In the movie “Twins”, Arnold Schwarzenegger got a car alarm to turn off by lifting one end of the car. I was wondering how heavy that is assuming that the car is 1500 kg and 4m long. Is there a formula to work it out?
Take the full weight of the car. Multiply it by y/x where
x=the horizontal distance from the point where the wheels contact the ground to the point where you are lifting, and
y= the horizontal distance from the point where the wheels contact the ground to the centre of gravity
What you are doing is using a lever arm of length x to lift a weight at a point y along the lever.
well, you could always drive one end of the car onto a scale.
I’ve done a push-up stand on a scale out of curiosity of how much weight it represents when you do a push up. From there you can compare to a bench press (of course, with pushups you’re using a lot more energy because you have to stabilize your core)
hibernicus:
Hi how much would that be assuming that the distance to the wheels is 4 metres…
That depends on where the center of mass is.
First, let’s consider the force it takes to get the end of the car off the ground. There are two things to consider: the weight of the car and the weight distribution. Let’s say that your 1500 kg car has a 60%/40% weight distribution, with the front being the heavier end. If you lift the front of the car, you’re lifting the equivalent of 900 kg. It will take about 8826 Newtons of force to get the front of the car off the ground.
The length of the car comes into play when figuring how much force it would take to hold one end at a given height. The car sits at an angle when you lift the end. The force to hold up the car at that angle is the weight of that end of the car times the cosine of the angle. For a given height, the shorter the car the closer the angle will be to 90 degrees, the closer the cosine will be to zero, and the less weight you’ll have to support.
Let’s consider your 1500 kg, 4 meter car. Let’s say it has a 60/40 weight distribution, so the front end weighs about 8826 Newtons. Now imagine that you lift the front end of the car by 2 meters. The car will be at an angle of 30 degrees. The cosine of 30 degrees is about 0.866, so it will take about 8826 * 0.866 ~= 7643 Newtons of force to hold the end of the car at this height. This is the same force it would take to hold a 779 kg weight. You could get the same answer by multiplying the equivalent mass of that end of the car (900 kg) by the cosine of the angle (0.866) to get about 779 kg.
I think in the movie he lifted it about 1 metre. I’m not sure what the record is of lifting something one metre but I suspect that the example in the movie couldn’t be lifted by a real person.
it has to be measured with a scale, otherwise it can only be very crudely estimated
Stand on a scale and see what you weigh. Call that number A. While still standing on the scale, lift one end of a car and see what the scale reads now. Call that number B. Subtract A from B to obtain the weight of one end of a car.
Ernie Pickett (1968 American Olympic weightlifter) picked up a Volkswagen once and dragged it in sideways (“like it was no more than a tonka toy”) into a parking spot it wouldn’t fit into the normal way. So it’s at least possible for some people for some cars. And Arnold was pretty strong. No Ernie Pickett but still…
http://startingstrength.com/index.php/site/article/george_ernie_pickett_pt_12/P4
I imagine that the engine end is much heavier than the trunk end.
The car Schwarzenegger supposedly lifted was a 1987 Cadillac Seville.
Front wheel drive
Wheelbase - 108 in (2,743 mm)
Length - 190.8 in (4,846 mm)
Curb weight = 3352 lbs.
Weight distribution = ?
There were four possible V8 engines:
4.1L, two 4.5L, and a 4.9L
I think we’ve uncovered the only thing that is not on the internet - the f/r weight distribution of an '87 Seville???
Given that this was before CGI was all the rage in movies, Schwarzenegger would have lifted a car that had been mounted on a hydrolic lift. All he had to do was “act” like he was straining to lift the rear of the car.
Assuming equal weight distribution and the centre of gravity is at the halfway point, and the overhang is equal at both ends.
Hibernicus’ formula gives
y = horizontal distance where wheels touch the ground to the centre of the car
From front of car to front wheel will be (Total length-wheelbase) /2
=1031mm
From front of car to centre of gravity will be (total length/2) = 2423
so Y =2423-1031=1392 mm
X will be Y plus total length/2
=1392+2423
=3815mm
So Y/X=1392/3815
= 0.3648
So Arnie would be lifting something like 1,200 pounds.
However all our assumptions are on the high side which could make things more possible.
The engine skews the weight distribution pretty far towards the front, I’d presume. Didn’t he lift it from the rear?
Arnie could probably deadlift 600 back in the 70s, but not 1988, right? I’m not saying he actually performed this feat for the movie, but supposing it’s half of lisate’s high estimate, it’s within the realm of possible and believable.
A 1993 El Dorado has a weight distribution of 64/36, and a 1994 Deville has a distribution of 61/39. Both of these cars are very similar to a Seville. And since the Seville was neither rear-wheel drive nor a sports car, assuming a 50/50 weight distribution doesn’t really make sense. A 60/40 distribution is both prudent and conservative.
Also, Arnold’s deadlift record is 710 lb.
Nice second post, dude(?) - welcome to the Dope!
(Anybody else noticing an uptick in newly, natural-born Dopers lately?)
Well…Yeah. But only after you remove all of the bodies.
In high-school, four of us snuck out of the building and moved a classmate’s VW onto a woman’s lawn. (she didn’t own a car and repeatedly left notes on the VW about how she didn’t want it parked there)
Of course, you don’t need CGI to have 2 identical looking Caddys, except one is drivable and and the other lacks an engine and transmission.
To everyone who said put the front wheels onto a scale and that will tell you how much force is required to lift the front end… you are wrong.
To everyone who said the car itself is acting as a lever, so you need to consider the distance from the car’s center of mass to the fulcrum (the rear axle in this case) and the distance from the fulcrum to the point where you’re lifting (the front bumper in this case)… you are right.
To illustrate why the second answer is right and the first answer is wrong, imagine a car which has a very very short wheelbase, so short that the front wheels and the rear wheels are almost touching each other. How easy would it be to push down on the front bumper, causing the rear bumper to lift off the ground an equal amount? How easy would it be to lift up the front bumper, causing the rear to go down? Very easy indeed, even if the car is very heavy. It’s all about leverage.
If you have a 1500 kg car with a 2.8 meter wheelbase and 4.8 meters overall length, and we assume that the center of mass is exactly in the middle and the front bumper and the rear bumper are the same distance from the middle, then…
- The front bumper is 3.8 meters from the rear axle.
- The center of gravity is only 1.4 meters from the rear axle.
- Lifting the front bumper 1 meter only lifts the center of gravity 37 cm.
- Lifting the front bumper 1 meter also causes the rear bumper to go down by 26 cm.
So you actually have a lever with a mechanical advantage of about 2.7, hence lifting 1500 kg 37 cm with this lever is just as easy as dead lifting about 550 kg a distance of 1 meter. The force required is about 5400 Newtons, or about 1200 pounds.
But if you tried to lift the exact same car at the front wheels instead of the front bumper, your lever would only have a mechanical advantage of 2.0 hence it would be like lifting 750 kg which would require 7350 Newtons (about 1700 pounds). That’s what you’d be measuring if you drove the front wheels of the car onto a scale, and it’s not the same as lifting it by the bumper (which is what Arnold did in the movie).