# Homework Help: - What is the tension in each wire?

1. Oct 24, 2005

### IB

Statics and Equilibrium

Hi. I read and re-read these problems but I don't get them. Could you please show me how to do them?
STATICS
A 700.0N window washer is standing on a uniform scaffold supported by a vertical rope at each end. The scaffold weighs 200.0N and is 3.00m long. What is the force in each rope when t he window washer stands 1.00 m from one end?
[This is a torque/rotational equilibrium problem]

A Street lamp weighs 150N. It is supported by two wires that form an angle of 120 degrees with each other. The tensions in the wires are equal.
- What is the tension in each wire?
- What is the minimum tension that each wire has to exert to support the lamp? Assume the wires exert equal forces.

I tried to do the second one. Here is my solution, but I don't know if it's right or not:
a) 120 divided by 2 we have 60 degrees. Calculate the complementary angle we have 90 - 60 = 30 degrees.
sin(30)*150 = 75.0 N
Answer: Tension in each wire is equal to 75.0 N.
b) Hmm, this one I'm not so sure, but perhaps it's also 75.0 N. It makes sense if each wire supports half of the weight of the lamp.
Thanks in advance for the help!

Last edited: Oct 24, 2005
2. Oct 25, 2005

### Diane_

I'd need to see some of your work on the first one before I could help.

For the second: One thing should be clear. If the wires were hanging straight down, each would be supporting one half of the weight of the lamp. In that case, the tension in each wire would be 75N.

The wires, however, are not hanging straight down. As the angle of the wires changes, the tension will have to increase - do you see that? If so, then it's clear your answer is not correct.

Do this: start by making a free body diagram of the situation. You can treat the lamp as a point, as the symmetry of the geometry of the situation will pretty much guarantee that there are no torques to worry about here. The tensions in the wires will be in the direction of the wires, and hence will have both horizontal and vertical components. Weight acts only vertically, so how will the vertical components of the tensions have to compare to the weight? How will the horizontal components of the weight have to compare with each other? And what does that tell you about the tensions in the wires?

A little simple geometry will give you the answer.

For the second part of the question, I assume what it means is that we should assume that we can change the angle between the wires. What angle would give you the minimum tension in the wires, and what would that tension be? You should already know the answer to that one.

3. Oct 25, 2005

### IB

2(Tcos60)=150
Tcos60=75
T=150

Is that right? :)

And for part b) I think 75.0 N is the most logical, yeah? It can't be less, cos otherwise it won't be able to hold 150.0 N.
----

First one:

Force x distance:
Forces going in one directon = 700x1 and 200 x 1.5
Forces going in other direction = F x 3

3F = 700*1 + 200*1.5
3F = 1000
F = 333.3333333 etc.

333.33333 + T = 700 + 200
T = 900 - 333.33333
T= 566.66666

Am I doing this right?

4. Oct 25, 2005

### Diane_

Problem 2, check. Good job.

Problem 1, mostly check. Keep track of your units and your significant digits. That's a quibble, though. Good job there, too. :)

5. Oct 25, 2005

### IB

Thanks so much, Diane_ :)