resultant force (r) will be directed upward, and have the magnitude equal to the difference between T and mg:

r = T β mg (1)

The larger mass is also subject to two forces: T (directed upward) and Mg (directed downward), where Mg is larger in magnitude. Therefore, the resultant force (R) will be directed downward, and have the magnitude equal to the difference between Mg and T:

R = Mg β T (2)

We now have two equations with 3 unknowns: r, R, T.

We can reduce the number of unknowns to 2 by remembering that the two masses have accelerations equal in magnitude. Indeed, applying Newton's 2nd Law, the magnitudes of the two resultant forces can be expressed as:

r = ma
R = Ma

So, we can substitute r and R in Eq. (1) and Eq. (2):

r = T β mg
R = Mg β T
β
ma = T β mg (3)
Ma = Mg β T (4)

We now have two equations with 2 unknowns (a and T), so we can solve them.

Let's solve Eq. (3) for T:

ma = T β mg
ma + mg = T
T = ma + mg (5)

Next, let's substitute T with ma + mg in Eq. (4):

Ma = Mg β T
Ma = Mg β (ma + mg)
Ma = Mg β ma β mg

And solve this equation for a:

Ma + ma = Mg β mg
a (M + m) = g (M β m)
 a = M β m g M + m
 a = 140 kg β 80 kg (9.8 m/s2) 140 kg + 80 kg
 a = 60 kg (9.8 m/s2) 220 kg
a = 2.7 m/s2

Finally, we can find the magnitude of the tension T using Eq. (5):

T = ma + mg
T = m (a + g)
T = (80 kg) (2.7 m/s2 + 9.8 m/s2)
T = 1.0 Γ 103 N

Therefore, the smaller mass has an acceleration of 2.7 m/s2 (which is also the magnitude of the acceleration of the larger mass), and the tension in the rope is 1.0 Γ 103 N.

## Tips & Tricks

• Remember that if two objects hang from a massless rope (or string, cable etc.) that runs over a frictionless pulley, the upward tensions exerted by the rope on the two objects will be equal in magnitude. And if the rope is also inextensible, then the accelerations of the two objects will be equal in magnitude (although opposite in direction).
• When you have two objects in a problem, to keep variables simple, you can use lower case letters for variables related to the smaller object, and upper case letters for variables related to the larger object, as we did in this problem. This way you avoid additional subscripts. For example, instead of writing R1 and R2, you write r and R. However, if you deal with many variables where the lower/upper case of a letter is already used by something else, just use subscripts.