# Transitional Equilibrium: Tension

A 60 kg tight-rope walker carries a long beam with a mass of 30kg accross a 10m 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 degrees to the horisontal. Calculate the tension in the wire.

This one is not that difficult but a lack a certain part of me to get the answer.

60+30 = 90 which is the combined weight Multiply this by 9.8 and we get the force the combined weight exerts on the wire. This is 882 Newtons. Now, from trigonometry i got the tension in the wire to be (2 * 882)/sin(5degrees) = 20239 Newtons of tension. This is 4 times the correct answer. So i though that maybe half the weight acts on half the wire to 822/sin(5degrees) might be the answer but it would leave you with double the correct answer. Argh.

I need help on this one :)

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Each wire supports half of the 90 kg.

The forces acting on one of the wires must be balanced. So, equate the x and y components of the tension T with the gravitational force acting on the tightrope walker (for one wire, 441 N).

It will probably be something like 441 = T sin(5), so solving for T gives T = 5059 N.

HallsofIvy
Homework Helper
futz is correct. Why in the world do you have that factor of 2 in
"(2 * 882)/sin(5degrees) "? The total weight is 882 Newtons and, since, each side supports 1/2 of that: you should have divided by 2, not multiplied by 2. That's why your answer is 4 times what it should be.

Ok, so my mistake was that i doubled the tension when i shouldnt have, and not divided the weight into halve for the 2 sides.

The main thing i did wrong was not knowing that once you find the tension for the side your working with, you dont double it to get the full tension of the wire.

So the tension on the side we found is the total tension :) if that makes sence to you.