# Calculating Tension in Cable

1. Oct 3, 2005

### DKTKD

Hello,

I'm having trouble trying to calculate the tension in a cable
when dealing with constant velocity and no acccel.

For example.

A block moving with constant velocity tied to a wire that is
connected to a fixed wall.

When the slack in the wire ends the blocks comes to
an immediate stop.

What is the tensions in the wire?

The only way I think i can answer this is if i make many other
time, distance, etc.. assumptions?

What do you think?

Thanks,
DK

2. Oct 3, 2005

### Fermat

In effect, you'll be getting an impulse applied to the block.

When the slack in the wire ends, the wire will be extended slightly.
This will do work on the wire.
And of course you willl do (the same amount of) work in stopping the block.

3. Oct 3, 2005

### bartieshaw

consider the acceleration required to stop the block moving from its initial velocity, from this you can claculate the force applied to it and hence the tension in the wire

4. Oct 3, 2005

### HallsofIvy

Staff Emeritus
You are not given enough information to determine the acceleration or force. You would need to know the time or distance in which the block decelerated from its initial speed to rest.

5. Oct 3, 2005

### El Hombre Invisible

Hmmm. If you knew the time the impulse acted for, then: <F> = p/t.

If you knew the spring constant k and the extension (delta)x, then I think it's just F = T = k(delta)x.

But we don't, so my guess is...

The impulse supplied by the wire to the block will be equal and opposite to the original momentum p of the block. Force is the derivative of momentum, so is F = d(delta)p/dx useable here? I can't think of any other way of doing it.

EDIT: aaaarrrggghhh! dx... dammit! Surely there's not enough info here!

6. Oct 3, 2005

### DKTKD

Hello All,

As El Hombre Invisible has stated, "aaaarrrggghhh! dx... dammit! Surely there's not enough info here!"
When I was working on this problem. I felt the same way.

For example for calculating the impulse force/forces you need the
time or distance in which the block decelerated from its initial speed to rest
as HallsofIvy mentioned.

How do you find this if you are starting from scratch. See my post on "Using physics to solve stunt challenge"

If you are assuming Worst case forces, the time to stop would be almost instance and the distance to stop would be almost zero.

Using these deltas would give you basicly infinite force.

Now you can make assumptions such as 1ms or 1ns. But where's the accuracy in that?

Also there is the approach of determining the wire extension (delta)x, to assist in determining distance to use F = d(delta)p/dx. However, you need the force acting on the cable to determine that?
Not to mentions that you are trying to resolve the forces so that you can determine what material and length of the wire that is required as not fail due to the forces.

I looked through all my physics books and online references. And this type of problem is never presented. Stopping distances and time and other that are necessary to solve the tension in the wire are usually provided but not explained as to where they originated.

I felt as if I had to make too many assumptions. Especially in regards to determining
tension on the wire.

What do you think?

7. Oct 3, 2005

### El Hombre Invisible

Is this an actual question you've been set? Could it be that you are supposed to state the tension in the wire as a function of some other variable? The problem here is that you state the block as coming to rest instantly. This just isn't feasible. There has to be some acceleration and this must take some time, however small, to reduce the velocity to zero.

8. Oct 3, 2005

### DKTKD

Yes, El Hombre Invisible.
My questions is based on my trying to solve a larger problem set to me.
I posted it in the General Physics section title Using physics to solve stunt challenge"

While trying to solve it, I felt i was also posed with the cart before the horse.

Like you mentioned There has to be some acceleration and this must take some time"

I agree. But the only way i thought I could solve the force would be to find out the
elongation in the wire. The elongation would give me the time and distance of decceleration. Knowing the time and distance should help me find the force, but I cannot resolve the elongation without knowing the force exerted on the wire.

I'll try it again and set it up with solving simultaneous equations.

Thanks.. let me know it you have anyother thoughts.

9. Oct 4, 2005

### Fermat

I read your post, DKDT, in the other forum about your stunt problem and got a better idea of what you're trying to do.

Is this right?

You have a 4000 lb truck moving at 40 mph and you want to see if it will stop within 5 ft by the stretching of two steel cables, where both cables are 5/8" diameter and 1000 ft long.

If the above is correct, then you can equate the work done in stretching the two cables to the work done in overcoming the kinetic energy of the truck (by bringing it to a complete rest).

You have steel cables, 5/8" diameter, and 1000 ft long.
You were given Young's modulus for the steel cables.

$$A = \pi \left(\frac{5}{8}\right)^2/4 = 197.93\ mm^2$$
$$A = 197.93*10^{-6}\ m^2$$
$$l = 1000\ ft = 304.8\ m$$
$$E = 210\ GPa = 210*10^9\ N/m2$$

=========================================

$$E = \frac{\sigma}{\epsilon}$$

$$\sigma = \frac{T}{A}$$
$$\epsilon = \frac{e}{l}$$

giving,

$$T = \left(\frac{AE}{l}\right).e$$
or,
$$T = k.e \mbox{ where } k = \left(\frac{AE}{l}\right)$$

Substituting in values for A,E,l we get the spring constant for steel as,

$$k = \frac{197.93*10^{-6}*210*10^9}{304.8}$$
$$k = 136.37\ kN/m$$
==============

$$M = 4000\ lb = 1814.4\ kg$$
$$V = 40\ mph = 17.882\ m/s$$

To overcome the kinetic energy in the car by bringing it to rest,

$$KE = \frac{1}{2}MV^2$$
$$KE = \frac{1}{2}*1814.4*(17.882)^2$$
$$KE = 290,091.65\ Nm$$
=======================

Work done in stretching both steel cables,

$$WD = 2*\frac{1}{2}k.e^2$$
$$WD = k.e^2$$

Equating Work done in stretching the cables to loss of energy in stopping the truck,

$$WD = KE$$
$$k.e^2 = \frac{1}{2}MV^2$$
$$136.37*10^3*e^2 = 290,091.65$$
$$e^2 = 2.12724$$
$$e = 1.458\ m$$
=============

So, it looks like the truck is going to stop within 1.5m and 5 ft = 1.524 m
So, you're just going to make it!