# Forces along a rope vs. along a stiff bar

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1. Nov 24, 2014

### richard9678

Hi. I'm trying to figure out what the forces will be in two differing circumstances.

ROPE
Let's say at one end a person is pulling a rope to the left with 120N force. At the other end a person pulling with 120N force. I know the tensile force inside the rope is 120N at any point along the rope.

Okay, well, let's say one third distance from the right end a third person begins to pull right, in line, gradually taking up the strain until he is pulling 120N.

I'm trying to figure out what will happen to the tensile force in the middle of the rope and between the person on the right and the third person.

I'm not quite sure whether the force at the middle remains at 120N and all that happens is the strain in the right third drops to zero as the third person takes up more strain, or something else happens.

BAR
Same scenario, but a solid bar.

2. Nov 24, 2014

### richard9678

ROPE: If the two end people were simply leaning to produce the force and they never altered their position in any way, I would say as the third person begins to pull, he begins to take the strain, and it gets lighter for the guy on the right. So, the tensile force in the middle stays at 120N. Just not 100% sure about that.

If the person on the right continues to lean, then I guess the forces on the right begin to add, and we see the tension in the middle of the rope begin to rise, if the person on the left can handle it.

Last edited: Nov 24, 2014
3. Nov 24, 2014

### SteamKing

Staff Emeritus
The rope can support only tensile loads. When the third person starts pulling on the rope from the sides, the tensions in the rope must change to keep the rope in equilibrium.

A metal bar can support tensile and compressive loads, bending moments and torsional moments.

4. Nov 24, 2014

### richard9678

My case is like taking a rope, affixing it at one point, pulling on it at the other end, putting it under tension, then clamping that end. Then going down a third way from the right, and pulling to right in line with the rope. As far as I can tell, the tension in the third part, the part that goes to the clamp point on the right, will reduce as the tension force exerted by the person a third way down increases. So, the force on the left fixing point is always 120N, left force on the right fixing point varies from 120N to zero. So, if the person pulls at 60N, we have: Left fixing point 120N, right fixing point 60N.

5. Nov 24, 2014

### SteamKing

Staff Emeritus
You may think that's how the rope reacts, but statics can tell a different story. Once you have a rope under tension and then you start pulling sideways, the original tension values in the rope are going to change. The sideways components of the tension in the rope must counteract the force of the sideways pull. The tension in rope acting on both sides of the pull point must be equal and opposite in order to maintain equilibrium.

6. Nov 24, 2014

### richard9678

In this case there are no sidesways pulls. All pulls are in line with the rope.

It's like placing a rope one third from the right in parallel with the original rope. As you tighten up that parallel rope you could get it taking 60N, When that condition is reached there is a total of 120N tension, shared at the right end by two ropes taking 60N each.

I'm thinking that is what will happen.

Last edited: Nov 24, 2014
7. Nov 24, 2014

### richard9678

You fix a rope at one end to a solid foundation, put the rope in tension at 120N and then affix the other end in a solid foundation.

There a 5 points on the rope:

*-----A-----B-----C-----D-----E-----*

At the start all points read 120N.

You pull with another rope at B to the right in line 60N.

A still shows 120N, To the right of B original rope all points show 60N. Because there is a rope in parallel attached at B which is taking the other 60N.

You then attach another rope at D, pulling left, in line.

What is the effect of doing that to the tension at these points?

I think A is 60N, B 0N and E 60N

But, does that compute, given that the rope was originally tensioned at 120N?

8. Nov 24, 2014

### Staff: Mentor

Why don't you just set up a calculation to remove all the guesswork. You have a metal bar under tension, with an initial tensile strain ε. The ends of the bar are fixed. The length of the strained bar is L. The area of the bar is A. The Young's modulus is E. What is the tension in the bar? You apply a displacement of δ at the location x along the bar. What is the new strain in the region from 0 to x? What is the new strain in the region from x to L? What is the tension in the region from 0 to x (including the effect of the initial strain)? What is the tension in the region from x to L (including the effect of the initial strain)? What is the change in tension from x- to x+? What force is being applied at x along the bar to bring about the displacement δ?

