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# Problem: A mass hanging from two ropes

A mass of 108 g is hanging from two massless ropes attached to the ceiling.

One rope makes an angle of 50Â° with the ceiling, while the other makes an angle of 29Â°.

Find the tensions in the two ropes.

## Solving the problem

Let's begin by drawing our mass hanging from the two ropes:

Looking at our sketch, we can infer that the mass is subject to 3 forces:

• the tension force exerted by the first rope, T1
• the tension force exerted by the second rope, T2
• and the force of gravity, mg

Here's the free-body diagram of our hanging mass:

We know the mass (108 g, which in kilograms is 0.108 kg), and the angles that the two ropes make with the ceiling (50Â° and 29Â°).

We are asked to find the tensions in the two ropes.

m = 0.108 kg
Î¸1 = 50Â°
Î¸2 = 29Â°

### We want to know

T1 = ?
T2 = ?

The mass is hanging.

What does this tell us?

This tells us that the acceleration of the mass must be zero.

And because the acceleration is zero, the resultant force acting on the mass is also zero. Indeed, for Newton's 2nd Law:

R = ma
R = m Ã— 0
R = 0

Now that we determined that the resultant force acting on the mass is zero, we can find the tensions of the two ropes using the following step by step process:

1. First, we find the x and y components of the resultant force, as a sum of the x and y components of all the forces that act on the mass.
2. Then, since we know that R is zero, Rx and Ry must also be zero, so we substitute them with 0 in the equations that we found in the previous step.
3. Finally, we use the resulting equations to find the tensions, T1 and T2.

We draw the coordinate axes on our free-body diagram. For convenience, we choose the x axis horizontal and the y axis vertical. Then, we determine the x and y components of all the forces that act on the mass.

We need to keep in mind that the angle between each tension force and its x component is equal to the angle that the rope, producing that tension, makes with the ceiling:

T1x = âˆ’T1 cos 50Â°
T2x = T2 cos 29Â°
mgx = 0
T1y = T1 sin 50Â°
T2y = T2 sin 29Â°
mgy = âˆ’mg

Thus, the x and y components of the resultant force will be:

x:

Rx = T1x + T2x + mgx
Rx = âˆ’T1 cos 50Â° + T2 cos 29Â° + 0
Rx = T2 cos 29Â° âˆ’ T1 cos 50Â° (1)

y:

Ry = T1y + T2y + mgy
Ry = T1 sin 50Â° + T2 sin 29Â° + (âˆ’mg)
Ry = T1 sin 50Â° + T2 sin 29Â° âˆ’ mg (2)

The next step is to substitute Rx and Ry in Eq. (1) and Eq. (2) with 0:

Rx = T2 cos 29Â° âˆ’ T1 cos 50Â°
Ry = T1 sin 50Â° + T2 sin 29Â° âˆ’ mg
â†“
0 = T2 cos 29Â° âˆ’ T1 cos 50Â° (3)
0 = T1 sin 50Â° + T2 sin 29Â° âˆ’ mg (4)

Using these two equations we can easily find T1 and T2.

There are multiple ways in which we can do this. One way would be to first solve Eq. (3) for T2:

0 = T2 cos 29Â° âˆ’ T1 cos 50Â°
T2 cos 29Â° âˆ’ T1 cos 50Â° = 0
T2 cos 29Â° = T1 cos 50Â°
 T2 = T1 cos 50Â° cos 29Â°
T2 = 0.735T1 (5)

Then, we substitute T2 with 0.735T1 in Eq. (4):

0 = T1 sin 50Â° + T2 sin 29Â° âˆ’ mg
0 = T1 sin 50Â° + 0.735T1 sin 29Â° âˆ’ mg

And we solve it for T1:

0 = T1 (sin 50Â° + 0.735 sin 29Â°) âˆ’ mg
0 = T1 (1.12) âˆ’ mg
mg = 1.12T1
1.12T1 = mg
 T1 = mg 1.12
 T1 = (0.108 kg) (9.81 N/kg) 1.12
T1 = 0.946 N

Finally, we substitute the value of T1 in Eq. (5) to find T2:

T2 = 0.735T1
T2 = (0.735) (0.946 N)
T2 = 0.695 N

Therefore, the tension in the rope at the 50Â° angle is 0.946 N, and the tension in the rope at the 29Â° angle is 0.695 N.

## Tips & Tricks

• Remember that when an object hangs from two ropes, the angle between the tension produced by a rope and the x component of that tension, is equal to the angle that the rope makes with the ceiling: