A tire sliding about a fixed point

[Zauce]

Homework Statement

There are two tires separated by a few feet, with a weighted beam attached on top of them. The beam's weight isn't distributed evenly. One of the tires is a fixed point. The other tire slides (doesn't roll) 90 degrees. How do you determine the force required to slide the tire all the way to this angle (or distance)?

 

[tiny-tim]

Hi Zauce!

(is that all the information … no numbers?)

There are two tires separated by a few feet, with a weighted beam attached on top of them. The beam's weight isn't distributed evenly. One of the tires is a fixed point. The other tire slides (doesn't roll) 90 degrees. How do you determine the force required to slide the tire all the way to this angle (or distance)?

What do you think? How many forces are involved?

 

[Zauce]

That's not all the numbers. I know the applied force on top of each tire and the coefficient of friction between rubber and asphalt, and the center of gravity of the beam. It's been a couple of years since I've taken Dynamics, so I'm having a tough time coming up with which equation to use. I know I need the normal force, frictional force, gravitational force. Can you help me with this?

 

[tiny-tim]

I know I need the normal force, frictional force, gravitational force.

ok, once you know the normal force, that tells you the friction force, and that tells you the work done.

So find the normal force first … how can you do that?

 

[Zauce]

The normal force would just be mass x gravity, I think. Correct?

 

[tiny-tim]

Nooo … what about the weight of the beam?

 

[Zauce]

The normal force would equal m x g + weight of beam??

 

[tiny-tim]

Don't forget that the beam is supported at two points …

so you'll have to take moments to find how much of its weight is supported at each tyre.

 

[Zauce]

Ok, I've taken the moment to find the weight of the beam at each tire location. So, I should have all the numbers to find the normal forces at both points. Whats the next step?

 

[tiny-tim]

(just got up :zzz: … )

Ok, I've taken the moment to find the weight of the beam at each tire location. So, I should have all the numbers to find the normal forces at both points. Whats the next step?

Can you clarify the original question …

Is the asphalt (that the tyre is on) horizontal? Is the beam fixed to a fixed point on the tyre, so that the beam increases in slope as the tyre turns?

Does the beam start horizontal, and at the top of the tyre (because if it does, I don't see why the tyre should start sliding)?

 

[Zauce]

Yes, the ashphalt is horizontal. The beam is attached to the top of each tire, and the back tire is a fixed point. I'm trying to figure out how much wind it took to blow the front tire 90 degrees. I drew up a couple of diagrams, and will attach both. The first one is a side view and I will post the view from the top in just a minute.

 

[Zauce]

Here is a view from above.

 

[tiny-tim]

Ok, I've taken the moment to find the weight of the beam at each tire location. So, I should have all the numbers to find the normal forces at both points. Whats the next step?

ok, from the normal force you can calculate the friction force, and from that you can calculate the work done.

 

[Zauce]

After I find the normal force and friction force, would I just add both of them up? Would that be what I'm looking for to find the force required by the wind to push the beam the distance? Thanks a lot for the help.

 

[tiny-tim]

After I find the normal force and friction force, would I just add both of them up?

No!

The normal force is vertical. The wind is horizontal. The friction force is horizontal.

So … ?

 

[Zauce]

So, the wind force needed to push the beam and tires would just have to be greater than the friction force.

 

[tiny-tim]

So, the wind force needed to push the beam and tires would just have to be greater than the friction force.

(just got up :zzz: …)

Yes, when the only opposing force is friction, the minimum force needed is just greater than the friction force.

But both tyres? You didn't say anything about them both moving.

Is the other tyre being forced to rotate "on the spot" also … because that will require extra force, won't it (if there's friction, and if the area of contact is not negligible)?

What exactly is the question?