Equilibrium Process for Calculating Tension in a Tight-Rope Walker's Wire

[barbiegirl42]

Homework Statement

A 60 kg tight-rope walker carries a long beam with
a mass of 30 kg across a 10 m long wire. When
she is at the centre of the wire (i.e. 5 m across),
each section of the wire makes an angle of 5° to
the horizontal. Assuming that the mass of the wire
is negligible, calculate the tension within it.

I would like to know if my process is right. I took forever just trying to figure out the problem and would like to verify that my process is correct or if there any omissions I may have left out.

Homework Equations

T=rf
Sum of the forces = 0
Sin=0 / H

The Attempt at a Solution

Sum of the forces = 0
F(Tightrope walker) + F(Beam) = F(Tension in one half of the rope) + F(Tension in the other half)
(60x9.8) + (30x9.8) = F(T1) + F(T2)
822=F(T1) + F(T2)

882 / (2 x Sin5 x 5)
= 1011.98
1011.98 / 2
= 505.98
= 506 N <~ My answer is right, I checked in the answers section from the book.

 

[Orodruin]

F(Tightrope walker) + F(Beam) = F(Tension in one half of the rope) + F(Tension in the other half)

The tension in the rope is what you are searching for, but here you are only interested in the vertical component of the tension.

882 / (2 x Sin5 x 5)
= 1011.98
1011.98 / 2
= 505.98

This is unclear. What are you computing? Where does the 5 come from in the first line? Where does the second division by two come from?

= 506 N <~ My answer is right, I checked in the answers section from the book.

You can check whether this is reasonable or not. With this answer, the total upward force on the walker+beam would be less than their gravitation even if the angle was 45 degrees ... 2*506/sqrt(2) = sqrt(2)*506 < 882.

 

[barbiegirl42]

The tension in the rope is what you are searching for, but here you are only interested in the vertical component of the tension.This is unclear. What are you computing? Where does the 5 come from in the first line? Where does the second division by two come from?You can check whether this is reasonable or not. With this answer, the total upward force on the walker+beam would be less than their gravitation even if the angle was 45 degrees ... 2*506/sqrt(2) = sqrt(2)*506 < 882.

Well the Sin5 is the 5 degree angle to the horizontal and second 5 is the distance of 5m since the tightrope walker is 5m across thus you get the distance from the pivot point. Well I was thinking that the weight force would be half as in the adjacent side = F/2 and well the answer is 506 so when going backwards, I'm thinking the question is asking what is the tension in each section maybe? It just seemed to work when I divided the answer by two. I don't get what you mean by the first and third part. Please explain a bit more detailed.

 

[Orodruin]

second 5 is the distance of 5m since the tightrope walker is 5m across thus you get the distance from the pivot point.

This has nothing to do with the tension in the rope. You are computing a force, not a torque.

If you assumed the angle to be 45 degrees instead, the tension of 506 N in each rope would result in an upward force of 506 N * sin(45) = 506 N/sqrt(2) on the walker+beam. This would give a total upward force of less than the weight of walker+beam - even if the angle was much larger than 5 degrees. The conclusion is that your tension must be larger.

 

[barbiegirl42]

why

This has nothing to do with the tension in the rope. You are computing a force, not a torque.

If you assumed the angle to be 45 degrees instead, the tension of 506 N in each rope would result in an upward force of 506 N * sin(45) = 506 N/sqrt(2) on the walker+beam. This would give a total upward force of less than the weight of walker+beam - even if the angle was much larger than 5 degrees. The conclusion is that your tension must be larger.

why would assume the angle to be 45 degrees? and where would I get this assumption from?
and why are you dividing by the square root of 2 in the formula? I never learned that aspect in any calculations I did. Through all the examples in the book, the distance was always times by either the sin or the cos of the angle which makes sense and you need the length really. Why should my tension be larger though? <<Mentor note: Edited to remove part in all caps>> I learned in Physics that if you over complicate things then it makes it worse and that simple is the way to go. Overcomplicating gets you no where from what learned.

 

[Orodruin]

why would assume the angle to be 45 degrees? and where would I get this assumption from?

It is a simple assumption to test the validity of the answer. It is not related to your problem in any other way than showing that even when a larger portion of the tension is giving a force in the vertical direction, the quoted tension is too small.

Through all the examples in the book, the distance was always times by either the sin or the cos of the angle which makes sense and you need the length really

No, the length here is an irrelevant piece of information. If you involve it the way you have, your answer will have units of N/m instead of N and thus not be a tension.

If you have a force T acting in a direction of 5 degrees from the horizontal, what is its vertical component?

 

[barbiegirl42]

Here is what my physics book says in two examples. As you can see, I used the examples as examples as to what I should do when I calculate the answer to this question. This is what I learned from the book and it made sense to me so I used some parts they did such as using the sin rule to calculate other things.

 

[barbiegirl42]

It is a simple assumption to test the validity of the answer. It is not related to your problem in any other way than showing that even when a larger portion of the tension is giving a force in the vertical direction, the quoted tension is too small.
No, the length here is an irrelevant piece of information. If you involve it the way you have, your answer will have units of N/m instead of N and thus not be a tension.

If you have a force T acting in a direction of 5 degrees from the horizontal, what is its vertical component?

But I'm not finding the vertical component of the tension neither the horizontal, I'm finding the tension along the wire. Why test the validity of the answer though. I would assume in a printed book that looks pretty valid to me that the answer would be right. I already know the force downwards don't I? It doesn't ask for the vertical component, it ask for the tension which is acting along an angle.

 

[Orodruin]

But I'm not finding the vertical component of the tension neither the horizontal, I'm finding the tension along the wire.

Yes, the tension along the wire is related to its vertical component through the sine of the angle and must equal to the weight of walker+beam. Thus, to find the tension you first find the vertical component and then relate this to the full tension with the sine.

I would assume in a printed book that looks pretty valid to me that the answer would be right.

Even if this is often the case, books are often full of misprints and mistakes. Just the other day I discovered a factual error in the seventh edition of one of the more well regarded books on mathematical methods in physics.

Also, in the images you linked, they are not using anything in a way different from what I have told you. In the hanging sign you are not even given the length of the ropes. The triangle there is a force triangle.

 

[barbiegirl42]

I managed to find the worked answers. You were right in not including the length. I only got mixed up with the dividing. Now I see that you first divide by 2 then divide by sin 5

 

[Orodruin]

So from the last line of that solution it is quite apparent that there is an arithmetic error. If it would be correct, then $\sin(5^\circ) = 4.41/5.06$, which obviously is not the case (it is 0.087, not 0.87). The answer is off by a factor 10.