# Forces in equilibrium. Tension, ring on string.

1. Feb 29, 2012

### afrocod

A ring of weight 2N is threaded on to a string whose ends are fixed to two points A and B in a horizontal line. The ring is pulled aside by a horizontal force P newton parallel to AB. When the ring is in equilibrium the two sections of the string are inclined to the vertical at angles of 40° and 20°. Find the two possible values of P.

The sum of all the forces = 0 and because it says the 2 possible values of P, I'm guessing the Cosine rule is relevant.

I tried to separate out horizontal and perpendicular vectors like : X = P + T1Sin20 - T2sin40 = 0 and Y = -2 + T1Cos20 + T2Cos40 = 0. and it doesn't work out. I'm also thinking the tension in both segments = 2N

I've followed lots of different tactics out but it just never comes out right and it's draining my will to live. I'm almost certain the problem is, I'm just not able to see what are the relevant forces in the problem, I'm having a hard time making a mental model of it.

Also I'm self studying this material, so I've got no one else to turn to. Please help me out.

2. Feb 29, 2012

### tiny-tim

hi afrocod!

(try using the X2 button just above the Reply box )
Well, that should work.

(Don't forget that it's a continuous string, so the tension is the same all the way along)

If it still doesn't work, show us how far you get.

3. Mar 1, 2012

### afrocod

So I set X = Y because they're both = 0

P + T1Sin20 - T2Sin40 = -2 + T1Cos20 + T2Cos40

P = T1Cos20 - T1Sin20 + T2Cos40 + T2Sin40 - 2

P = T1(Cos20 - Sin20) + T2(Cos40 + Sin40) - 2

P = 0.6T1 + 1.41T2 - 2

But since you say then tension is the same for the whole string then T1 = T2

P = 2.01T - 2

And I'm guessing the tension is equal to the weight of the ring = 2N

So, P = 2.02

Which is not the right answer. What baffles me more is why is there 2 values for P?

4. Mar 1, 2012

### tiny-tim

sorry, afrocod, but this strategy is crazy :yuck: …

you had two equations, with two unknowns (P and T)

you've just reduced it to one equation, with two unknowns!

that's not solvable!!

solve Y = 0 first (for T), then use that solution in X = 0 …

what do you get?

5. Mar 1, 2012

### afrocod

It all seems like perfect logic till someone calls you crazy.

Y = 0.94T + 0.64T = 2

T = 1.27

X = P + (1.27)Sin20 - (1.27) Sin40 = 0

P = 0.39N

Which is close enough to one of the answers (0.35N), if I took a few more decimal places I probably would have got that.

But now I'm still left with the question what is the other possible value for P, and why does it even have two values?

6. Mar 1, 2012

### tiny-tim

hmm

what makes you think that the two strings have to be sloping in opposite directions?

7. Mar 1, 2012

### afrocod

All I can think of is that when P = 0 but the other answer is actually 1.15N

8. Mar 1, 2012

### tiny-tim

you're reading too much into the question …

how many different ways are there of drawing this?

9. Mar 1, 2012

### afrocod

Oh, I see... facepalm... Sloping the same direction.

Thank you for all your patient help Tim.

10. Oct 31, 2012

### sareba

I am having problems with the same question can you explain a bit more on how you reach the answer?

11. Oct 31, 2012

### tiny-tim

welcome to pf!

hi sareba! welcome to pf!

show us what you've tried, and where you're stuck, and then we'll know how to help!

12. Oct 31, 2012

### sareba

I came up with the exact same strategy i.e. resolving the firces into components but you guys lost me on afrocod's third post where he seems to be using T cos 20 + T cos 40 = 2 but cos 40 = .766 not .64 which is cos 50

EDIT: Oh i put in cos 40 and it gave .35 the correct answer... Thanks a lot tim for the quick reply!!

But still how do i get to the other answer? I mean i understand what you mean by same slopes. We have toadd the forces instead of subtracting them but that does not give the right answer...

Last edited: Oct 31, 2012
13. Oct 31, 2012

hi sareba!
it should