# Contact force in an ideal situation

1. Jan 10, 2006

### TheDonk

Assume I am holding a bar parallel to the ground. I let go of the bar and it lands on something at an arbitrary point on the bar before it hits the ground. How much force is dealt to the object (and bar)?
This is the problem, but I'm going to simplify it to make a calculation possible. First imagine the situation in two dimensions. A bar falling and landing on a small circle with no air resistance. Also lets not bother with gravity and just say that the velocity is perpenticular to the bar (downwards) at contact.
I think the major factor to calculate this force is the composition of the bar and triangle. So assume that the objects are completely solid, with absolutely no compression. The bar has uniform density.

Bar's length = L
Bar's width = W
Bar's mass = M
Bar's center point = C = (Cx, Cy)
Bar's contact point = H = (Hx, Hy)

So if you know an equation for the contact force or I haven't given enough information to come to an exact answer, please reply.

2. Jan 10, 2006

### HallsofIvy

Staff Emeritus
You don't have enough information- You can calculate the speed of the bar when it hits the object but force= mass*acceleration. How long, from time of contact, does it take the bar to come to 0 speed? That is what you need to find the acceleration (or, more correctly, decelaration) of the bar and from that the force. That depends upon things like the softness or resiliency of the object and bar. An iron bar striking a steel cylinder will experience more force than one striking a sack of feathers.

3. Jan 10, 2006

### TheDonk

I see what you mean. But what if the bar doesn't stop? Is there a way based on density or something to calculate the acceleration change on contact? I've also said that they are completely solid, so imagine something closer to steel than a sack feathers. It seems that to get the acceleration change I need the force applied, but this is my original problem. So I guess I'm asking for a different way to find the contact force or the acceleration change on contact.

I was starting to think that absolutely no compression implies infinite acceleration and so infinite force. But I realised that it would also be applying the force over a smaller amount of time the harder an object gets, so a rod that didn't compress at all would apply its force over an infinitesimal amount of time. This allows for a finite force.

Also the collision is ellastic. Basically all variables involved are positions, rotations, velocity, and mass (which would stay constant).

4. Jan 10, 2006

### lightgrav

Yes, "Force multiplied by collision time" is constant, to stop the bar's fall.
It will have to equal negative of "weight multiplied by fall time".

Seems that you have to choose a collision time (or a compression distance).

5. Jan 10, 2006

### TheDonk

I'm not exactly sure that I do, but for your sake (you're more likely right than me in this case) why don't we make amount of time for the collision to end the variable T. If it matters we can assume that T is very short.