How much force/energy is required to lift a 2000lb car off the ground?

[wwwebster]

I'm a mathemetician trying to test out a theory of mine. I have never taken any physics before, so i need help from the physicist and the engineers here. How much force/energy/work is required to lift a 2000lb car off the ground? What is that transalted into newtons and into rpms?

[GutZ]

My guess would be F=MxA

 

where:

M = 908kg

A= 9.8m/s

 

(908 kg) x 9.8 (m / s^2) = 8 898.4 m kg / s^2

 

kg/s^2 = 1 N

 

So..anything great than this:

 

8.8984 KN

 

I probably got that wrong though.

[Bignose]

I'm a mathemetician trying to test out a theory of mine. I have never taken any physics before, so i need help from the physicist and the engineers here. How much force/energy/work is required to lift a 2000lb car off the ground? What is that transalted into newtons and into rpms?

 

As a mathematician, you know how important is it to be very precise with your terminology. Therefore, you need to be careful when you use the words force and energy and work. Force comes in units of Newtons, and energy and work come in units of Joules. Two very different quantities there.

 

Do yourself a favor and pick yourself up a good calculus based physics textbook. Wikibooks has an incomplete book here: http://en.wikibooks.org/wiki/Physics_with_Calculus

 

But any university-level calculus-based physics text will help you a lot. Not to seem snide about it, but it is incredibly unlikely that you will discover anything about basic Newtonian physics that hasn't been studied yet. And asking how much force is required to lift a car is very basic Newtonian physics, as is knowing the difference between force and energy. As a mathematician, it will actually probably be pretty easy stuff for you, you just need to learn the correct terminology/theories.

 

In other words here, I am suggesting that you not try to reinvent the wheel with your own theories, when all of that work has been done for you. Study the great theories that are there, and then build up on that foundation. And, you'll know the terminology and be able to speak the same language as the other very knowledgeable physicists on this forum and around the world. Just like other people need to learn a significant amount of math to speak the language of math with you, you need to learn physics so you can speak the language of physics with other physicists.

[Klaynos]

Force due to gravity:

 

F=m*9.81

 

Any force greater than this will lift the car, the bigger the force the faster it'll lift it.

 

Energy:

 

E=m*9.81*h

 

Work is just the change in energy.

[MrMongoose]

The tyres will have an elasticity, which I'm going to foolishly assume is linear for this purpose. Assuming the overall spring constant of the four tyres to be k, when the car is sitting on the ground in equillibrium,

kx=mg,

 

where x is the deflection of of the tyres, m is the mass of the car, and g is the acceleration due to gravity.

 

For the car to be just lifted off the ground, it should be raised by x. In doing so, the increase in gravitational potential energy will be mgx, whereas the decrease in elestic energy will be (kx^2)/2. Thus the total energy change (the energy provided to the car) is

 

mgx-(kx^2)/2.

 

Using x=mg/k from the equillibrium equation above,

 

Energy required= ((mg)^2)/k-((mg)^2)/2k

 

=((mg)^2)/2k

 

Edit: I forgot that the mass was given...

 

Of course, tyres never behave linearly, so being lazy enough to assume that a pound is half a kilogram and g is 10 m/s/s would be the least of our worries and doing so gives (50/k)MJ.

[swansont]

The force can be vanishly small, because you can use systems that give you a mechanical advantage (e.g. pulleys, levers). The work done will be the same, because energy must be conserved, and that depends on how high you lift the car, as Klaynos indicated.

[Sisyphus]

A mathematician with intro to physics homework?

[Rote Learner]

are you saying to lift the ENTIRE car off the ground?

 

or pulling from lets say the wheel well...or the hood or trunk?

 

let's assume you will lift it from the front of the car. apply the e-brake for good measure too. the force you will apply will be in the upward direction. Since you are not lifting all 2,000 lbs due to the back tires planted on the ground, it will be very hard to find the exact force.

 

During this "lift" your forearm respectively rotates about the elbow. This will be the "hinge point" for the applied force. Lets say the distance from the hinge point to the lift is 12 inches. Say you are lifting 1200 lbs of that car. 12X1200 =14400 in lb. Don't worry though, once you get the car lifted, the distance will get smaller and the car will be easier to lift. So at the peak of you lift it should be around: 5 in X 1200 = 60,000 in lb

[MrMongoose]

Its not that hard to find the force required if we know the centre of mass of the car.. Its just a matter of moments.

[tvp45]

I'm a mathemetician trying to test out a theory of mine. I have never taken any physics before, so i need help from the physicist and the engineers here. How much force/energy/work is required to lift a 2000lb car off the ground? What is that transalted into newtons and into rpms?

 

The amount of force required is 2000.00000001 pounds (you can make the decimal part as large or small as you want - it only has to do with how much time you want to spend doing the lifting).

