Tension w/ rope hanging from a ceiling

[iJamJL]

Homework Statement

The ends of a massless rope are tied to a ceiling. Two identical 4.21 kg masses are now hung along the length of the rope, dividing the rope into three segments of equal length. Segment 2 (the central segment) is horizontal (parallel to the ceiling). The ends of segments 1 and 3 are attached to the ceiling, making an angle of 50.2° with the ceiling. (The distance between the points at which the rope is hung is greater than the rope segments.) Find the magnitude of the tension in rope 2.

Homework Equations

Pythagorean and components

The Attempt at a Solution

I attached my diagram of how I drew everything. I'm hoping that this is correct. If it is, then I assume rope 2 is the middle section of the rope where it is parallel with the ceiling (my professor doesn't seem to phrase his questions that well, or I'm just not used to it). After solving for the tensions in rope 1 and 3, which are equal to each other, we could move the x-component vector to where I made it in red. These two vectors act in opposite directions, causing the tension of the middle rope to be twice as much. That is my assumption.

If my assumption is correct, then the tension in rope 2 should be:

T*cos(50.2)*2=68.82

This answer is incorrect, so could someone tell me where I went wrong?

 

[LawrenceC]

Why do you multiply by 2?

 

[iJamJL]

Why do you multiply by 2?

I thought since there is tension from both weights that go in opposite directions (like tug-of-war), then the tension would be twice as great.

 

[LawrenceC]

What about Newton's law?

 

[tiny-tim]

hi iJamJL!

These two vectors act in opposite directions, causing the tension of the middle rope to be twice as much.

noooo …

do a free body diagram for a small horizontal bit of string starting at the right-hand end …

Tcos50.2° to the right, and … to the left? ​

 

[LawrenceC]

If you tie a rope to a doorknob and pull on it with a force of 10 N, is the tension in the rope 20 N? I don't think so.

 

[iJamJL]

What about Newton's law?

Newton's third law says that for every action, there is an equal and opposite reaction. Therefore, if we created those reactions where the masses are, then we would have two equal forces going toward each other (one horizontally to the right of the left mass, and one horizontally to the left of the right mass), resulting in 0. That can't be because tension exists, making the string parallel. I suppose I'm just not thinking in the correct manner.

hi iJamJL! noooo …

do a free body diagram for a small horizontal bit of string starting at the right-hand end …

Tcos50.2° to the right, and … to the left? ​

Isn't it the same force, but in the opposite direction?What about the left-end rope?

 

[tiny-tim]

do a free body diagram for a small horizontal bit of string starting at the right-hand end …

What about the left-end rope?

the left-end rope isn't anywhere near the small bit at the right-hand end, is it?

what is the force on the left of that small bit?​

 

[iJamJL]

the left-end rope isn't anywhere near the small bit at the right-hand end, is it?

what is the force on the left of that small bit?​

It's equal to the force on the right side, Tcos(50.2°). I saw this before. When we look at the center of the middle section, then the net force would be equal to 0, wouldn't it?

 

[tiny-tim]

It's equal to the force on the right side, Tcos(50.2°).

i meant, what's it called?

it's called the tension ​

 

[iJamJL]

i meant, what's it called?

it's called the tension ​

Oh, I knew that, lol. The whole difficulty of the problem for me is the fact that I see two forces of tension that act in opposite directions due to the two weights. I thought we were supposed to find the net force, or net tension, due to those weights. It would either be twice (that's why I multiplied by two: Fnet = F-left + F-right) the force of one, or 0, which didn't make sense at all. The answer seems to be that there is only one force of tension acting on the central section of the rope..but I don't think I fully, 100% understand it.

 

[LawrenceC]

Oh, I knew that, lol. The whole difficulty of the problem for me is the fact that I see two forces of tension that act in opposite directions due to the two weights. I thought we were supposed to find the net force, or net tension, due to those weights. It would either be twice (that's why I multiplied by two: Fnet = F-left + F-right) the force of one, or 0, which didn't make sense at all. The answer seems to be that there is only one force of tension acting on the central section of the rope..but I don't think I fully, 100% understand it.

Think about the rope and doorknob example I mentioned. The knob pulls on the rope with 10 N but in the opposite direction from your pull. But the tension in the rope is 10 N not 20 N.

 

[tiny-tim]

You can't find the tension in the middle section by looking at the whole middle section …

you can only find it (or any other force) when it's an external force …

that means you have to "cut" the string, so that you can consider the tension as an external force

 

[PeterO]

I thought since there is tension from both weights that go in opposite directions (like tug-of-war), then the tension would be twice as great.

In your tug-of-war example.
Team 1 pull on the rope with a combined force of F Newtons.
Newtons 3rd law requires the rope to pull back on them with a force of F Newtons.

Meanwhile Team 2 pulls on their end of the rope with a combined force of F Newtons.
Newtons 3rd law requires the rope to pull back on them with a force of F Newtons.

All that is easily achieved by the Tension in the rope being F Newtons - exactly the required force.

Note: What is impossible is for one team to pull more strongly on the rope that the other team. The team that wins the tug of war is the team that pushes on the ground the hardest!