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New Friction Problem

by Nquisitive
Tags: friction
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Nquisitive
#1
May2-12, 02:35 PM
P: 3
I want to know how much force I am exerting on the top of a post with a chain hooked up to my truck. 2 Conditions:

1. With a running start of 15 feet, some impulse action happening here, (Assume chain is unbreakable)
2. Just at a steady pull with no running start, up until the wheels slip from overcoming friction.

Given:

- A Certain truck's specs; i.e. torque at a certain rpm, mass, 0-60 specs, etc.
- Certain friction coefficient

Did I miss anything?
It's been a while since ENGR school.
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Travis_King
#2
May2-12, 03:58 PM
P: 820
If you know the torque at the wheels at any given rpm / speed, and you have a reliable coefficient of friction for the tire/ground interface, then you can get a reasonable estimate.

Make sure you account for the angle of the chain when taut.

However a real analysis will be difficult as you will likely encounter rocks, or the nose of the post will dig into the ground, stuff like that.
AlephZero
#3
May2-12, 04:20 PM
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P: 6,953
You missed the change in weight distribution between the front and rear wheels of the truck. This may work either for or against you on a 2WD truck, depending on the geometry of the situation.

jack action
#4
May3-12, 04:51 PM
P: 536
New Friction Problem

In your second case, the maximum force exerted by the truck will be the tire coefficient of friction times the weight portion of the truck on the driven wheels (hence, if it is a 4WD, the total weight of the truck).

For example, tires on gravel (CoF = 0.60) with a 2WD truck weight 4000 lb and 45% of its weight on the rear axle:

0.60 X 0.45 X 4000 lb = 1080 lb

Like AlephZero said, if the chain pulls the truck up (chain goes up from the truck to the post), the effective weight will be decrease; if it pulls the truck down (chain goes down from the truck to the post), the effective weight will be increase.

For the first case, in addition of the previous force calculated, you need to add the force produced by the energy of the truck (= ½ X truck mass X (truck velocity)²). You could find its speed after 15 ft of acceleration here to determine the energy available. The force produced could be defined as the energy divided by the displacement. Unfortunately, that displacement is the key variable that is almost impossible to determine. Assuming your truck and the post are also perfectly rigid, you would need to know how much the chain would elongate with that much energy. Your chain is then looked at as a spring with some elasticity. The equations would be as follow:

[itex]K=\frac{2E}{x^2}[/itex]

and

[itex]F=Kx=\frac{2E}{x}[/itex]

Where [itex]E[/itex] is the energy stored in the moving truck, [itex]x[/itex] is the chain elongation, [itex]K[/itex] is the chain «spring stiffness» and [itex]F[/itex] is the added force produced by the moving truck.

If you knew the «spring stiffness» of your chain, then you could also determine the force with [itex]F=\sqrt{2EK}[/itex].

Unfortunately, your truck and post are not rigid and it will lower the equivalent «spring stiffness» of you set-up (or increase the total displacement, if you prefer). By how much? that's the «almost impossible» part of the equation.
Nquisitive
#5
May7-12, 01:02 PM
P: 3
Those are all great replies, thank you for your help. I think I am pretty much on track then.

One thing more, if the front does lift up, is not the weight transfered to the rear and the force at the rear increased?

So in other words, the total force is the total force no matter how it is distributed to the axles??

Thanks again.
jack action
#6
May7-12, 05:30 PM
P: 536
Quote Quote by Nquisitive View Post
One thing more, if the front does lift up, is not the weight transfered to the rear and the force at the rear increased?
Yes.
Quote Quote by Nquisitive View Post
So in other words, the total force is the total force no matter how it is distributed to the axles??
For a 4WD, basically, yes.
Nquisitive
#7
May8-12, 04:06 PM
P: 3
I did forget to specify 4WD.

Thanks everyone.


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