How to find the normal force of a car jack lifting a car

[Protium_H1]

Homework Statement

A car is lifted vertically by a jack placed at the car's rear end 40cm off the central axis, so that the weight of the car is supported by the jack and the two front wheels. The distance between the front wheels is 1.60m, the distance from the axis connecting the two wheels to the center of mass of the car is 80cm, and the distance from the rear of the car to the center of mass is 2.10m. What fraction of the car's weight is carried by each of the wheels, and what fraction is carried by the jack? (Note: the weight on the wheels will not be symmetrically distributed).

Homework Equations

[/B]
Net Torque: \Sigma \vec{\tau} = I \alpha = \tau_{1} + \tau_{2} + ...
Net Force: \Sigma \vec{F} = ma = F_{1} + F_{2} + ...

3. The Attempt at a Solution

1. Made my coordinates like so, plus I made clockwise rotation be positive.
2. 
   
   
3. Wrote the restriction \vec{F}_{NL} \neq \vec{F}_{NR} and that \vec{F}_{NL} > \vec{F}_{NR}
4. Wrote the net force formula in vector form.
5. \Sigma \vec{F}=\hat{j}Ma=\hat{j}(F_{NL}+F_{NR}+F_{NJ} - Mg)
6. Wrote the net torque formula in vector form,
7. \Sigma \vec{\tau}_{about CM} = I_{car} \alpha = [\hat{j}F_{NJ} \times (\hat{k}2.10 + \hat{i}0.40)] - [\hat{j}F_{NL} \times (\hat{-k}0.80 + \hat{-i}0.80)] - [\hat{j}F_{NR} \times (\hat{-k}0.80 + \hat{i}0.80)]
8. Simplified torque to,
9. \Sigma \vec{\tau}_{about CM} = \hat{-i}(2.10F_{NJ}+0.80F_{NL}+0.80F_{NB}) + \hat{k}(0.40F_{NJ}+0.80F_{NL}-0.80F_{NR})
10. Note: So far this makes sense since \vec{\tau} can't be parallel to the \vec{F}_{applied} (i.e.: \Sigma \vec{\tau} has not \hat{j} component.)
11. After this, I'm totally stuck since I have 3unknownss and only 2 equation (net force and net torque). I don't even know if my choice of using the center of mass (CM) of the car to find torque was a good idea, since the car is pivotted in the position of the two wheels. I don't even know if this system is \Sigma \vec{F} = 0 and \Sigma \vec{\tau} = 0

 

[haruspex]

3unknownss and only 2 equation (net force and net torque).

Your torque equation is a vector equation with two degrees of freedom. It counts as two.

 

[Protium_H1]

Your torque equation is a vector equation with two degrees of freedom. It counts as two.

You mean something like this?
\Sigma \vec{\tau}_{x-direction} = -\hat{i}(2.10F_{NJ}+0.80F_{NL}+0.80F_{NB})
\Sigma \vec{\tau}_{z-direction}= \hat{k}(0.40F_{NJ}+0.80F_{NL}−0.80F_{NR})
\Sigma \vec{F}_{y-direction} = \hat{j}(F_{NL}+F_{NR}+F_{NJ}−Mg)
But how can I combine these if they have different directions? Also, With these 3 equations, I would have 4 unkowns (Mg and the 3 Normal Forces). Do I need to use the m\vec{a} portion?

 

[haruspex]

how can I combine these if they have different directions?

For equilibrium, each must have zero net torque (or net force, as appropriate).

With these 3 equations, I would have 4 unkowns (Mg and the 3 Normal Forces).

You are only asked for the ratio of forces. The actual forces are all proportional to M, so M must cancel out in taking the ratio.

 

[Protium_H1]

For equilibrium, each must have zero net torque (or net force, as appropriate).

Do you mean that, in this system, either net torque or net force is zero?

 

[haruspex]

Do you mean that, in this system, either net torque or net force is zero?

In equlibrium, the net force in any selected direction is zero, and the net torque about any selected axis is zero. If not, something is going to accelerate.

 

[Protium_H1]

In equlibrium, the net force in any selected direction is zero, and the net torque about any selected axis is zero. If not, something is going to accelerate.

You’re actually a beauty. I just solved it, thanks!

 

[haruspex]

You’re actually a beauty.

Sadly, my resemblance to my avatar is only slight.