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

[SignaturePF]

Homework Statement

See:
http://www.aapt.org/physicsteam/2010/upload/2010_FmaSolutions.pdf
Question 11

Homework Equations

T_net = 0
F_net = 0

The Attempt at a Solution

Start with F_net = 0:
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.

 

[tms]

There is no tension in question 12 at that link.

 

[SignaturePF]

Sorry I meant question 11.

 

[tms]

Let T1 be the force of tension on the left, T2 on the right
T1cosθ=T2cosθ

What is \vartheta? If you are assuming that the angle on each side is the same, you have to justify that.

I'm really confused about how to find the torque:

You don't need torque in this problem.

 

[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?

 

[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.

 

[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?

 

[tms]

so T1y + T2y = T0 = T3

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?

How could we relate this to a,b?

Trigonometry.

 

[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

??

 

[tms]

I don't see why T1y + T2y = mg is wrong.

It isn't. I don't know what I was thinking.

Using the trig:
T2sinθ=b
T1sinθ2 =a

??

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.

 

[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?

 

[tms]

We know that 2Tcosθ = mg

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.

Could you just show me?

That's against the rules of PF.