# Math Help - How much force to lift a car?

1. ## How much force to lift a car?

Hello,

I was wondering how much force it would take to lift a small hatchback car off the ground by the back axle? (i.e. so its just leaning on its front axle).

Assuming the car is 4.0 metres long, and has wheels at 0.5m from the back end, and 1 m from the front end, has mass 1000 kg, and the back end needs to be lifted till my hands are level with my hip, 1.00 metres off the ground. How much force would I need to lift it?

I calculated that i would be lifting 3.0m from the pivot, a mass of 10,000N, so 30,000 Nm, can anyone help me?

2. Originally Posted by Undertaker
Hello,

I was wondering how much force it would take to lift a small hatchback car off the ground by the back axle? (i.e. so its just leaning on its front axle).

Assuming the car is 4.0 metres long, and has wheels at 0.5m from the back end, and 1 m from the front end, has mass 1000 kg, and the back end needs to be lifted till my hands are level with my hip, 1.00 metres off the ground. How much force would I need to lift it?

I calculated that i would be lifting 3.0m from the pivot, a mass of 10,000N, so 30,000 Nm, can anyone help me?
As you correctly surmised this is a torque problem.

I am going to assume that the center of mass of the car is at the midpoint of the axle and rear of the car. This assumption is clearly ridiculous, but I don't know that much about cars. Obviously the CM is going to be somewhere between the geometric center of the car and the front axle. But this is what I'm going to assume and you can adapt the problem accordingly.

I am going to assume that the car has been lifted to an angle $\theta$. Call the distance from the front axle to the rear of the car (where it's being lifted from) to be D. Take a counter-clockwise torque to be positive and take the axis of rotation to be about the front axle.. Then we have
$\displaystyle \sum \tau = F(D)sin(90 - \theta) - w(D/2)sin(90 + \theta) = 0$

Noting that $sin(90 - \theta) = sin(90 + \theta) = cos(\theta)$ we get, upon solving for F:
$\displaystyle F = \frac{w \left ( \frac{D}{2} \right )~sin(90 + \theta)}{(D)~sin(90 - \theta)} = \frac{mgD~cos(\theta)}{2D~cos(\theta)} = \frac{mg}{2}$.

What surprises me about this is that the applied force is constant. I would have guessed that it would depend on $\theta$.

-Dan

3. Originally Posted by topsquark
As you correctly surmised this is a torque problem.

I am going to assume that the center of mass of the car is at the midpoint of the axle and rear of the car. This assumption is clearly ridiculous, but I don't know that much about cars. Obviously the CM is going to be somewhere between the geometric center of the car and the front axle. But this is what I'm going to assume and you can adapt the problem accordingly.

I am going to assume that the car has been lifted to an angle $\theta$. Call the distance from the front axle to the rear of the car (where it's being lifted from) to be D. Take a counter-clockwise torque to be positive and take the axis of rotation to be about the front axle.. Then we have
$\displaystyle \sum \tau = F(D)sin(90 - \theta) - w(D/2)sin(90 + \theta) = 0$

Noting that $sin(90 - \theta) = sin(90 + \theta) = cos(\theta)$ we get, upon solving for F:
$\displaystyle F = \frac{w \left ( \frac{D}{2} \right )~sin(90 + \theta)}{(D)~sin(90 - \theta)} = \frac{mgD~cos(\theta)}{2D~cos(\theta)} = \frac{mg}{2}$.

What surprises me about this is that the applied force is constant. I would have guessed that it would depend on $\theta$.

-Dan
the angle to which the car is lifted to the ground would be tan-1 ( 1.00m/3.00m) = 18.43 degrees

also the centre of mass would be nearer the front than the back obviously, call the centre of mass 1.33m from the front and 2.67 m from the back?

how does this change things? cheers

4. Originally Posted by Undertaker
the angle to which the car is lifted to the ground would be tan-1 ( 1.00m/3.00m) = 18.43 degrees

also the centre of mass would be nearer the front than the back obviously, call the centre of mass 1.33m from the front and 2.67 m from the back?

how does this change things? cheers
The only thing it changes is where the weight force is acting from. In my example it was located at D/2.

My point about the angle is that the needed force does not change from start to finish, so we don't even need to worry about the angle.

-Dan