What are the tensions in the strings?

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AB is a uniform bar of length 5m and weighs 60N. It is supported in a horizontal position by two vertical strings P and Q. P is 0.5m from A and Q is 1.5m from B.

(a) What are the tensions in the strings? (Ans= 20N and 40N)

(b) What weight must be applied to A if the tensions in the strings are to be equal? (Ans= 15N)

(c) What is the biggest mass which can be hung from B if the bar is to remain horizontal? (Ans= 4.1kg)

My teacher gave us the answers to these questions. I have gotten part (a), but I truly am lost with (b) and (c). For some reason I keep getting (b) to be 0N... I would really appreciate any help and explanations. Thanks.
 

Doc Al

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Show us how you trying to solve (b) and we can see what's going on.
 
19
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Ok... I'm just working based on the only example my teacher gave us, which is as far from this problem as possible, but here it goes. This time I got 58N, uh.

F*5 - (60*2.5) - (40*3.5) = 0
5F - 150 -140 = 0
5F = 140+150
5F = 290
F = 58
 
19
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Well, now I was just trying something and got 15N but I have no idea if this is the way to do it. I considered the tension to be 30N.

5*F = 30 * 2.5
5F = 75
F = 15
 

radou

Homework Helper
3,085
6
fickle said:
Well, now I was just trying something and got 15N but I have no idea if this is the way to do it. I considered the tension to be 30N.

5*F = 30 * 2.5
5F = 75
F = 15
Regarding (b). The first equation is the sum of moments around B. This will give you an equation with two unknowns - F and P, where P is the tension in the strings. The second equation is the equation of equilibrium of all vertical forces. This equation will give another equation with two unknowns - F and P. Now all you have to do is solve this system of 2 equations with 2 unknowns, i.e. elliminate P from the first equation to obtain F. I got the right answer, so this should work.
 

Doc Al

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To solve (b), start by identifying all the forces acting on the bar (including the unknown weight, F). Hint: I count four forces.

Then you will apply the conditions for equilibrium:
(1) Sum of the forces must equal zero.
(2) Sum of the torques (about any point) must equal zero.
 
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Um... I tried what I thought you were telling me to do, and I got 30N.

120 -3P + 0.5F
60 -2P + F

240 -6P + F
180 -6P + 3F

60 = 2F
F = 30N

What went wrong?
 

Doc Al

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fickle said:
120 -3P + 0.5F
I don't understand where this comes from. I assume P = tension; F = added weight?

60 -2P + F
I assume you mean:
60 -2P + F = 0

That's good.

Now find the torques about point B to get a second equation.
 
19
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Wow I got it after all... thanks! I'll try (c) now... can I still post in this thread for help on that one?
 

Doc Al

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fickle said:
can I still post in this thread for help on that one?
Of course!
 
19
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Do I take a similar approach as I did with (b)? And do I treat the tension in the strings as equal forces?
 

Doc Al

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fickle said:
Do I take a similar approach as I did with (b)?
You will use the same general principles that I outlined in post #6.
And do I treat the tension in the strings as equal forces?
Absolutely not! That's part of the trick with this question--how will you know if the bar remains level? Try to picture what's going on as you load extra weight on the bar.
 

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