Physics - Force of a Pushup

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Physics - Force of a Pushup

Postby philsov » Tue Apr 24, 2012 3:52 pm UTC

Some random question that I've long since forgotten how to solve.

How much force is necessary from a pushup to take a body from about 5 degrees flat up to 45 degrees? Or up to perfectly vertical?

Random variables/simplifications:
- 160 lb load, evenly distributed over 6 feet (72.57 kg spanning 182.88 cm)
- Feet are flat and stationary; act as the pivot point
- Pushup is a purely vertical force

Anyone help sate my curiosity?
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Re: Physics - Force of a Pushup

Postby Giallo » Tue Apr 24, 2012 4:59 pm UTC

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Re: Physics - Force of a Pushup

Postby philsov » Tue Apr 24, 2012 5:13 pm UTC

Ah, yes. Thanks for the clarification.

Energy. I wish for the body to work.

Through browsing I suspect acceleration would affect the amount of work required since you're constantly trying to overcome gravity. I suppose the necessary distance traversed occurs over the span of 1 second?
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Re: Physics - Force of a Pushup

Postby starslayer » Tue Apr 24, 2012 5:54 pm UTC

Getting the work done against gravity is fairly simple; it's just the difference in potential energy between the end and starting points.

Taking the feet to be at height z = 0, and the body to have mass m and length l, I have the potential energy of a particular body segment dU as dU = gzdm = r\sin(\theta)gdm = r\sin(\theta)g\mu dr, where r is distance along the body measured from the feet, \theta is the angle of the body from the floor, and \mu is the linear density m/l of the body. Then the total potential energy of the body is
U = \int_0 ^l r\sin(\theta) g \mu dr = \frac{\mu l^2}{2} g \sin(\theta) = \frac{ml}{2}g \sin(\theta)


Substituting in and taking the difference of potential energies, I get ~400 J to rotate the body to 45 degrees, and ~600 J to get it vertical.
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Re: Physics - Force of a Pushup

Postby idobox » Wed Apr 25, 2012 9:46 am UTC

45º for a push up looks like a lot.

The math you used is kinda complex for what we need, and is correct only if the mass of the body is equally distributed along its length (which is a decent approximation). You just need to measure/compute the difference in altitude of the body's center of mass, which is usually a bit below to the bellybutton. The energy is then m*g*delta z
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Re: Physics - Force of a Pushup

Postby philsov » Thu Apr 26, 2012 4:11 pm UTC

I get ~400 J to rotate the body to 45 degrees, and ~600 J to get it vertical.


Wonderful! Thanks.

I had initially thought it'd take even more energy by relation for the second half of the trip, but I guess I was (improperly) using the cosine function.

The math you used is kinda complex for what we need, and is correct only if the mass of the body is equally distributed along its length (which is a decent approximation).


And if we were to compare a male versus a female, the male needs more energy because the broader shoulders and narrower hips takes the load farther from the pivot point?
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Re: Physics - Force of a Pushup

Postby Giallo » Thu Apr 26, 2012 7:32 pm UTC

Yes. Also, men normally weight more than women.
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Re: Physics - Force of a Pushup

Postby krogoth » Fri Apr 27, 2012 3:38 am UTC

center of gravity being higher in males always reminded me of this 'game' where you sit on your knees (Japanese style) arms behind your back and put a matchbox on the floor in front of you and try tip it over with your nose.

your belly button would normally be just above the knees and for males with big enough shoulders they can't lean down far enough without falling over.

(I used to be a runner so my big calves/smaller shoulders allow me to do it)
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