Optimizing Center of Gravity for Inclined Driving: Solving a Friction Problem

[SheldonG]

A car with a load equal to 1/4 the total mass of the empty car has its center of gravity halfway between the front and rear axles. The drive wheels (rear) start slipping when the car is driven up a 20 degree incline. How far back must the load be shifted for the car to get up a 25 degree slope. (The distance between the axles is 10 ft.)

I am really stuck on this one. I write the force equation as

mg sin 20 - u mg cos 20 = 0

And it seems to me that the additional frictional force that must be supplied by moving the load would then be given by sin 25 - sin 20. But I have no idea how to turn this value into a distance. I have tried using torques around the center, like this:

mgd/4 = (sin 25 - sin 20)mg

But this is so small it can't be right. Any suggestions that would help shake me loose would be appreciated.

 

[whozum]

I think were missing some information, is that the whole problem? It looks like a torque problem about the rear axle.

 

[SheldonG]

No, that's the entire thing.

 

[Dorothy Weglend]

I am really stuck on this one. I write the force equation as

mg sin 20 - u mg cos 20 = 0

And it seems to me that the additional frictional force that must be supplied by moving the load would then be given by sin 25 - sin 20. But I have no idea how to turn this value into a distance. I have tried using torques around the center, like this:

mgd/4 = (sin 25 - sin 20)mg

But this is so small it can't be right. Any suggestions that would help shake me loose would be appreciated.

I think your approach isn't quite right, because the difference wouldn't just be sin 25 - sin 20 on the higher incline. This doesn't take the change in the frictional force on the higher incline into account.

You could calculate the coefficient of friction using the first incline, that should be the same in both cases.

And then I think you should just focus on the back wheel. I always get confused at this point. If you consider torques, then the back wheel is applying 5 mg cos 25. But it seems that the force should really just be (mg/2) cos 25. I am not sure about this.

If I work this problem out this way, (using the torque values for both inclines) I get 5.6 feet. But I am not at all sure I am right, since this seems to close to the middle to make a difference. But maybe this will help.

Dot

 

[Dorothy Weglend]

I always get confused at this point. If you consider torques, then the back wheel is applying 5 mg cos 25. But it seems that the force should really just be (mg/2) cos 25. I am not sure about this.

I made a mistake there, and wanted to clear it up. In the case of the 20 degree incline, the back wheel is applying mg/2 cos 20, whether you consider the torques or just the center of gravity of the car. I've finally cleared up my own confusion on this subject, which came from forgetting to use 10N in the torque equation (I was just using N). Use the front wheels as the pivot, and it all works out nicely.

I'd like to know what you get, and if you know what the answer is supposed to be, I am curious.

Dot

 

[mukundpa]

I think...the normal reactions on front and rear wheels are equal in first case and both are (5mg/8)cos20
hence coe. of friction

mu = (5mg/4)sin20 / (5mg/8)cos20

In second case the angle is 25 so the car will slip. To avoid slipping the load is shifted by x in such a way that the normal reaction at rear wheel increases to provide necessary friction.

With shifted load the torque about the center of gravity should be zero.

Have a try..

 

[SheldonG]

Thank you both so much for your helpful thoughts. To answer your question, Dorothy, the answer given is 7 ft. But I think you are on the right track, and thanks so much for your help.

I think...the normal reactions on front and rear wheels are equal in first case and both are (5mg/8)cos20
hence coe. of friction

mu = (5mg/4)sin20 / (5mg/8)cos20

I don't understand how you get this, I'm sorry. When I calculate the torque, using the front wheel as the pivot, I get:

10N - 5mg cos 20 = 0

So N = (mg cos 20) / 2 ?

And the force of gravity on the rear wheel is (mg sin 20)/2? If I work it out this way, I also get 5.6 ft, so I think I am still missing something.

I tried it with (5mg/4)sin 20/(5mg/8) cos 20, but it leads to such a weird equation for the case of 25 degrees, I think I must be missing something here too.

But this is much better than I had.

 

[Dorothy Weglend]

Yes, I get about 5.6 feet too. So either we are still both missing something, or your book has the wrong answer. Maybe someone else will help out.

Your calculation for the torque around the front wheel looks right to me.

Dot

 

[mukundpa]

I also think it is not possible to hold the car by just shifting the load.
The load must be greater then the value given.

sol. is attached, please check for errors.


The height of the COM is not given so it is considerd near the surface and hence the torque due to friction is taken zero about the center of mass.

You may take the torque about any point to get the same answer.

 

[SheldonG]

You may take the torque about any point to get the same answer.

Mukundpa, thank you so much for your help, and going to the trouble to write out a complete solution for me. I really appreciate it.

The only difference between our solutions is the way you computed the torques, and yet I do not understand where I went wrong here. If I could trouble you for one more bit of help, I would appreciate it.

I took the torques around the front axle, and using clockwise torques as positive, I got

Nd - x (mg/4) cos 25 - (d/2) mg cos 25 = 0

which led me to N = (mg cos 25)(x/4d + 1/2)

Your N = (mgcos25/8)(5 + 2x/d) is different enough to get the correct answer, so I don't know where I went wrong. If I multiply mine by 8/8, I see that

N = (mgcos25/8)(2x/d + 4),

So I am guessing that my error must be in the term (d/2) mg cos 25, and yet that seems correct to me. Torques are a problem for me, so if you (or anyone) can explain what I did wrong, it would help a lot.

Thanks again for your wonderful assistance.

 

[mukundpa]

Nd - x (mg/4) cos 25 - (d/2) mg cos 25 = 0


.

Nd - (d/2 + x) (mg/4) cos 25 - (d/2) mg cos 25 = 0