# Calculating Pushup Force: Torque and Work Considerations

• Dave27
In summary, the conversation discusses the calculation of force needed to do a push up, taking into account the individual's height, weight, and arm length. The formula for torque is used and it is clarified that torque cannot be converted to force or mass. The concept of center of mass is introduced and it is determined that the weight of the arms is negligible in this context. The calculation of work done is also discussed, taking into account the length of the arms and the height difference of the center of mass. The conversation ends with confirmation that the calculations are correct.
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.

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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.

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:

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.

Dave27
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?

Last edited by a moderator:
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.

Dave27
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

Uh? Is it correct?

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.

Thank you good lemur

## What is the difference between torque and work?

Torque is a measure of rotational force, while work is a measure of linear force. Torque involves the application of a force at a distance from a pivot point, while work involves the application of a force over a distance.

## How do pushups involve torque?

Pushups involve torque because they require the use of muscles in the arms, shoulders, and chest to push against the ground at a distance from the pivot point, which is the shoulder joint. This creates rotational force, or torque, that allows the body to move in an upward motion.

## Do pushups involve work?

Yes, pushups involve work because they require the use of muscles to exert a force over a distance, which is the movement of the body from a lower to a higher position. This work is necessary in order to overcome the force of gravity and lift the body off the ground.

## How does the body maintain balance during pushups?

The body maintains balance during pushups by using multiple muscle groups to stabilize the body. The core muscles, including the abdominal and back muscles, play a crucial role in maintaining balance and preventing the body from tipping over during the pushup motion.

## Are there any other factors that affect torque and work during pushups?

Yes, there are several other factors that can affect torque and work during pushups. These include body weight, body position, and the surface on which the pushups are performed. For example, performing pushups on an unstable surface, such as a stability ball, can increase the amount of work and torque required to complete the exercise.

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