# Drawbridge cable tension

(not a physic person so hope my description is clear enough)

bridge mass = 3150 newtons (center of mass is in center)
length = 5 meters
cable is attached to wall 3 meters straight up from bridge hinge
cable is attached to bridge 3.5 meters from hinge

situation: bridge is suspended 14 degrees below horizontal

Question: what is the calculation I use to determine the tension on the cable?
(I'm physically going to build this and the numbers may change)

Thanks.

## Answers and Replies

could you make a diagram an post it here.....

drawing attached

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sophiecentaur
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This looks awfully like a homework / School question. It should. perhaps, be in the other (appropriate) forum.
These problems are easy by using Moments. The only thing is that you need to be able to identify the 'perpendicular distance' correctly. That often stumps students. Then take moments about the 'hinge'.

graduated HS in 1980. Building a ramp for a semi trailer for my daughter's band. Do not know anything about physics except that they apply and I am concerned that the cable&winch is strong enough to support the load.

sophiecentaur
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OK you non-schoolboy. You have my full attention.
I did some sums on your setup but, because that triangle is very nearly right angled and the two distances (3m and 3.5m) are much the same, the answer comes out that the tension in the cables (shared between them) is more or less exactly equal to the weight of the ramp.

I could display my workings out but they are on the back of the proverbial fag packet. It boils down to the fact that the distance from the hinge to the c.m. of the ramp is almost exactly equal to the (shortest) distance from the cable to the hinge. This means that the weight force and tension are more or less equal. The numbers would need to change quite drastically for the requirements to change much. The main thing is to ensure that the top pulleys are not taken any lower. Higher, and the tension is less - the limit being half the weight when the pulleys are very high or vertically above the ends of the ramp.

If you choose a winch that can handle at least twice the weight of the ramp and similar cable (not forgetting suitable pulleys and fixings for all the pulleys), you should be OK. I am assuming that there will be no other loading on the ramp when it's being raised and lowered. Go for something that is on the chunky side and that will be able to cope with wear and extra friction.

Have you considered how you will wind the cables both sides at the same time?

haruspex
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Building a ramp for a semi trailer for my daughter's band. Do not know anything about physics except that they apply and I am concerned that the cable&winch is strong enough to support the load.
I assume (hope) that the cable only needs to support the flap, no load on the flap. (If there is, it won't always be in the centre.) But if the weight is 3150 N, that means the mass is over 300kg, right? Seems kind of heavy.

My next concern is the position when the flap is drawn up. Since you have 3m v. 3.5m, it won't reach vertical. In fact, if the "wall" leans out in the slightest you could get very high tension in trying to reach vertical.

W = weight of flap
L = distance to mass centre of flap
A = height of cable top from hinge
B = distance from hinge to anchor on flap
α = angle of fully open flap to vertical (104° in your diagram):
C = length of cable at full extension = √(A^2 + B^2 - 2.A.B.cos(α))
Tension in each cable, T = W.L.C/(2.A.B)

With the figures you give, T =1924N.

(Sophiecentaur, following the approximations you made, I think you miscalculated by a factor of √2. The tension in each of 2 cables would be W*√2/2, or 2227N.)

sophiecentaur
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Good point about the position of the pulleys at the top. It would obviously be better to have equal distances from the hinge. This could be done by having the cable fixings brought in by 0.5m on the ramp or by moving the pulleys higher.

As for my geometry, I used both cos rule and sin rule to find the perpendicular distance and it came out ajmost exactly 1.75m. This is because the triangle is not 45 degree equilateral. The root2 factor is too approximate. If the maths is hard then a scale drawing would work as an accurate check.

haruspex
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As for my geometry, I used both cos rule and sin rule to find the perpendicular distance and it came out ajmost exactly 1.75m. This is because the triangle is not 45 degree equilateral. The root2 factor is too approximate. If the maths is hard then a scale drawing would work as an accurate check.
Strange. The angle at the hinge is 1.82 radians. By the cosine rule I get 5.13 as the length of the cable. That makes the perpendicular distance from hinge to cable 1.99, but you don't need to work that out because of cancellation:

If β is the angle of the cable to the upright,

W.L.sin(α) = 2.A.T.sin(β)
By the sine rule:
B/sin(β) = C/sin(α)
Thus T = W.L.C/(2.A.B)

sophiecentaur
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1.99 is what I got and that' s pretty near 1.75 (half 3.5). That means the tension must be almost the same as the weight??

haruspex
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1.99 is what I got and that' s pretty near 1.75 (half 3.5). That means the tension must be almost the same as the weight??
OK, I see what you've done. You've taken the 3.5 as the end of the flap. It's only the point at which the cable attaches. The flap as a whole is 5m, making the mass centre nearly 2.5m from the hinge.

sophiecentaur
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Oh boy, what a div. Yes, you've hit the nail on the head there. At least I can say that, if I were to be actually designing something like that for real, I'd have done the sums more than once!

http://puu.sh/pF94 [Broken]

hope it explains but the problem is that the tension is more than the weight of the rod

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haruspex
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hope it explains but the problem is that the tension is more than the weight of the rod
But there are two cables sharing the load, so you have to halve, giving 1924N - the same value I got.

Awesome information and thanks for all input. Here is a really bad drawing of how I intend to do the cabling. For the pulleys, I intend to use roller fairleads (unsed on winches). (still need to find a good deal on 2 I-beams for the ceiling. Ramp weight has gotten a little lighter (making it narrower) and I also have room to use extension springs if needed to support 'some' of the weight. Sufficient hinge strength is also of a concern (in my mind when the ramp is stowed upright most of the weight will not be on the hinge. Hope to get approval by the Booster board () to get started on this soon.

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I'm looking at using extension springs for safety and support so I tried to plot out the cable tension as the ramp lowers. The tension seems to increase evenly as it lowers and that is not what I thought would happen. Am I missing something? (see attached) Thanks.

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haruspex
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Well, that's what the formula I gave you predicts.

T = W.L.C/(2.A.B)

where A and B are fixed lengths (2.9m now), L is a fixed length (half flap length) and C is the cable extension. (Is that what you used?)
Imagine allowing the flap to hang almost vertically downwards. In that position, the cable is supporting the load W directly, plus it is working against the hinge, which is pushing downwards on the flap. So it's not surprising that's where the maximum tension is.
What curve would you expect?

Found a better calculation that looks better graphed out.
Next Problem: I know using a single pulley on an extension spring divides the weight it will hold. But, will 2 pulleys (one attached to the spring) still only divide by 2 since I am still attached only to the beam and the load? (diagram attached)

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haruspex
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Not sure how this relates to the original question, but looking at the diagram:

If the load is W, the tension in the string is W, so the downward force on the spring and (thence) the upper beam is 3W, and the upward force on the lower beam is 2W.
(Just count the strings at each attachment.)

update: finally finished the project. Mechanical actions did not play out exactly as I wanted but the results are working well. Thanks for all the help so that I could assure the people I made this for that the cable was well under it's rated working tension.

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