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March 4, 2015

March 4, 2015

Posted by **Leah** on Saturday, January 5, 2008 at 7:27pm.

PLEASE GIVE ME SOME HINTS TO DO IT...THANKS A LOT!

- physics! -
**Damon**, Saturday, January 5, 2008 at 7:39pmThe total force on the four tires is the weight of the car:

m g = 1160 * 9.8

Now do a see saw

You have m g down at 1.02 meters behind the front axle

You have 2Fb up at the rear tires,2.54 meters behind the front axle (I called it two Fb so I will not forget to divide by two to get each of the rear tires.)

take moments (torques) about the front axle

2Fb * 2.54 = m g * 1.02

solve that for Fb, the force on the back axle

then the total force up = total force down

so

2Fb + 2Ff = m g

You know mg and your know 2Fb so now you know 2Ff

divide 2Ff by two to get the force on each front wheel

divide Fb by two to get the force on each rear wheel

- riddled with tpos :( -
**Damon**, Saturday, January 5, 2008 at 7:47pmThe total force on the four tires is the weight of the car:

m g = 1160 * 9.8

Now do a see saw

You have m g down at 1.02 meters behind the front axle

You have 2Fb up at the rear tires,2.54 meters behind the front axle (I called it two Fb so I will not forget to divide by two to get each of the rear tires.)

take moments (torques) about the front axle

2Fb * 2.54 = m g * 1.02

solve that for 2Fb, the force on the back axle

then the total force up = total force down

so

2Fb + 2Ff = m g

You know mg and your know 2Fb so now you know 2Ff

divide 2Ff by two to get the force on each front wheel

divide 2Fb by two to get the force on each rear wheel

- physics! -
**Arno**, Thursday, April 3, 2014 at 7:48pmOk, so after struggling through the above explanation I decided to repost a (imho) better worded, generalized explanation:

Firstly let me define my variables: Fb=force of BOTH back wheels on the ground=2F(1backwheel)

Ff=force of BOTH front wheels on the ground=2F(1frontwheel)

d1= distance of back wheels from front

d2= distance of CM from front wheels

We should first recognize that the force of the ground on both back wheels is equal in magnitude (but opposite indirection) to the force of the wheels on the ground or Fb=F(ground.on.back.wheels)and the force of the ground on the front wheels is equal to the force of the front wheels on the ground or Ff=F(ground.on.front.wheels).

We know that the force of the ground on all four wheels is equal to the force of weight of the car (mg)because the car isn't falling through the ground.

So F(ground.on.back.wheels)+F(ground.on.front.wheels)= mg = Fb+Ff

Define down as positive:

Now if we act as though the car could rotate around the front axle and recognize that the net torque is 0 then we know that Torque at CM + torque at back caused by ground=0=d2*mg+ d1*(-F(ground.on.back.wheels) so F(g.o.b.w)=mg(d2/d1)= magnitude of Fb

which is twice the magnitude of the force of each back wheel.

Use Fb+Ff=mg to find Ff which is twice the magnitude of the force of each front wheel

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