Tension of 10 m rope hung across 6 m horizontal distance with 50 kg box hanging

[TRed503]

Homework Statement

A rope is 10.0 m long and is attached at points A and B, which are at the same elevation 6.0 m apart. If a 50.0 kg mass is attached to the rope 4.0 m away from point A, what is the tension in the two segments of the rope?

Homework Equations

W = ma
ƩFy = 0 = T1 + T2 - W
ƩFx = 0 = T1 + T2

The Attempt at a Solution

W = 50.0 kg * 9.81 m/s^2
I'm stuck after finding the weight of the hanging mass as no other angles or forces are given? I tried using 6.667 m and 3.333 m as the respective lengths for T1 and T2 assuming that the ratio of 2/3 distance of the hanging mass between point A and B would be the same ratio on the rope length. Then tried to use trig to get the vertical hanging distance but came up with 2 different distances so these numbers don't work. I'm lost and stuck as to were to go. Any help is much appreciated as this homework assignment is already late and I've spent about 5 hours reading and looking up stuff online with no success! Feel as though I'm trying to extract blood from a rock!

 

[SteamKing]

Not only must your forces balance, you must also have 10 m of rope between A, the load, and B.

 

[TRed503]

Yes that is correct. Not sure if that was a statement or a question, but I'm past that part, drew my diagram and am stuck trying to suck Newtons out of meters and no angles somehow?

 

[TRed503]

I think I figured it out... I was interpreting the problem to mean that the mass is 4 m horizontal distance from point A, but I think that the 4 m was supposed to be the length of T1 or the hypotenuse for my right triangle! If only story problems came with pictures! Thanks for you're reply SteamKing I'll be back if I hit another wall but I should be good for this problem now.

 

[DeeNos]

I would recommend approaching this problem by breaking it down into smaller parts and using the equations provided. First, let's label the forces acting on the rope: T1 and T2, the tensions in the two segments of the rope, and W, the weight of the hanging mass. We can also label the distances from point A to the hanging mass as x and from the hanging mass to point B as 6-x.

Next, we can use the equation ƩFy = 0 to find the relationship between T1 and T2. Since the rope is hanging vertically, we can assume that the net force in the y-direction is zero. This means that T1 and T2 must be equal in magnitude. Therefore, we can write T1 = T2.

Using the equation ƩFx = 0, we can set up an equation to find the tension in the rope. We know that T1 and T2 are equal, so we can write T1 + T2 = 2T1. The only other force acting in the x-direction is the weight of the hanging mass, which we can calculate using W = ma. Since we know the mass (50 kg) and the acceleration due to gravity (9.81 m/s^2), we can calculate the weight.

Now, we can set up the equation 2T1 = W and solve for T1. Once we have T1, we can use the relationship T1 = T2 to find T2. This will give us the tensions in the two segments of the rope.

As a final check, we can make sure that our calculated tensions make sense. Since the rope is hanging vertically, the tension should be greater at the bottom (where the hanging mass is) and decrease as we move up the rope. Also, the total tension should be greater than the weight of the hanging mass, since the rope is also supporting its own weight.

I hope this helps guide you towards a solution for this problem. Remember to always break down the problem into smaller parts and use the equations provided to help you solve it. Good luck!