A racing car has a mass of 1600 kg. the distance between the front and rear axle

Question

A racing car has a mass of 1600 kg. the distance between the front and rear axles is 3m. If the center of gravity of the car is 2m from the rear axle, what is the normal force on each tire?

So the rear tire is chosen as the pivot point. and sum torque =0 gives

2Fr (0) -W (2m) + 2Ff (3m) =0 (1)

where Fr is the force on each rear tire and Ff is the force on each front tire. From (1)

we get Ff= W/3 .

The rear force cancels due to 0 distance. that leaves

-W(2m) + 2Ff (3) =0

moving 2Ff to the right makes sense,

-W(2m) _ (3m) = -2Ff

but that leaves 2W/3 not W/3?



For the sum Fy=0

2Fr -W -2Ff =0

Since Ff= W/3, we find Fr = W/6 (where W = 15680N)

So substitute W/3 for 2Ff

2Fr - 15680 - 15680/3 = 0

-15680 - 15680/W = -2Fr.

I know the negatives will cancel out. but I'm missing how we came up with W/6.
I'm certain it's a simple manipulation I'm missing, and I'm also almost certain that it's the same one both time. Thanks!

Explanation / Answer

about Ff = W/3 :
in your solution, we have: -W(2m) + 2Ff (3m) =0
we can divide the equation by 2; then we have: -W + Ff (3) =0
it results in: Ff = W/3

(Please note that there is a mistake in your solution, please look at this mistake:

moving 2Ff to the right makes sense,
-W(2m) _ (3m) = -2Ff)

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About Fr = W/6:

there is anpther mistake in your solution. this equation is not true: 2Fr -W -2Ff =0

the right equation is: 2Fr - W + 2Ff = 0 (Note that this equation is for the sum Fy=0)

Since Ff= W/3, we find Fr = W/6

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