Centre of mass problem

[John O' Meara]

A station wagon of mass 2000kg has a wheelbase (distance between front and rear axles) of 3.0m. Ordinarily 1100kg rests on the front wheels and 900kg on the rear wheels. A box of mass 200kg is now placed on the tailgate, 1.0m behind the rear axle. How much total weight now rests on the front wheels? On the rear wheels?
X=(m1x1 + m2x2 + ...)/(m1 + m2 +...)
X=position of centre of mass. Vector sum of torques must be zero.
900kg/2000kgx3m = 1.35m; is the c.m., of the 2000kg car from the front wheels. I cannot think what to do next! Any one any ideas. Many Thanks.

[Chi Meson]

It is not necessary to find the position of the cm. Treat the system as three point objects of 1100 kg, 900 kg, and 200 kg.

Balance torques from there choosing one of the axels as the pivot (I don't see why you wouldn't).

[John O' Meara]

Let x be the centre of mass of the car plus the 200kg weight. And let y = mass at the front wheels. And let z = mass at the rear wheels. Taking moments about x, we have :

yx = z(3-x) + 200 (4-x),
yx = 3z -zx + 800 -200x,
but z=(2200-y-200), therefore
yx = 3(2000 - y) -2000x + yx + 800 -200x,
therefore
x = (6800 -3y)/2200.
The question is, how do you get y?

[Saketh]

Let x be the centre of mass of the car plus the 200kg weight. And let y = mass at the front wheels. And let z = mass at the rear wheels. Taking moments about x, we have :

yx = z(3-x) + 200 (4-x),
yx = 3z -zx + 800 -200x,
but z=(2200-y-200), therefore
yx = 3(2000 - y) -2000x + yx + 800 -200x,
therefore
x = (6800 -3y)/2200.
The question is, how do you get y?
If you use x as your pivot point, you will be unable to get an answer. Do what Chi Meson said and choose one of the axles as the pivot point - that way, you can eliminate one of the variables from the start (torque at that point is zero). Then you can go back and make the other axle the pivot point.

[John O' Meara]

When I try to do what Chi Meson says, I get the following; taking moments about the 900kg point first, I get an inconsistant equation namely: 1100*3 = 200*1, but if I do the following; taking moments again about the 900 kg mass: 1100(3-x)=200*1 => x=2.8182. I must not understand Chi Meson? although what he says looks simple enough.

[Chi Meson]

pivot=front axel (on the left)
cw torque: 1100*0 + 900*3 + 200*4

ccw torque: rear axel force*3

[John O' Meara]

AS a matter of interest what does "cw" and "ccw" stand for. Thanks very much for your help.

[Chi Meson]

"clockwise" and "counterclockwise."

[WyldFyr]

CW and CCW stand for Clock Wise and Counter Clock Wise.
I just got done doing this stuff a few weeks ago, so i know what you are going through. What they are trying to tell you is that the normal force of each wheel (and the associated torque) counters the load on the wheel from the car. You can use that to create the equation with the unknown being the normal force of the wheel (which equals the load on the wheel). Hope this helps.

[John O' Meara]

If I do the following: equate the cw torque with the ccw torque as given by Chi Meson in reply no., 6( although I didn't think that you could equate them because they were from two different pivot points), the force = 1666.67N. If you connot equate them, I don't see any ccw torques for the front axle or any cw torques for the back axle?