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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.

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, T
_{1} - the tension force exerted by the second rope, T
_{2} - 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°

T_{1} = ?

T_{2} = ?

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 2 ^{nd} 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:

- 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.
- Then, since we know that R is zero, R
_{x}and R_{y}must also be zero, so we substitute them with 0 in the equations that we found in the previous step. - Finally, we use the resulting equations to find the tensions, T
_{1}and T_{2}.

Let's start with the first step.

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:

T_{1x} = âˆ’T_{1} cos 50Â°

T_{2x} = T_{2} cos 29Â°

mg_{x} = 0

T_{1y} = T_{1} sin 50Â°

T_{2y} = T_{2} sin 29Â°

mg_{y} = âˆ’mg

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

x:

R_{x} = T_{1x} + T_{2x} + mg_{x}

R_{x} = âˆ’T_{1} cos 50Â° + T_{2} cos 29Â° + 0

R_{x} = T_{2} cos 29Â° âˆ’ T_{1} cos 50Â° (1)

y:

R_{y} = T_{1y} + T_{2y} + mg_{y}

R_{y} = T_{1} sin 50Â° + T_{2} sin 29Â° + (âˆ’mg)

R_{y} = T_{1} sin 50Â° + T_{2} sin 29Â° âˆ’ mg (2)

The next step is to substitute R_{x} and R_{y} in Eq. (1) and Eq. (2) with 0:

R_{x} = T_{2} cos 29Â° âˆ’ T_{1} cos 50Â°

R_{y} = T_{1} sin 50Â° + T_{2} sin 29Â° âˆ’ mg

â†“

0 = T_{2} cos 29Â° âˆ’ T_{1} cos 50Â° (3)

0 = T_{1} sin 50Â° + T_{2} sin 29Â° âˆ’ mg (4)

Using these two equations we can easily find T_{1} and T_{2}.

There are multiple ways in which we can do this. One way would be to first solve Eq. (3) for T_{2}:

0 = T_{2} cos 29Â° âˆ’ T_{1} cos 50Â°

T_{2} cos 29Â° âˆ’ T_{1} cos 50Â° = 0

T_{2} cos 29Â° = T_{1} cos 50Â°

T_{2} = T_{1} | cos 50Â° |

cos 29Â° |

T_{2} = 0.735T_{1} (5)

Then, we substitute T_{2} with 0.735T_{1} in Eq. (4):

0 = T_{1} sin 50Â° + T_{2} sin 29Â° âˆ’ mg

0 = T_{1} sin 50Â° + 0.735T_{1} sin 29Â° âˆ’ mg

And we solve it for T_{1}:

0 = T_{1} (sin 50Â° + 0.735 sin 29Â°) âˆ’ mg

0 = T_{1} (1.12) âˆ’ mg

mg = 1.12T_{1}

1.12T_{1} = mg

T_{1} = | mg |

1.12 |

T_{1} = | (0.108 kg) (9.81 N/kg) |

1.12 |

T_{1} = 0.946 N

Finally, we substitute the value of T_{1} in Eq. (5) to find T_{2}:

T_{2} = 0.735T_{1}

T_{2} = (0.735) (0.946 N)

T_{2} = 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.

- 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:

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