Calculating Wheel Forces on a Car's Center of Gravity

[robertmatthew]

Homework Statement

An automobile with a mass of 1321 kg has 3.40 m between the front and rear axles. Its center of gravity is located 1.78 m behind the front axle.
(a) With the automobile on level ground, determine the magnitude of the force from the ground on each front wheel (assuming equal forces on the front wheels).
(b) Determine the magnitude of the force from the ground on each rear wheel (assuming equal forces on the rear wheels)

Homework Equations

I don't think there are any?

The Attempt at a Solution

The way I attempted to approach this was splitting up the mass according to the center of gravity. I assumed that because the center of gravity was 1.78m behind the front axle, it would have 1.78/3.4 of the mass and the rear wheels would have (3.4-1.78)/3.4 of the mass, so I got 691.58kg for the front of the car and 629.42kg for the back of the car. Then I multiplied these by 9.8 to get force and got 6777.507N and 6168.293N, respectively. I then divided each of these values by two, because I assumed the force would be split between the two wheels. For the front of the car, I got 3388.754N per wheel, but WebAssign's telling me this isn't right. That was my only idea, so now I have no idea what to do.

 

[SteamKing]

Set up your problem in a rational way.

you can draw a free body diagram of the car, showing the location of the c.g. in relation to the position of the wheels. Use the equations of statics to determine the reactions at the front and rear wheels.

 

[robertmatthew]

Isn't that just the sum of the forces and torques = 0? How could I use that?

 

[SteamKing]

You would be surprised how useful the equations of statics are.

By the way, how did your approach work?

 

[PeterO]

Isn't that just the sum of the forces and torques = 0? How could I use that?

If you consider roataion about the contact point of the front wheels, the c of m is 1.78 m away, and the back wheels are 3.4m away.
The torque effect of the whole weight through the centre of mass must equal the torque of the rear tyre support through the tyres.

 

[robertmatthew]

I still don't really understand. My teacher hasn't been focusing on this very much, but it's pretty much the only thing on my homework, so I'm really lost here. I'm not looking for anyone to give me the answer, just a push in the right direction so I'll understand it.

 

[PeterO]

I still don't really understand. My teacher hasn't been focusing on this very much, but it's pretty much the only thing on my homework, so I'm really lost here. I'm not looking for anyone to give me the answer, just a push in the right direction so I'll understand it.

How much is torque effect of the weight of the car, with reference to the contact point of the front tyres?

 

[robertmatthew]

Well τ=rFsinθ, so is r the distance from the front tire to the center of gravity, and F just the mass of the car times gravity?

 

[PeterO]

Well τ=rFsinθ, so is r the distance from the front tire to the center of gravity, and F just the mass of the car times gravity?

Correct - that is what r is, and that is what F is, but I asked what the torque was.

 

[PeterO]

Well τ=rFsinθ, so is r the distance from the front tire to the center of gravity, and F just the mass of the car times gravity?

Perhaps your formula might look more hepful if you wrote it is τ = F.rsinθ

 

[robertmatthew]

Well Fr is 23043.524, but I'm a little confused on the angle. The angle between the distance r and the force due to the weight is just 90, right? So would the torque then be 23043.524?

 

[PeterO]

Well Fr is 23043.524, but I'm a little confused on the angle. The angle between the distance r and the force due to the weight is just 90, right? So would the torque then be 23043.524?

I don't see how you can calculate Fr - you don't know r, but you can determine rsinθ [or should it actually be rcosθ? depends how you define the answer.

The CofM may be 178 cm behind but 1 m above the front wheels, so r = 204 cm.

The CofM may be 178 cm behind but 1.5m above the front wheels, so r = 233 cm

Of course the CofM may be 178 cm behind, and level with the front wheels, so r = 178cm

What value of r did you use - and why? What significance does your choice have on the value of θ ?

 

[SteamKing]

It might help to pretend the car was like a seesaw. The board of the seesaw is 3.4 m long. The balance point is located at the center of gravity of the car, or 1.78 m from one end.

What forces, when applied to the ends of the board, will cause the seesaw to balance?

 

[PeterO]

Well Fr is 23043.524, but I'm a little confused on the angle. The angle between the distance r and the force due to the weight is just 90, right? So would the torque then be 23043.524?

Yes the torque would be, I was just trying to get you to explain why.

So the torque from the rear wheels has to be that same value, but in the opposite direction.

SO what Force has to act through the rear wheels [you know the radius].

To find the force through the front wheels you can either repeat the torque calculations using the rear wheels as the pivot point, or just consider the total up force has to match the total down force.

 

[robertmatthew]

I had my physics class today so I got there a little early and went to ask my teacher, we revisited what SteamKing was saying about the static equilibrium. He talked me through how the two torques had to balance each other out, like what PeterO just said. I realize I appear pretty clueless here, my teacher hasn't been lecturing much during this unit so I was pretty lost. After he marked my free-body diagram a little, I understood what you guys were saying much more easily. Many thanks to both of you!