# Thread: Tension on rope

1. ## Tension on rope

Ropes 3m and 5m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 5kg. The ropes, fastened at different heights make angles of 52 degrees and 40 degrees with the horizontal. Find the tension in each wire and the magnitude of each tension.

I started with an example in the book and then realized that the lengths of the rope where the same. This is what I did:

T1 = -T1cos(52) + T1sin(52)
T2 = T2cos(40) + T2sin(40)

T1 + T2 = 5

(-T1cos(52) + T2cos(40))i +( T1sin(52)+ T2sin(40))j = 5

-T1cos(52) + T2cos(40) = 0
T1sin(52)+ T2sin(40) = 5

T2 = (T1cos52)/cos40

T1sin(52)+ ((T1cos52)/cos40)sin(40) = 5

T1 = 3.83kg

T2 = ((3.83)cos52)/cos40
T2 = 3.1kg

and this is where I realized I forgot about the different lengths of the ropes. Can anyone help? I don't have any idea how to go about this problem.

Thanks

2. Originally Posted by kenshinofkin
Ropes 3m and 5m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 5kg. The ropes, fastened at different heights make angles of 52 degrees and 40 degrees with the horizontal. Find the tension in each wire and the magnitude of each tension.

I started with an example in the book and then realized that the lengths of the rope where the same. This is what I did:

T1 = -T1cos(52) + T1sin(52)
T2 = T2cos(40) + T2sin(40)

T1 + T2 = 5

(-T1cos(52) + T2cos(40))i +( T1sin(52)+ T2sin(40))j = 5

-T1cos(52) + T2cos(40) = 0
T1sin(52)+ T2sin(40) = 5

T2 = (T1cos52)/cos40

T1sin(52)+ ((T1cos52)/cos40)sin(40) = 5

T1 = 3.83kg

T2 = ((3.83)cos52)/cos40
T2 = 3.1kg

and this is where I realized I forgot about the different lengths of the ropes. Can anyone help? I don't have any idea how to go about this problem.

Thanks
The length of the rope is irrelevant. Lami's theorem gives the answer quickly:

Lami's theorem - Wikipedia, the free encyclopedia.

By the way, if you take the weight force as 5, the unit of force is kg wt, not kg.

3. Originally Posted by kenshinofkin
Ropes 3m and 5m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 5kg. The ropes, fastened at different heights make angles of 52 degrees and 40 degrees with the horizontal. Find the tension in each wire and the magnitude of each tension.

I started with an example in the book and then realized that the lengths of the rope where the same. This is what I did:

T1 = -T1cos(52) + T1sin(52)
T2 = T2cos(40) + T2sin(40)

T1 + T2 = 5

(-T1cos(52) + T2cos(40))i +( T1sin(52)+ T2sin(40))j = 5

-T1cos(52) + T2cos(40) = 0
T1sin(52)+ T2sin(40) = 5

T2 = (T1cos52)/cos40

T1sin(52)+ ((T1cos52)/cos40)sin(40) = 5

T1 = 3.83kg

T2 = ((3.83)cos52)/cos40
T2 = 3.1kg

and this is where I realized I forgot about the different lengths of the ropes. Can anyone help? I don't have any idea how to go about this problem.

Thanks
I didn't check the numbers (or the algebra) but the setup looks good. By the way, tension is a force. What's the unit for that?

-Dan

4. Originally Posted by kenshinofkin
Ropes 3m and 5m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 5kg. The ropes, fastened at different heights make angles of 52 degrees and 40 degrees with the horizontal. Find the tension in each wire and the magnitude of each tension.

I started with an example in the book and then realized that the lengths of the rope where the same. This is what I did:

T1 = -T1cos(52) + T1sin(52)
T2 = T2cos(40) + T2sin(40)

T1 + T2 = 5

(-T1cos(52) + T2cos(40))i +( T1sin(52)+ T2sin(40))j = 5

-T1cos(52) + T2cos(40) = 0
T1sin(52)+ T2sin(40) = 5

T2 = (T1cos52)/cos40

T1sin(52)+ ((T1cos52)/cos40)sin(40) = 5

T1 = 3.83kg

T2 = ((3.83)cos52)/cos40
T2 = 3.1kg

and this is where I realized I forgot about the different lengths of the ropes. Can anyone help? I don't have any idea how to go about this problem.

Thanks
Assuming the system is in equilibrium :-

$\displaystyle F_y = T_1 sin52 + T_2 sin40 = 5g$

$\displaystyle F_x = T_1 cos52 = T_2 cos40$

Solve these simultaneously and you you should get the tension. The length of the rope is not relevant to the calculation of tension.

5. The length of the ropes doesn't matter for the tension but what about the magnitude of each tension? Would I multiply the vectors for each of them by their length?

So

T1 = -2.36i + 3.02j
T2 = 2.37i + 2j

would be

T1 = 3(-2.36i + 3.02j)
T2 = 5(2.37i + 2j)

6. Originally Posted by kenshinofkin
The length of the ropes doesn't matter for the tension but what about the magnitude of each tension? Would I multiply the vectors for each of them by their length?

So

T1 = -2.36i + 3.02j
T2 = 2.37i + 2j

would be

T1 = 3(-2.36i + 3.02j)
T2 = 5(2.37i + 2j)
The tensions do not depend on the length of the ropes. Period. End. Never. Finito. Nunca, nada, jamas. Nein. Nyet.

Think about it. You have a tension in Newtons. That means a tension is a force, because it has the units of force. You want to multiply it by a length, which will give it a unit of Nm, a unit of energy.

The only thing that the tensions have to do with the ropes is that they are directed along them. We occasionally will use the length of the rope to find the angle that the tension is at, but never for the magnitude.

-Dan

7. Originally Posted by topsquark
The tensions do not depend on the length of the ropes. Period. End. Never. Finito. Nunca, nada, jamas. Nein. Nyet.

[snip]
-Dan
You forgot bupkis

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### ropes 3m and 5m in length

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