How much force is required to lift the back of a 2,480 lb car in the air from the bumper?

I don't know the EXACT measurements, but I can closely guess that the bumper is about 18 inches from the ground, and the engine is in the front, yes(the engine is about 20 inches from the front bumper inside the hood, and about 23 inches centered between the hood on both sides). I don't know what the engine weighs, but the car's overall curb weight (whatever that means) is 2,480 lbs. Convert to any other weight measurements if you wish, but I'm most used to pounds.

I can assume you can easily find the specs of the size ... it's a Toyota Corolla 2002 blue SE limited-edition model.

Specs:

It is about fourteen feet and six inches long.

It is five feet and size inches wide from both doors on the front left to right seats.

The wheelbase is around eight feet.

The car is in total four feet and five inches from the floor to the roof in height.

I don't know if this question is relevant or meets the standards here, but I would really appreciate someone decent at physics and maths to help me find the weight here that is lifted from the back bumper's height, and the total force necessary to raise both wheels in the air about one inch.

Some have said that since there's shocks, the weight is favorable to the lifter somewhat, but to what degree I'm uncertain.

Any help would be greatly appreciated in your findings on this issue.

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To answer with precision would require data about the location of the front axle and the center of mass of the car. You can expect an answer vaguely on order of 1000 pounds (because these cars have a forward center of gravity) to get the rear wheels clear of the ground, but it could be a hundred pounds or more in either direction. Either the owner's manual or the tire data plate inside the drivers door may provide some more useful information. –  dmckee Jan 31 '13 at 21:52
Between 40% and 60% of the total weight depending on the weight distribution (which you can look it up). This should be close enough in general. –  ja72 Jul 11 '13 at 19:06

Cars are normally designed so that the weight is evenly distributed between the front and rear wheels. That means the centre of gravity is close to halfway between the axles.

The above diagram shows the forces involved in lifting the front of the car so it pivots around the rear wheels. $d_1$ is the wheelbase and $d_2$ is the distance from the front wheels to the point where you're lifting. To work out the force required to lift the car you just take moments so:

$$mg \frac{d_1}{2} = F(d_1 + d_2)$$

and a quick rearrangement gives the force required to lift the car:

$$F = mg \frac{d_1}{2(d_1 + d_2)}$$

Just feed in the dimensions and weight of your car to calculate the force. I couldn't find the distance from the front wheels to the bumper online, so you'll have to measure it then do the sum.

It doesn't matter whether you use imperial or metric units as the units of length cancel out. Replace $mg$ by the weight, i.e. 2,480 lbs, and you'll get the answer in lbs. As dmckee says, it's going to be around 1,000lbs.

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Of course if you want to lift it briskly, then additional force is needed to provide for acceleration. The full treatment needs to include the inertial forces of the motion. –  ja72 Jul 11 '13 at 19:05
"Cars are normally designed so that the weight is evenly distributed" is more aspirational than achieved for many small and inexpensive cars. It gets more and more true as the cars get bigger and more expensive. –  dmckee Jul 11 '13 at 21:10

protected by Qmechanic♦Jul 11 '13 at 19:31

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