Solve Balancing Force: Find the Tensions & Identify String Breakage

[Johnny Leong]

I have one question about the balancing force. Please help.

Find the tension (T1, T2) of each string. Please see the figure.

My answer:
Set T2 as pivot.
30 * 0.2 = 20 * 0.3 + 20 * 0.6 - T1 * 0.4, solve for T1

Set T1 as pivot.
30 * 0.6 - T2 * 0.4 + 20 * 0.1 = 20 * 0.2, solve for T2

If any problem with my answer, please tell.

Now, if the breaking tension of each string is 50 N, find what weight of A should be when one of the strings breaks. State which string will break first and why?
Don't know how to solve and explain. Please give me some help.

 

[HallsofIvy]

Your equations for T1 and T2 are correct.


It is not clear from your picture what "A" has intended to be.

It looks like it is intended to replace the 20 N force. If that is true, then just replace the "20N" by the variable A. Replace T1 and T2 by 50N and solve each equation for A. The smaller value for A is the "breaking" force and it is that string that breaks.

 

[Johnny Leong]

Originally posted by HallsofIvy
Replace T1 and T2 by 50N and solve each equation for A. The smaller value for A is the "breaking" force and it is that string that breaks.

Why the smaller value for A is the "breaking" force and it is that string to be broken?

 

[HallsofIvy]

Why the smaller value for A is the "breaking" force and it is that string to be broken?

? What does A MEAN?

You original post said: "If the breaking tension of each string is 50 N, find what weight of A should be when one of the strings breaks."

I suggested that you replace the tension in each string by 50 N (it's maximum possible) and solve for A in each equation. That means that each value of A is the weight that causes that string to reach its breaking tension.

If the two values for A were, say, 30 and 35 N, that means that when A= 30 N one of the strings will reach its breaking tension and break. Given that, there is no way to add more weight to get to 35 N.

 

[Johnny Leong]

I get stuck in the problem. Thank you for your explanation. I understand it now.

 

[Johnny Leong]

Sorry, I have a silly question to ask.
I have carried out the calculation. When I substitute T2 = 50 N and then calculate for A in the second equation of my answer, I get A should be 0 N. Does it mean if A has just a small weight, then T2 will break?

 

[Raiden]

Wait, I'm confused. When you put in T2=50N, does that mean how much weight is ON it, or how much weight it can hold? Because if it's currently holding 50N, then it would break if you put a small weight on.

 

[Johnny Leong]

It should be how much weight it can hold.