Calculating Tension in Multiple String Systems

[kaffekjele]

Homework Statement

A weight with mass m is strung up by two strings which are attached in A and B. The mass m is considered a point.
Calculate the tension i in string A and in string B.

Figure is here http://tinypic.com/r/1y1jrr/6

The mass m is 11,1kg
Distance a is 3,2m
Distance b is 3,4m
Distance c is 2,6 m
Distance d is 3,7m

(I'm translating directly from Danish, so I might not have gotten all the terms into correct English...)

The Attempt at a Solution

The first thing I did was to calculate the force G for the weight using G= mg
= 11,1*9,81 = 1089N.

I then tried to find the angles between the strings and the imaginary x-axis by using the measurements provided.

The angle between string A and the x-axis(angle v1): invers tan 3,2/2,6≈51°

The angle between string A and the x-axis(angle v2): invers tan 6,6/3,7≈61°

I then set up two equations, one for the horisontal force and one for the vertical force:

Horisontally: A*cosv1-B*cosv2=0
Vertically: A*sinv1+B*sinv2=108,9


And this is where I get stuck... I don't even know if I'm going about this the right way so I'd appreciate it if someone could have a look at it and point me in the right direction.

 

[voko]

So far you have done everything correctly. Why are you stuck? You should just solve the system now.

 

[kaffekjele]

I wasn't sure I'd done it correctly since I ended up with two equations with two unknowns(up until now I've done tasks where either A or B would have been given from the beginning.)

To continue on from my first post, would this be the correct way to solve the system?

A*cosv1-B*cosv2=0
A*sin v1+B*sinv2= 108,9

A*0,6293-B*04848=0
A*0,7771+B*0,8746=108,9

(add up A)

1,4064A = 108,9 -->A=774kN

Then put value for A into the first equation which gives B = 100,5kN

 

[voko]

This is not entirely correct. If you simply add together the two equations, you won't eliminate B; you will have

1.4064 A + 0.424 B = 108.9, which leads you nowhere.

What you should do is multiply the first equation by 0.8746/0.4848 and then add the resultant equation to the second one; the result will not have B. Can you see why?

 

[kaffekjele]

Ah, I see my mistake. In order to do what I did the two B values would have had to be the same value in order to cancel each other out?(one being - and the other being +)

What you should do is multiply the first equation by 0.8746/0.4848 and then add the resultant equation to the second one; the result will not have B. Can you see why?

I'm not sure if I understand you correctly, but do you mean using 0,8746/0,4848 as a fraction like this:

A*0,6293-B*04848=0 *\frac{0,8746}{0,4848} and then cancel 0,4848 against 0,4848?

I'm pretty sure what I just wrote is "illegal" in the world of math, but I'll leave it for now as I don't have anything better to write. I know most calculators could solve equation systems like these, but I would like to learn how to do it by hand as well, so I'd be grateful if anyone could recommend some good videos on the topic.

 

[voko]

I am not sure about videos, but what you have here is a system of linear algebraic equations - the simplest case, two equations for two unknowns. I am pretty sure your high school curriculum should cover this.

When you multiply an equation such as aA - bB = 0 by some c, you end up with the equation acA- bcB = 0. In this case, c = 0.8746/0.4848, and, indeed, it would transform b = 0.4848 into bc = 0.8746. But mind what happens to a: it becomes ac.

 

[kaffekjele]

Thank you so much for your help. I appreciate it.