 

The amount of work (which in this instance is equal to the increase in potential energy) depends on how far you lift it and can be as low as about 1000 ft-lb (remember that you must lift a real car about 6 inches in order to get everything up). Kinetic energy can be vanishingly small if you lift slowly.

[ydoaPs]

Am I the only one thinking that since the car is 2000lbf, it will take any force greater than 2000lbf to lift it(assuming you lift from the centre of gravity with no pulleys and such)?

 

 

edit: I somehow missed the post directly before mine.

[chilehed]

Am I the only one thinking that since the car is 2000lbf, it will take any force greater than 2000lbf to lift it... ?....

No, you're not.

[swansont]

If you're trying to deadlift it, yes. But that's a restriction not present in the OP; it's apparently being assumed, either explicitly or implicitly. Go ahead and get that hernia, but until the question is better defined, I'm going to use physics to make the job easier. A hydraulic lift, or a simple machine such as the lever, the ramp or a block and tackle (pulley system)

 

The restriction is energy, so the work will be the measure — if you lift the car 1 foot, you will do 2000 lb-ft of work regardless of the system you use.

[MrMongoose]

If you're trying to dealift it, yes. But that's a restriction not present in the OP; it's apparently being assumed, either explicitly or implicitly.

 

The question was how much force is required to lift a car, not how much force is required to operate a mechanism that can provide enough force to lift a car.

 

Also, I was assuming he meant 2000lbm as thats a far more realistic weight for a car.

[swansont]

The question was how much force is required to lift a car, not how much force is required to operate a mechanism that can provide enough force to lift a car.

 

The question doesn't specify if it's the force a person must exert, or what must be exerted on the car. As I said, it depends on what assumptions you've made.

[D H]

Also, I was assuming he meant 2000lbm as thats a far more realistic weight for a car.

 

2000 lbm is not a realistic weight for a car. Weight has units of force. 2000 lbf is a reasonable weight for a car. What is the mass of a car that weighs 2000 lbf? 2000 lbm, of course.

 

I'm going to use physics to make the job easier. A hydraulic lift, or a simple machine such as the lever, the ramp or a block and tackle (pulley system) ...

 

Regardless of whatever simple machine you choose to use to lift the car, there will still be a 2000 lbf (plus epsilon) upward force at the business end of the machine -- the end attached to the car.

[MrMongoose]

Oh yeah.. sorry. I'm not much good with silly foreign units. I was thinking 200lbf was 1000N whilst 2000lbm was 1000kg (or something of that order of magnitude)

[iNow]

Oh yeah.. sorry. I'm not much good with silly foreign units. I was thinking 200lbf was 1000N whilst 2000lbm was 1000kg (or something of that order of magnitude)

 

The funny thing about this is that the US system is what the rest of the planet calls those "silly foreign units."

[Mr Skeptic]

The funny thing about this is that the US system is what the rest of the planet calls those "silly foreign units."

 

Most of the scientists living in the US also think the American (sorry, "English") units are silly, as do anyone who learned metric first. Where they live has nothing to do with it, really.

[ydoaPs]

I use the imperial system in my job and I think they're silly. IMO, metric is way better.

[MrMongoose]

Just to clarify, I'm from england and I was calling Imperial units silly. Though I do regularly drive a few miles home after having several pints and measure myself in feet... And measure my mass in stones and pounds... which apparently most Americans don't know about.

[swansont]

And measure my mass in stones and pounds... which apparently most Americans don't know about.

 

The ones who have studied physics will recognize that those are weight, not mass.

 

(and I consider lb-mass to be an engineering abomination, the equivalent of bad grammer being adopted just because everyone mis-uses it so much)

[tvp45]

Most of the scientists living in the US also think the American (sorry, "English") units are silly, as do anyone who learned metric first. Where they live has nothing to do with it, really.

 

Amen. I normally work in American units and routinely use slugs as a mass unit because it works, i.e., lets me use the same equations as the rest of the world. I don't think it's proper and it drives everybody else crazy. Then, there's the Troy or Avoirdupois question. Can we please change? Soon?

[MrMongoose]

The ones who have studied physics will recognize that those are weight, not mass.

 

UGGGGHHHH

 

Thank god everything of importance that I do is metric!

[enapp]

I think you may be trying to determine the electrical energy it may take to move x weight.

The problem, if this is the case, is more about understanding a great number of laws such as Joules, Newtons etc.

Things like:

1. One joule is equal to the energy used to accelerate a body with a mass of one kilogram using one newton of force over a distance of one meter. One joule is also equivalent to one watt-second.

But if you are doing a classroom demonstration there is no better way than to take a ton or x weight attached to your lifting energy source ie. a spoon full of petrol, bucket of water, equal counter weight, pump or whatever else you have in mind and get the students to write their reports (on time) to give you the answer. Would also be good to ask them how much they can recover and design new ways to extract said energy and you may trip over the world's next big storage solution