# A weight suspended by two strings

1. Oct 19, 2014

### LLR

1. The problem statement, all variables and given/known data
A weight is suspended by two strings at two different angles (the picture is included in the attempted solution)
a.) Find the expressions for the tension in each of the strands
b.) Assuming the tension, T1, does not change, what will be the acceleration of the weight if the second string is released?

2. Relevant equations
ΣF = ma

3. The attempt at a solution

This is my attempted solution for the problem. A) I think I'm good with but if there's something wrong please point it out. B) is where I really have trouble. I'm not sure I did accel in the Y correctly (I had substituted the equation for T2 into my B equation, sorry I didn't show that).

I just feel like I'm missing something here, like it should be simpler than that. Also I am unsure if there will be acceleration in the X. But if there was I imagine I would figure it similarly to the way I did Y

2. Oct 19, 2014

### voko

A is not finished. The two expressions that you obtained refer to each circularly.

In B, you have to assume that tension 2 disappears, while tension 1 is entirely present.

3. Oct 19, 2014

### LLR

Okay, let me start with A and see if I'm on the right track

Lets say I keep what I got for T2. Would I substitute that in for the second equation and solve for T1? My algebra is admittedly kind of rusty, and I'm having trouble isolating both T1s by themselves when I do that

4. Oct 19, 2014

### voko

In A, you had two starting equations. You used one of the to obtain the relationship between tensions. Use the other one to determine the tensions completely.

5. Oct 19, 2014

### LLR

Kind of like this? Or am I not doing this correctly?

EDIT: Obviously I didn't solve this yet, but when solved would this be correct? I havent solved it because I have trouble getting T1 by itself since there is two of them in the equation

6. Oct 20, 2014

### haruspex

Yes. How would you rewrite 5x+3x so that there is only one x?

7. Oct 20, 2014

### LLR

This is what I end up getting but it really doesn't seem right. Once again I apologize for my incredibly rusty algebra

8. Oct 20, 2014

### voko

No that is not correct. You need to brush up on your algebra.

9. Oct 20, 2014

### BvU

Rusty is a good term. Dissolve some of the rust by
* taking one step at the time
* doing on the right the same what you do on the left
* rewriting things like T1 bla + T1 da as T1 (bla + da)
* gradually bringing what you want to know (T1) to one side
so

T1 bla + T1 da - mg = 0
T1 ( bla + da) - mg = 0 $\ \ \ \ \$ group terms with T1
T1 ( bla + da) = mg $\ \ \ \ \$ add mg to left and right
T1 = mg/( bla + da) $\ \ \ \ \$ divide by ( bla + da) left and right (make sure it's never zero)

10. Oct 20, 2014

### LLR

Something like this? I have two versions as I was unsure whether to combine the terms or divide the sin

Thank you all for your replies though, it has been a while since any sort of math class and I know I'm still really rough on it

11. Oct 20, 2014

### voko

12. Oct 20, 2014

### BvU

Version 2 step 2: You should divide both terms on the right by sin theta2

13. Oct 20, 2014

### LLR

Alright, so using the same methods as above I solve for T2 in a similar manner. This should take care of part A, assuming this is correct?

14. Oct 20, 2014

### LLR

I went ahead and plugged in the equations from part A to solve part B, let me know if I messed up somewhere please

15. Oct 21, 2014

### haruspex

In your ay calculation you failed to cancel an m. There should be no mass terms in the answer.
All your answers can be simplified by multiplying top and bottom by cos theta2. It's a bit ugly having a division inside a division.
There is a quicker way to get (b) having got (a). Since T1 doesn't change, the only difference in the system of forces is the removal of T2. This is equivalent, as far as the mass is concerned, to adding a force -T2. So the acceleration is -T2/m.

16. Oct 21, 2014

### LLR

So the final answer would look something like this?

Or am I multiplying the top and bottom wrong?

I aplogize I forget to remove the extra m from this one, but I'm a bit confused as to why it would cancel. The starting equation has 1 m on one side, and 2 on the other. Wouldn't only one m on each side cancel then?

And I'm afraid I'm not quite following why it would only be a = -T2/m. Why would T1 not be factored into it?

17. Oct 22, 2014

### haruspex

The m should cancel as a common factor. If you can't get that to happen, post all your steps.
Since the system is in equilibrium before T2 is removed, the other forces must add up to -T2. When T2 is removed, they still add up to -T2 (since T1 doesn't change).