Force in wire rope pulled by opposing hydraulic cylinders

[Bill O'Donnell]

A wire rope is stretched horizontally between two opposing hydraulic cylinders. One cylinder pulls to the left with a force of 100 pounds, the opposing cylinder pulls to the right with a force of 150 pounds. What is the force imparted to the wire rope?

Please show me how this is calculated.

Thank you,

Bill O'Donnell

 

[Mech_Engineer]

A wire rope is stretched horizontally between two opposing hydraulic cylinders. One cylinder pulls to the left with a force of 100 pounds, the opposing cylinder pulls to the right with a force of 150 pounds. What is the force imparted to the wire rope?

Please show me how this is calculated.

I have a sneaking suspicion you're making this problem much harder than it needs to be...

 

[Bill O'Donnell]

I'm just trying to find out how to do the calculation

No, Sir, I'm not making the problem harder than it needs to be. If you know how to calculate the tension in the cable, I'll be grateful if you will show me how to do it.

 

[russ_watters]

Is this a homework problem?

Since a rope can only have one tension on it, how can the two cylinders pull on it with different forces...?

 

[Mech_Engineer]

Is this a homework problem?

Since a rope can only have one tension on it, how can the two cylinders pull on it with different forces...?

I wondered about that as well, it's not in equilibrium.

 

[Bill O'Donnell]

Hydraulic cylinder forces

Let's presume the annular piston areas of the two cylinders are equal to each other (say, 1 square inch). Let's presume, as well, that each cylinder is powered by an independent pump (and prime mover). The pressure relief valve of one pump is set at 100 psi and the pressure relief valve of the other pump is set at 150 psi. The force output of a hydraulic cylinder is calculated as: pressure x area. So, one cylinder is pulling with a force of 100 pounds (1 sq.in. x 100 psi) and the force of the other cylinder (pulling in the opposite direction) is 150 pounds (1 sq.in. x 150 psi). Anyway, that explains the question of how the cylinders can be pulling with unequal forces.

Really, if someone can tell me how to calculate the tension in a cable that is being pulled from opposite ends (as opposed to being anchored at one end and pulled from the other) I really don't care about the numbers. It's the method I'm trying to find out.

Thanks for any help.

 

[FredGarvin]

Bill,
It is well understood how a hydraulic cylinder produces a force. The issue here is one of basic statics. You can not have an unbalance in forces and not have the rope stay stationary. You have a net force of 50 Lbf in one direction. That will create an acceleration. If this is not an issue, that the rope indeed is intending to be moving, then, neglecting dynamic effects, the max tension will be what the lowest reaction force will be, which is 100 Lbf.

 

[Bill O'Donnell]

Hydraulic force

Sir:

I was asked how two hydraulic cylinders can pull a cable in opposite directions with unequal forces, so I answered the question. I presumed, since the questioner asked the question, that he did not know the answer. It was not my intention to imply that others don't know how the force of a hydraulic cylinder is calculated.

Also, nothing in my originally posted question implied that the wire rope (and the two cylinder pistons) are not moving. It stands to reason, it seems to me, that they have to be moving, but that fact does not tell me how to calculate the tension in the wire rope.

I think you may have answered my original question, although I'd still like to see a mathematical representation of the answer. Sorry for my ignorance, but my "education" (such as it is) is nearly all quite informal.

Again, any help with my original question (especially a mathematical formula) will be appreciated.

 

[AlephZero]

You don't say if this is a real situation or a theoretical question, but if it's real situation, most likely one of the hydraulic cylinders is at the end of its travel and the tension in the rope will be either 100 or 150, depending which cylinder hit the stop first. The tension will correspond to the pull from the cylinder that's not on its end stop.

If it's a theoretical question, bear in mind that you changed the question part way through the post. Originally you asked "what is the force imparted to (the ends of) the wire rope". The answer to that is straightforward, the forces are equal and opposite to the cylinder forces, 100lb at one end and 150lb at the other end.

Then you changed the question to "what is the tension IN the (complete length of) cable". That's a harder question, and since the system probably isn't in equlibrium, we don't have enough data to answer it. I say it "probably" isn't in equilbrium, because if the rope was hanging vertically and weighed 50lb, with a pull of 150lb at the top and 100lb at the bottom, life would be simple again...

But that's just speculating of course.

 

[jasc15]

The cylinders cannot push without a reaction (by saying that they have different forces initially, it is implied that a reaction force is not needed). since there is only one thing in this system to create the reaction, the rope, and since it can only have one tension, the cylinders cannot push with different forces.

 

[Bill O'Donnell]

Thanks for the responses

Gentlemen:

Although I still don't have a mathematical formula to use for calculating the tension in a cable, I have figured out the flaw in my thinking. At least I know the forces now.

Thanks, again.

 

[stewartcs]

Bill,

The formula you are looking for can be found from a simple free-body diagram and applying Newton's Second Law...F=Ma. Remembering that F must be the net force, we can write the sum of the tensions in the wire rope as T2+T1. However, since they are pulling in opposite directions, this becomes T2-T1. Now, set this equal to Ma and we get,

T2-T1 = Ma

where,

T2 = tension from cylinder pulling from the right
T1 = tension from cylinder pulling from the left
M = mass of the string
a = acceleration of the string

(Refer to the attached drawing)

For an accelerating string the tension is not the same throughout the string. The front end of the string (T2) has a higher tension than the back end (in this drawing).

Mathematically stated as,

T2 = Ma + T1

This seems to be the case for your system. So the max tension in the wire is T2 given from the equation above.

In general the mass of the wire is much smaller than the mass of the other objects in a problem. Hence, the mass of the wire can be ignored. We often use a "massless" wire (or string) in real life for this reason. For a "massless" wire the tension is always the same in magnitude at both ends.

The way you have stated the problem it appears that the wire is accelerating (since there is a net force of 50 lbs in one direction), thus the tension will vary throughout the wire as previously stated. If it is not accelerating, then the tension in the wire is the lesser of T1 or T2 (the reactive force, or 100 lbs in this case).

CS

 

[Bill O'Donnell]

Thank you for the answer

Stewartcs:

Thank you for giving me the explanation I was looking for.

Bill O'Donnell