Chet

9. Nov 24, 2014

### richard9678

Here is the situation regarding the rope. First diagram shows the rope under tension at 120N. Fixed both ends. Below it the diagram shows a force at B applied in line, of 60N to the right. Notice I've said all that does is reduce the tension at points C D and E, but tension at A stays the same.

Last diagram shows another force applied in line at D, this time pulling in the left direction with force 60N. I show tension at A 60N, at C 0N and at E 60N.

I'm probably wrong about the tensions (I think). But, I don't know what to Google for to get the calculations. I'm not dealing with weights hanging off ropes, horizontally tensioned ropes with weights hanging off them, or block and tackle calculations.

https://flic.kr/p/pU9J4r [Broken]

Last edited by a moderator: May 7, 2017
10. Nov 24, 2014

### Staff: Mentor

I guess you didn't like my suggestion.

I can tell you that the tension changes in both sections of the rope.

Chet

11. Nov 24, 2014

### richard9678

I just thought I'd try to understand the rope situation first, thinking it simpler. Because a rope can only pull, it cannot be in compression.

The forces at B and D in the diagram can be imagined as being caused by weights dangling on ropes. But all tension forces are in line with the main rope affixed at both ends.

I'll come back tomorrow. it's 1:45am here. :)

12. Nov 24, 2014

### Staff: Mentor

If the rope is initially under tension, then its behavior is the same as a bar.

If the force of 60 N is applied at the center of the rope, then one section of the rope goes up to 150 N, and the other section of the rope drops to 90 N. If the force is applied at other locations, the split is different.

Chet

13. Nov 25, 2014

### richard9678

I actually got into this considering truss forces. I know, we have method of joints. That tells you the (approximate) horizontal forces applied to the horizontal members by the off-vertical members. However, I got to thinking that the entire horizontal top part and bottom part of a truss, are like bars or rods experiencing horizontal forces at the ends and along their lengths.

Yes, for trusses we might do method of joints to find forces,, but I was considering a horizontal bar entirely on it's own.

14. Nov 25, 2014

### Staff: Mentor

OK. So, getting back to my previous post, are you comfortable with it or not?

I can also provide the detailed analysis I was asking you to do in post #8 if you're interested. That would cover all the cases you have been asking about.

Chet

15. Nov 26, 2014

### richard9678

I'm learning this as I go.

The bar is under direct stress and the bar is in tension. If the force responsible for the tension was 120KN, and the area of the bar was 100mm2 the stress would be 1.2 GPa. This is before any other forces are applied.

16. Nov 26, 2014

### Staff: Mentor

OK. One end of the bar is at x = 0, and the other end is at x = L, and both ends of the bar are fixed. At an arbitrary location x along the bar, we apply a force F, in the direction going from x = 0 to x = L. The application of this force causes a displacement at the location where the force is applied equal to δ. As a result of all this, the strain in the bar in the region between x = 0 and location x increases uniformly by δ/x. In addition, the strain in the bar in the region between location x and x = L decreases uniformly by δ/(L-x). So the tension within the bar between x = 0 and location x increases by EAδ/x, and the tension in the region between x and x = L decreases by EAδ/(L-x), where E is the Young's modulus and A is the cross sectional area. So, the change in tension across the cross section at x is given by:
$$ΔT=EA\left(\frac{δ}{x}+\frac{δ}{L-x}\right)=\frac{EAδL}{x(L-x)}$$
From a force balance on the cross section, we have:
$$F=ΔT=\frac{EAδL}{x(L-x)}$$
So, from this:
$$EAδ=\frac{Fx(L-x)}{L}$$
From this, it follows that the tension increase in the region between x = 0 and x is $F\frac{(L-x)}{L}$ and the tension decrease in the region between x and x = L is $F\frac{x}{L}$.

Chet

17. Dec 3, 2014

### richard9678

Hi. Thanks. I will get back to this when I'm finished with something.