arfgabdk Posted February 20, 2008 Posted February 20, 2008 I would like to have some help figuring out how many people that is needed in order to lift a car (and how you would calculate it). Here is an example of the problem: Mass: 500 kg Length of the bottom of the car: 2,5 m Width of the bottom of the car: 1 m Height of the car: 1,5 m Let's assume the car is shaped as a regular box. Let's assume that the mass is distributed evenly throughout the box. All the people to lift are placed around the car with equal distance between each other (if you drew a polygon connecting everyone, all the sides of the polygon would have the same length). Everyone is holding underneath the bottom of the car. Let's assume they all lift vertically. Everyone lifting is able to lift the same amount of weight in a "dead-lift" (like when the bodybuilders lift a heavy ball, where they use their thigh-muscles to lift the weight). They can all lift 80 kg from ground level to 1 m above ground level this way. Let's assume that the tires are pumped so hard, that they only touch the ground with 1 cm^2 of tire each. We can forget about the spring effect in the tyres. No tools can be used. How many people will it take to lift the car 0,2 m off the ground? If I have forgotten to supply some information, please let me know, and I will. Thanks a lot for the help.
swansont Posted February 20, 2008 Posted February 20, 2008 80 kg/person. You have N people. 80*N is the mass they can lift. It must be >= 500 kg. Solve for N. Make sure it's an integer.
Psycho Posted February 20, 2008 Posted February 20, 2008 Isn't half that information completely irrelevant, it is just weight of the car divided by the number of people lifting it. Unless you want the energy they have to exert, which would be something completely different.
5614 Posted February 20, 2008 Posted February 20, 2008 I suppose you might consider the dimensions of the car if you were to need loads of people, who might not, potentially, be able to fit around the car. But this is clearly not the case. A good thing in questions like this is to single out the information you need. In this case all you need is that the car weighs 500kg and that 1 person can lift 80kg (so they're strong people!). As for the fact that the people can lift 80kg up by 1m, and you only want to raise the car by 0.2m... you can't really use that information because you can't relate lift height to lift weight (you would need more information. For example you cannot say that because it's only 1/5 as high they can lift 5x more weight or anything of the sort. As you cannot use this you cannot really account for it in the maths. Although if this is an estimation problem then maybe if it were, say, 10.3 people can lift the car up 1m, then round that down to 10 people can lift it 0.2m... but that's really iffy and you can't really justify it, so I wouldn't say it. Incidentally if it were 10.3 people needed, and obviously you can't have 0.3 people, then 10 would not be enough, so you would need to round up to 11.
Fuzzwood Posted February 21, 2008 Posted February 21, 2008 I suppose you might consider the dimensions of the car if you were to need loads of people, who might not, potentially, be able to fit around the car. But this is clearly not the case. A good thing in questions like this is to single out the information you need. In this case all you need is that the car weighs 500kg and that 1 person can lift 80kg (so they're strong people!). As for the fact that the people can lift 80kg up by 1m, and you only want to raise the car by 0.2m... you can't really use that information because you can't relate lift height to lift weight (you would need more information. For example you cannot say that because it's only 1/5 as high they can lift 5x more weight or anything of the sort. As you cannot use this you cannot really account for it in the maths. Although if this is an estimation problem then maybe if it were, say, 10.3 people can lift the car up 1m, then round that down to 10 people can lift it 0.2m... but that's really iffy and you can't really justify it, so I wouldn't say it. Incidentally if it were 10.3 people needed, and obviously you can't have 0.3 people, then 10 would not be enough, so you would need to round up to 11. Sure you can: use a midget
CaptainPanic Posted February 25, 2008 Posted February 25, 2008 I thought that the correct question is: "What is the color of the car?"
insane_alien Posted February 25, 2008 Posted February 25, 2008 I thought that the correct question is: "What is the color of the car?" Blue of course. its the SFN party car. hence the big dent at the back from blikes parking.
John Cuthber Posted February 25, 2008 Posted February 25, 2008 One person can lift the car. He just has to stand next to it (he may need to jump if his CoG is lower than that of the car, but that seems unlikely). Gravity will atract the car towards his CoG; provided that's above the car's, the car wil rise on its springs. Just don't ask me how to measure the extent to which this stretches the car's suspension.
YT2095 Posted February 25, 2008 Posted February 25, 2008 0.2 metres off the ground would have to be one hell of a FAT BA$TARD! hanging over it
YT2095 Posted February 25, 2008 Posted February 25, 2008 nada, not until I find a suitable Belly-Wheel on eBay for him *sigh*
TWJian Posted March 5, 2008 Posted March 5, 2008 "They can all lift 80 kg from ground level to 1 m above ground level this way." Doesn't the appearance of force (through 80kg*g) and displacement suggest that this is a work-energy problem? 80*9.81N of force and 1 meter of vertical displacement mean each person can do 80*9.81J of work, or 80*9.81n J of work, where n is the number of people doing the lifting. Lifting the car would increase the gravitational potential energy of the car, and therefore you need to input work to achieve that. Therefore, n(80kg*9.81m/s^2)= 500kg*9.81 m/s^2*0.2m Cancel out g or 9.81m/s^2 since it appears at both sides of the equation. (I would have removed it in the first place before I started calculating, but it seems clearer this way) (Note: J=kg*m^2/s^2) Solve for n. Note that other information are irrelevant unless you want to determine if you can conceivably fit in the n amount of people needed to lift the car. (i.e.: Is it possible to have enough people to lift the car vertically by 0.2m?)
swansont Posted March 5, 2008 Posted March 5, 2008 No, I think the work is a misdirection, just as other data that are given, such as the dimensions, are. If you cannot exert a force larger than the weight, you will not lift the car.
TWJian Posted March 6, 2008 Posted March 6, 2008 Assuming the 'correct' way is to solve as the force needed: 80g*n=500g Solve for n, and you get the force needed to counteract the weight of the car (the force exacted by gravity) Which is the force you need to apply to make the car effectively weightless. This just means the car will be just barely touching the ground (seems more like it's floating) with F(normal)=0. Problem is, the car is not going anywhere at all, and you are not actually lifting the car. The question asked "How many people will it take to lift the car 0,2 m off the ground?" Try to solve for the 0.2 m part with Newton's 2nd Law. (You cannot) Even if you put in more force to accelerate the car(initial acceleration), the time required for the car to arrive at z=0.2 m is not given, and you can't solve it at all. Unless of course, you want to launch the car as a projectile and set the maximum vertical displacement as 0.2m... but.. Isn't assuming the whole question is a work-energy question far simpler? I am going to apply Occam's razor to this problem, and hence I'm going to use the law of conservation of energy to solve this, unless there is a better method of computing the amount of people needed to lift the car (a given mass) by a given height. (a given displacement). Newton's second law cannot do it, and therefore the Law of conservation of energy should be used, sans the availability of a better solution.
swansont Posted March 6, 2008 Posted March 6, 2008 The 0.2 m is another misdirection. As long as it's less than a meter, the actual value doesn't matter. I interpret the 1 meter as the physical limit, not a work value: it's a dead-lift — that's how far you can lift using your legs. Any higher and you have to use your arm muscles, which tend to be weaker.
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