# F = MA 2010 # 11 (Static equilibrium, 3 masses on a string)

1. Jan 28, 2013

### SignaturePF

1. The problem statement, all variables and given/known data
See:
Question 11

2. Relevant equations
T_net = 0
F_net = 0

3. The attempt at a solution
Let T1 be the force of tension on the left, T2 on the right
T1cosθ=T2cosθ
Y components:
T1sinθ+T2sinθ = mg
Since T1 = T2, as established by the x-component force equation:
2T1sinθ= mg
T1sinθ = mg /2

Now let's look at T_net = 0
Take the leftmost point to be the PoR
This is where I got stuck.
T1cosθb = T2sinθa
T2sinθ=T1sinθ = mg/2
T1cosθb = mg/2a
a/b = mg/2T1cosθ

I'm really confused about how to find the torque:
I get that it is rxF or rFsintheta but I'm struggling to make the equation.

Last edited: Jan 28, 2013
2. Jan 28, 2013

### tms

There is no tension in question 12 at that link.

3. Jan 28, 2013

### SignaturePF

Sorry I meant question 11.

4. Jan 28, 2013

### tms

What is $\vartheta$? If you are assuming that the angle on each side is the same, you have to justify that.
You don't need torque in this problem.

5. Jan 28, 2013

### SignaturePF

Huh? How would you do the problem then?
Ok, instead of justifying that, I'll just split it into components.
T1x = T2x
I'm not at all sure how to proceed. Could you give me a hint?

6. Jan 28, 2013

### tms

Start with a free-body diagram. Look at the leftmost mass. What are the forces on it? What are the forces on the other masses? What can you say about the tensions in the two strings?

The next step is, as you say, breaking the forces on the center mass into components.

7. Jan 28, 2013

### SignaturePF

Left-most mass:
Down: mg
Up: T0

Middle Mass:
Left: T1x
Right: T2x
Up: T1y, T2y
Down: mg

Right-most mass:
Down: mg
Up: T3

Thus we know T1x = T2x and that T1y + T2y = mg
Also T0 = T3 = mg
so T1y + T2y = T0 = T3

How could we relate this to a,b?

8. Jan 28, 2013

### tms

Up to here everything is okay, and $T_0 = T_3$ is okay, too, [strike]but the first term is a problem[/strike]. If the tension at one end of a string is $T$, what is the tension at the other end?
Trigonometry.

Last edited: Jan 28, 2013
9. Jan 28, 2013

### SignaturePF

I don't see why T1y + T2y = mg is wrong.
The two y-components of the two tensions need to add up to equal the weight of the middle sphere.

Using the trig:
T2sinθ=b
T1sinθ2 =a

??

10. Jan 28, 2013

### tms

It isn't. I don't know what I was thinking.
By now you know that all the tensions are equal, so you can drop the subscripts.

Next you should look at the horizontal components of the forces on the central mass.

11. Jan 28, 2013

### SignaturePF

We know that 2Tcosθ = mg
This implies that Tcosθ = .5mg
So θ is logically 60 degrees.
I'm still stuck. Could you just show me?

12. Jan 28, 2013

### tms

You are again assuming that the angles are equal. They are, but you have to demonstrate it first; you can't just assume it.

Which angle is θ? Between the string and the horizontal, or between the string and the vertical? You should look at the horizontal forces first; that will be a big help.
That's against the rules of PF.