Calculating Pushup Force: Torque and Work Considerations

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The discussion focuses on calculating the force required to perform a push-up, considering factors like height, weight, and arm length. The initial torque calculations suggest that lifting approximately 37kg is needed when starting flat, while adjustments for body position reduce this to about 25kg. Participants emphasize the importance of determining the center of mass to accurately calculate the necessary torque and force. The final calculations for work done during the push-up yield around 349 Joules, confirming the accuracy of the approach. Overall, the thread highlights the complexities of biomechanics in exercise physics.
Dave27

Homework Statement


I wanted to calculate the force I need to do a push up. I'm 1.7m tall and weight 72kg. The distance between the top of my head and my shoulders (where the shoulder joint would be) is around 0.3m and my arm length is 0.63m.

Homework Equations

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The Attempt at a Solution

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So:
(T as in torque)
T = 0.63m * 140/170 * 72kg * 9.8m/s2 = 366.093Nm.
Which would be like lifting about 37kg.

That would be asumming I start completely flat on the floor. If I was on my feet with my torso completely perpendicular to the floor and the distance between the floor and my torso is 0.2m. I'll have to substract that to my arm length (0.43) and I'll get 249.873Nm which would be like lifting about 25kg

Im confused. Is this the same Energy (as in Work) I'm expending??

Also, I went a little more in depth and tryed to calculate the same but just for a part of my arm. If the distance between my shoulder joint and my elbow joint is 0.27m and the angle between them is 45 degrees when I start doing the push up. Then the distance between my elbow joint and my torso should be 0.27m * sin(45) = 0.19m which is around 44% of the total distance which yields:

T = 249.873Nm * 0.44 = 109.9Nm which would be like lifting 11kg for that part of my arm. I also thought about calculating the distance that the arm has to rotate over the elbow joint but I'm not sure.

Tell my If my calculations are correct please. Thanks in advance.
 
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Dave27 said:
T = 0.63m * 140/170 * 72kg * 9.8m/s2 = 366.093Nm.
Which would be like lifting about 37kg.
I don't understand what you are doing here but that cannot be right.
The arm length shouldn't appear there, and with that formula your torque would go down if you are taller, that doesn't seem right.
Dave27 said:
Which would be like lifting about 37kg.
You can't convert torque to a force or mass like that.
 
mfb said:
I don't understand what you are doing here but that cannot be right.
The arm length shouldn't appear there, and with that formula your torque would go down if you are taller, that doesn't seem right.You can't convert torque to a force or mass like that.
I was visualizing something like this:
p3.png


If is not like I'd put it then I honestly have no clue.
 
First you'll have to figure out where your center of mass is. Then you can calculate how much torque that provides with your feet as rotation axis. Your arms will counter this torque, which you can convert to a force afterwards.
 
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Well, that seems a bit complicated. How would I go about finding my center of mass? THe only idea I have is viewing my body as a system where I can separate each limb, find its center and then joining everything together
 
EDIT:
Ok, let's suppose my CM is a little bit over mid section, say... 1.1m from my feet. Then

Tfeet = 1.1m * 72kg * 9.8m/s^2 = ~776Nm
Tarm = 1.4m * F;

And Tarm has to be greater than 776Nm because otherwise there would be no acceleration so..
1.4m * F > 776Nm
F > ~554N

If I tryed to find how much acceleration do I need to produce over 554N, do I have to take into account the weight of my arms?
 
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There are studies where the center of mass typically is, but in this context "a bit above the center" is probably a good approximation.

The weight of the arms is probably negligible. They accelerate differently, but typically acceleration will be small compared to the gravitational acceleration of Earth anyway and the dominant force comes from holding yourself up.
 
mfb said:
There are studies where the center of mass typically is, but in this context "a bit above the center" is probably a good approximation.

The weight of the arms is probably negligible. They accelerate differently, but typically acceleration will be small compared to the gravitational acceleration of Earth anyway and the dominant force comes from holding yourself up.
Interesting. Well I supose my calculations were correct.

If I wanted to calculate the work done then I justo have to multiply 554N by the distance the CM traveled, right?
 
No, multiply it by the length of your arms, that's where the 554 N act.

You can cross check this by multiplying your body weight by the height difference the center of mass makes.
 
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  • #10
So
Warm = 0.63m * 554N = ~349J

Calculating for the CM. First I find the angle (in degrees) the arm lengh makes with the torso. Get the sine of that angle and multiply it by the distance from my heels to the CM Then the heigth from floor to CM should be:

h = 1.1m * sin( arcsin(0.63/1.4) ) = ~0.495m

and of course the force for the CM should be 776/1.1. So then

Wcm = 0.495 * (776/1.1) = ~349J

There slightly off by some decimal points but that just might be some rounding errors I made. Anyway,if you tell me this is correct then everything should be clear for me now
 
  • #11
Uh? Is it correct?
 
  • #12
Dave27 said:
So
Warm = 0.63m * 554N = ~349J

Calculating for the CM. First I find the angle (in degrees) the arm lengh makes with the torso. Get the sine of that angle and multiply it by the distance from my heels to the CM Then the heigth from floor to CM should be:

h = 1.1m * sin( arcsin(0.63/1.4) ) = ~0.495m

and of course the force for the CM should be 776/1.1. So then

Wcm = 0.495 * (776/1.1) = ~349J

There slightly off by some decimal points but that just might be some rounding errors I made. Anyway,if you tell me this is correct then everything should be clear for me now
Yes, that all works.
 
  • #13
Thank you good lemur
 
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