# Homework Help: How do you work out the tension and angle of this block in equilibrium?

1. Apr 1, 2009

### xllx

1. The problem statement, all variables and given/known data
A block that has a weight of 20N, is in equilibrium. It is attached to 3 strings, A, B and C. String A has a tension of 15N to the right of the block. Between String A and B is the angle (a) and between B and C is 90degrees. The tension in String B and C is the same. Calculate the tension and the angle (a).

2. Relevant equations

sum of the forces=0

3. The attempt at a solution

For the horizontal forces I got:
15+Tcos(a)=Tcos(90+a)

And then for the vertical:
20=Tsin(a)+Tsin(90+a)

But I don't know how to figure it out from there, or whether they are even right. Any help at all would be greatly apprecaited!

2. Apr 1, 2009

### rootX

cos(90+a) = sin(a)
sin (90+a) = -cos(a)

But, why the angle is 90+a?

3. Apr 1, 2009

### xllx

The angle of String C is 90degrees plus (a) away from the 15N to the right. I've attached a diagram of the situation, sorry I aren't very good at expalining. Am I totally wrong with the angles??

Thanks by the way!

File size:
273.9 KB
Views:
139
4. Apr 1, 2009

### rootX

It is easier to visualize if you use 90-a (triangle left to the force vector C).

I would recommend to divide both equations and eliminate T. Then you will get Acos (a) = B sin(a) relationship..which is easier to solve

5. Apr 1, 2009

### xllx

So my two equations are:

15+Bcos(a)=Ccos(90-a)
20=Bsin(a)+Csin(90-a)

Which because the tension is the same in both means the A and C can be replaced by T:

15+Tcos(a)=Tcos(90-a)
20=Tsin(a)+Tsin(90-a)

From this do I rearrange to get Tcos(a) and Tsin(a) as the subjects? And if they are divided they equal tan(a), or is this wrong?

6. Apr 3, 2009

### Redbelly98

Staff Emeritus
I would next simplify the cos(90-a) and sin(90-a). You can use the trig formulas for sin(x-y) and cos(x-y) to do this.

Last edited: Apr 3, 2009
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook