# Homework Help: Finding Tension in 2 strings

1. Nov 2, 2008

### swede5670

1. The problem statement, all variables and given/known data
A 119 mass hangs from two string Calculate the tension in both strings. Let θ1 = 77.0° and θ2 = 26.0 degrees

2. Relevant equations
Fnet=ma

3. The attempt at a solution
This is the last question of my homework packet (so its the most challenging) and I have no clue how to solve it or where to begin. I'd appreciate any and all help.

2. Nov 2, 2008

### hage567

Have you drawn a diagram?

3. Nov 2, 2008

### swede5670

Yes I have, basically two strings hanging from a ceiling of sorts, with the weight hanging to the right. The angle in the left corner is 77 degrees and the one on the right is 26 degrees.

4. Nov 2, 2008

### hage567

Do you know how to resolve vectors into their components?

5. Nov 2, 2008

### swede5670

Im not sure how to do anything after I have diagrammed the situation out. Please help! This is due tomorrow.

6. Nov 2, 2008

### swede5670

I broke the free body diagram down into two portions and now I'm trying to solve for θ1 = 26 degrees. I need to find the Upwards component and the sideways component. I created a triangle with T1 (tension 1) as my hypotenuse and then for the sideways component I have cosine(26)*T1 = adjacent and then sine(26)*T1 = Opposite

I just have no idea what I need to do in order to find the adjacent value so that I can solve for T1 in the vector.
Since I have an object with a mass of 119N (I actually don't know if the object is in kg or N because my teacher decided not to include that piece of vital information) I can assume that the strings must be exerting a total of 119N up right? The Object isn't accelerating. So can I use this value as the opposite portion of the triangle?

I said that since gravity is pulling down with 119N on the system then I can use that value for upwards component of my tension. Then to find the sideways component I took 119 x tan(26) = 140.27N for my adjacent and then to find the hypotenuse I used cosine. 140.27/cosine(26) = X and then the value I got was 216.83N. This is wrong however and once again I don't know what happened.

Last edited: Nov 3, 2008
7. Nov 3, 2008

### hage567

You don't need to directly find the "adjacent" value. You need to find the components of each tension and then sum them up in the x and y directions. So you should have two horizontal forces in one equation and then three vertical forces in another equation.
I'm not sure what you've done here.

This site might help: http://hyperphysics.phy-astr.gsu.edu/hbase/fcab.html