# Deadlifting the rear end of a car with a lever. What it equates to in actual weight?

I checked, and these cars (PT Cruiser) weigh $1417\:\rm{kg}$. What would the equation be to estimate how much weight he is actually deadlifting? The lever would obviously be assisting as he is translating a smaller movement to a larger one. Is the $1417\:\rm{kg}$ is also not being completely lifted as the car is tilting on the axis of it's front wheels? What would the equation look like if the inputs were the

1) length of the lever,

2) the weight of the car, and I guess

3) length of the car.

I'm guessing the longer the car the more weight he would be lifting as he is getting less assistance as that unstable centre of gravity helps him pull up the car. I also checked and the length of the car is $4.2\rm{m}$.

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I will use data for the manual 2007 PT Cruiser found here (pdf). The mass is 3,076 lbs and the front/rear weight distribution is 58/42.

Gross estimate

Let's do a gross estimate and then be a little more careful. We consider only the very start of the pull. Roughly half the mass of the car is on the rear wheels, so about half the mass of the car is lifted. The weight is lifted using a lever, and the weight is placed roughly halfway along the lever. Torque is moment times force. The torques will sum roughly to zero, so $$r F - \frac{r}{2} \frac{m g}{2} = 0,$$ where $F$ is the force applied by the lifter, $r$ is the length of the lever, and $m$ is the mass of the car. Therefore, $$F = \frac{1}{4} m g \approx 769 \ \mathrm{lb}.$$ Deadlifting 769 for 10 would put Pudzianowski's 1RM at about 1025 lbs, so this is an overestimate of the weight lifted.

Second estimate

Let's be slightly more careful. Since the engine is in the front of the car, the center of mass will be shifted forward. We will use the weight distribution from Chrysler's site, given above.Thus, only forty two percent, or 21/50, of the weight of the vehicle is being lifted. In addition, the mass of the vehicle is actually about 5/11 of way up the lever. This can be found by studying the video. Thus, $$r F - \frac{5 r}{11} \frac{21 m g}{50} = 0$$ and so $$F = \frac{21}{110} m g \approx 587 \ \mathrm{lb}.$$ Deadlifting 587 for 10 would put Pudzianowski's 1RM at about 783 lbs. Pudzianowski can deadlift more, but of course this is in the context of a competition with many events. Thus, this strength feat is roughly equivalent to deadlifting six plates for 10 in a fatigued state. Impressive indeed!

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Wow - a physics mind with knowledge of strength athletics. Formidable haha! – Mike S Apr 11 '12 at 6:14
@MitchRobertson: Glad to help. Thanks for the bounty and keep lifting! – user26872 Apr 11 '12 at 6:26
sure will. I'm going for a 240kg deadlift this month – Mike S Apr 13 '12 at 12:45

For a first approximation you can consider to calculate the initial lift, so the angles are about 90 degrees and you dont need to worry about them, it is only the position of the levers and the center of gravity of the car, and rule-of-three simple proportionality. As it is said, there are two levers in play: the car itself and the real lever. The reduction factor for the real lever is easy, as in the 3rd comment of Manishearth. The reduction factor in the "car lever" depends on locating the center of gravity of the car, and for a front engine it is going to be very near of the front wheel, so the reduction is going to be big. Perhaps 1::5, perhaps more. But composing both, a reduction of 1::15 could do for a starting guess.

In Valencia it is not strange to see the dustmen moving a bad parked car by raising it from the side of the blocked (hand-braked) wheel; usually this is done by two men, but no extra lever is involved.

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Yep, cars aren't that heavy if you only lift them from one side--cos the car itself becomes a lever :) – Manishearth Apr 7 '12 at 14:36

The equation is the conservation of moments/torque.

Basically, a torque is the product of the force applied at a point on a lever, and the distance of the point from the center of rotation (there's an angle factor as well). Forces on the same side which go in opposite directions have an opposite sign of torque, as do forces on different sides having the same direction. Basically the sign of torque can be said to measure the direction (clockwise/anticlockwise) that the force is trying to push the lever.

At equilibrium (or at a quasi-static situation where the process is slow), the sum of torques is zero.

Note that $L_{car}$ is not the length of the car, but the length of the point of contact of the front wheel from the fulcrum. Let the weight of the car be $F_{car}$. Now, we need to calculate the effective weight picked up by the man, $F_{man}$. ($\theta$ is the angle made by lever against ground)

Adding torques, $$0=\sum\tau=\underbrace{F_{man}L_{man}\cos\theta}_\text{clockwise}-\underbrace{F_{car}L_{car}\cos\theta}_\text{anticlockwise}$$

$$\therefore F_{car}L_{car}=F_{man}L_{man}\implies F_{man}=F_{car}\frac{L_{car}}{L_{man}}$$

Looking at the video, $L_{car}$ is $\approx$ a third of $L_{man}$, so the weght that he's lifting is a third of the weight of the car.

Actually, no. The fact is, the entire weight of the car isn't even on the lever.

Now, the car itself becomes a lever. And we have to solve this:

...And you'd have to solve torques for this. As well as throw in a bunch of force equations. Without knowing all the exact parameters, this becomes rather tedious.

Since this is tagged , I'll let you use the previous answer as a template to solve this(and its better to draw your own diagram--this is slightly hard to use as I haven't separated the stuff into FBDs)

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I'm not sure who added the homework tag - I'm out of school and university and I'm just a amateur bodybuilder. Thanks for your detailed answer. You reckon considering the 1/3rd of the weight the lever is assisting with and the car being a lever itself would 1/4th the weight of the car be a conservative estimate? I just want to roughly see how strong this guy is if he can do 10 reps. – Mike S Apr 5 '12 at 12:38
@MitchRobertson: Homework tag $\neq$ homework. It just means that its a sort of "give me the answer"-type question. Mouse over the tag, you'll see Applies to questions of primarily educational value - not only questions that arise from actual homework assignments, but any question where it is preferable to guide the asker to the answer rather than giving it away outright. – Manishearth Apr 5 '12 at 12:40
@MitchRobertson: $1/3$ was slightly off anyway. I just did a rough estimate of the bar length--it may be $2/5 (>1/3)$ or something. So yeah, the final result will be slightly less than a third. Not sure-- $1/4$ seems OK. You can do this experiment at home, take two sticks, tie a rock to the center of one of the sticks. The lengths of the sticks should be in the ratio of the cars width to the lever's width. Now, attach a "foot" to the leve, and place the foot on a weighing machine. Get it angled properly, and find the measured weight:rock weight ratio. – Manishearth Apr 5 '12 at 12:44