# Moment of force and tension

1. Aug 5, 2005

### marooned

hello,
i have a problem i have having trouble solving, it reads as follows:

a vertical pole is supported by two ropes, each 5m from the base of the pole. The rope to the right of the pole extends 3m up the pole, and the rope to the left of the pole extends 4.5m up the pole. The ropes and pole lie in the same vertical plane. The mast cannot provide any moment of force about its base. What is the ratio of the tension of the two supporting ropes, where the tension is marked as the hypotenuse of the two triangles formed on each side of the pole?

I realise that for the pole to remain in equilibrium, the forces acting on it must have a vector sum of zero, and the sum of the moments of force must also equal zero. I have found the forces acting on the pole at the point of each ropes attachment, and are 10N for the lower rope on the right, and 6.67N for the higher rope on the left. Is the ratio between these forces the same as the ratio between the tension on the ropes?

Any help would be greatly appreciated, thanks.

Last edited: Aug 5, 2005
2. Aug 5, 2005

### Brad Barker

i'd try and help but... i have no idea what that's supposed to look like.

3. Aug 6, 2005

### sniffer

i made a lazy sketch here. maybe this is what you meant?

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4. Aug 6, 2005

### marooned

Sorry, here is the diagram... my description skills arent the best..

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5. Aug 6, 2005

### marooned

I have been working on this problem to no avail... can anyone help? So far i have tried to find the horizontal component of the forces using the formula F1+F2=0 (since the pole is in equilibrium, the sum of the vector forces will be zero, as will the sum of the vector moments of force) so also: rxF1+rxF2=0. Since the distance from the axis of rotation to the force is just the distance up the pole i got: 4.5F1+4.5F2=0, but have no idea what to do with this equation. Any one help?

6. Aug 6, 2005

### VietDao29

I dunno why you need to use moment to solve this problem.
Let $$\alpha$$ be an angle between the left rope and the pole and let $$\beta$$ be an angle between the right rope and the pole.
You will have:
$$T_{1} \cos \alpha = T_2 \cos \beta$$
$$\Leftrightarrow \frac{T_{1}}{T_{2}} = \frac{\cos \beta}{\cos \alpha}$$
Viet Dao,

Last edited: Aug 7, 2005
7. Aug 7, 2005

### marooned

Thank you thank you thank you. I can´t believe i didnt see that before, setting both horizontal components to eachother, because they have to add to zero? I didn´t even think of disregarding the horizontal components because they are the same distance from the pole, right? That leaves only the vertical components to deal with, thus the cosine relationship. Thankyou very much Viet Dao,

Regards,
marooned

8. Aug 7, 2005

### VietDao29

Delete (This is wrong...)
Viet Dao,

Last edited: Aug 7, 2005
9. Aug 7, 2005

### marooned

So it is the same equation as posed before, but with the sin relationship instead of the cosine relationship? That way the horizontal forces are resolved? But i dont see how the forces can equal each other, if they are at different heights on the pole, one force shall be bigger than the other force, will setting them equal to eachother give the correct ratio?

10. Aug 7, 2005

### marooned

I see now, thanks it is much clearer, I should be able to get it out now. I see the different angles mean the forces are not equal, so setting them to equal eachother makes sense. (However clumsy that sounded I do understand).

Thanks again Viet Dao,
marooned

11. Aug 7, 2005

### VietDao29

I am very very sorry, I forget the friction force. My posts above are wrong. Sorry...
Let's say the pole base is O. The moment of 2 tensions relative to the point O must be the same to keep the pole from falling. The P, N, and friction force go through O therefore create no moment.
Call r1 the distance between O and T1, r2 the distance between O and T2.
You have:
$$T_1r_1 = T_2r_2$$
Call the length of 2 rope are l1, and l2. You will have:
$$r1 = \frac{5 \times 4.5}{l_1}$$
$$r2 = \frac{5 \times 3}{l_2}$$
So you will have:
$$\frac{T_1}{T_2} = \frac{r_2}{r_1} = \frac{3 l_1}{4.5 l_2} = \frac{\cos \beta}{\cos \alpha}$$
I am sorry if I did make you confused...
--------------------
P.S : I felt sooo ashamed of myself . Please read this post, the posts above are sooo wrong... I think I'll get some sleep now. I forgot to consider the friction force. The friction force also has the horizontal component. Without the friction force, this pole can never stand. Sorry again. Hope you are not mad at me...
Viet Dao,

Last edited: Aug 7, 2005
12. Aug 7, 2005

### marooned

I am just curious as to what the moment arm is; what is r1 and r2? Are they the distances from the ground where the ropes meet to their point of joining on the pole? No i am not mad at you, and thankyou very much for your help Viet Dao. I am also curious where you get the equation $$r1 = \frac{5 \times 4.5}{l_1}$$ from?

13. Aug 7, 2005

### VietDao29

It's the distance between the point O (the pole base), and the 2 ropes.
In the right triangle ABC, angle BAC = 90, AH is its altitude, you will have:
$$2S_{ABC} = AB \times AC = BC \times AH$$
So $$AH = \frac{AB \times AC}{BC}$$
Viet Dao,

Last edited: Aug 7, 2005
14. Aug 7, 2005

### marooned

Are you suggesting the diagram as shown below? If so the relationship stated does not work, i am having trouble understanding where your H point is located, sorry to cause you so much work, i have trouble interpreting things sometimes...
Thanks in advance,
marooned
C H
|\
| \
|_ \B
A

Last edited: Aug 7, 2005
15. Aug 7, 2005

### VietDao29

AH is the line that's perpendicular to BC, and $H \in BC$
Hope you get it.
Viet Dao,

16. Aug 8, 2005

### marooned

Thats very much Viet Dao, i got it out with no problems after your help,
Thanks again,
marooned

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