1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple Physics Question

  1. Jan 15, 2006 #1
    It's been a while since I took Physics in college and I have a simple gravity-based question that should be really simple for this forum.
    Two of my friends are having a disagreement over the difficulty of doing a pushup. One claims it is harder to do a pushup when your hands and toes are on the floor and the other claims it is harder when your hands are on an aerobics step which puts them a 8 inches above the toes. I tend to agree with the latter position but I don't know how to prove (or disprove it).
    This seems like a work calculation based upon the angle of the person while elevating his plane upward at one end.
    Any help would be appreciated.
  2. jcsd
  3. Jan 15, 2006 #2


    User Avatar
    Science Advisor

    You are right if it were simply an issue of summing the moments about a point. However, when dealing with anything in the human body, it is not that easy. By raising your feet, say, 4 inches or so, you're really not changing the moment that needs to be exerted by your upper body. If you look at only your upper body, in both positions, they are pretty much the same. You are SLIGHTLY moving your total body's CG closer to your shoulders.

    What does really change is the muscle groups being called into play. By elevating your feet, your center section has to work quite a bit harder to maintain a rigid frame. There's an increase in energy expediture right there. The next thing is that the force from your shoulders and chest is being moved from the whole chest and shoulder muscles to the upper pecs and the front part of the shoulders. That's where the real increase in effort comes in.
  4. Jan 15, 2006 #3
    Push-up Mechanics

    Yes, I agree this could really get messy if I start bringing in the various muscle groups involved in the push-up and the angle of the motion on these various muscles. :grumpy:
    If I take out the impact on the core muscles, etc, and assume the body is totally flat and does not require any work to maintain that, how do I calculate the work required just by the arms in the 2 positions? Two pushups done with the same body length, toes at the same point and arm length the same? The only difference is the point at which the ends of the arm touch the floor/step.
  5. Jan 15, 2006 #4


    User Avatar
    Science Advisor

    Work is = Force*Distance. The force and distance are pretty much in the same direction in this case so you can simplify it as such.

    Power would be work/time.

    Just in case you are going to go there...remember a calorie in food terms is actually 1000 calories in terms of mechanical work.
  6. Jan 15, 2006 #5
    Push-up Mechanics


    The distance is exactly the same but the force will be "marginally" different because of the different angle. And that's what I need help identifying.

    Can you help me with the formula for the different angles (one absolutely flat 180 degrees) and one slightly less than that. What is the force required to rais the body a constant distance in these two scenarios?
  7. Jan 16, 2006 #6


    User Avatar
    Science Advisor

    You can think of it this way: As the feet progress from horizontal to it's lowest point the arms (held straight out) will form the hypotenuse of a right triangle (with some assumptions being made). So, as the body is at 180°, the force exerted is multiplied by a factor of 1. As the feet drop, the angle caused by the arms roatating slightly will cause the force exerted in the direction of the motion to be multiplied by a factor of cosβ where β is the angle created by the feet moving downwards, measured at the shoulder (see attached beautiful rendition of FBD).

    If you want to assume a static condition and calculate the force required just to hold one's self up, then sum the moments about the feet and set them equal to 0. If you want to arbitrarily assume an acceleration, then sum the moments again, but set them equal to either m*a (linear) or I*α if you want to look at it from a rotational standpoint. I think that since the movement is through a relatively small angle that the linear assumption is valid.
    Last edited: Mar 11, 2006
  8. Jan 17, 2006 #7
    Thanks Fred...

    Thanks very much Fred. Excellent explaination.

    You should be a teacher :smile:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Simple Physics Question