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I came up with a random question which looked simple, but I can't work out the answer. The question is, why are inclined push-ups harder? And by inclined push-ups I mean push-ups with your legs on a bench or something.

I did try various ways to solve this, including moments and energy arguments, which I won't post here, but they are of no success. I made assumptions that the centre of mass of your body is half-way between your feet and your arms.

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  • $\begingroup$ You're CoM is rather higher than that, but as usual one has to be very careful that you understand all the details before trying to apply physics 101 principle to biological systems. You should be asking yourself about the geometry of bones and muscles in the two postures as well as basic static load issues. $\endgroup$
    – dmckee --- ex-moderator kitten
    Oct 31, 2015 at 22:55
  • $\begingroup$ @dmckee is right. One place to start is to ask yourself if you can bench as much on an inclined bench as on a flat bench (I know I can't). That seems to suggest that your body is just "better" at one type of motion--the muscles are better designed to push out than push up. $\endgroup$
    – Jahan Claes
    Oct 31, 2015 at 22:57
  • $\begingroup$ @dmckee Yeah I was thinking if the muscles that are used for declined push-ups are generally weaker than those for regular push-ups. But could anyone confirm that in terms of physics and the assumptions I've made, declined one shouldn't be any harder? Or have a made a mistake somewhere.. $\endgroup$
    – RelativisticDolphin
    Oct 31, 2015 at 22:57
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    $\begingroup$ @VincentL if you want a physics-101 treatment, treat the body as a straight rod pivoted at the feet, and ask how much torque is needed by your hands to push your head upwards a given distance. You'll find that declined pushups are harder. $\endgroup$
    – Jahan Claes
    Oct 31, 2015 at 22:59
  • $\begingroup$ @JahanClaes I just get Wsin(theta) = 2 Fsin(theta) and the sin(theta)'s cancel so it's independent of the angle.. have I made a silly mistake? $\endgroup$
    – RelativisticDolphin
    Oct 31, 2015 at 23:01

3 Answers 3

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I decided to get out the bathroom scale and measure my weight at 3 different inclines; level, 16", and 30". I used a digital scale and I weighed in at 198 lbs. With 3 measurements at each hight here are the results I came up with.

Everything in lbs

30" incline 159.6 157 152.4.

16" incline 144.8 143.2 144.6

Level 137.4 139.6 144.8

Obviously they vary a decent amount likely due to my straining and slight variations in posture. Another possible factor in the accuracy is how my feet were positioned on the surface they wee on. This would include If they were hooked and some of my weight were "hanging" or the fact that the scale was always level and not in line with the incline. One more factor to consider is that the muscle you are using to lift may not be experiencing its own weight, only everything above or beyond.

I think it would be safe to say that as your feet rise, the weight you have to lift increases. Imagine slowly increasing the height of your feet to a hand stand position where you experience your full bodies weight.

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  • $\begingroup$ I see your last point, but what's the physics explanation for that? $\endgroup$
    – RelativisticDolphin
    Nov 1, 2015 at 14:16
  • $\begingroup$ Wait, you put your feet on the scale for this? You're trying to measure the force your arms need to exert, no? BTW, I thoroughly approve of the experimental approach here. $\endgroup$
    – dmckee --- ex-moderator kitten
    Nov 1, 2015 at 16:12
  • $\begingroup$ Initially I took measurements at my feet and hands, but only posted my hand measurements which was most relevant. The comment about my feet on the scale slipped through and should have simply been worded as the surface my feet were on. I will edit. Thanks. $\endgroup$
    – hiddenharmonizer
    Nov 1, 2015 at 17:05
  • $\begingroup$ I think it might make more sense if we say, the center of gravity of your body at a specific incline. Think of tying a rope around your waste so that your suspended so that your shoulders are level with your feet when outstretched. If you want to be suspended at a certain incline you would have to adjust the position of the rope on your body. So your center of gravity has not changed, your just doing push-ups off center. Perhaps someone with better formula experience can give us a mathematical representation. $\endgroup$
    – hiddenharmonizer
    Nov 1, 2015 at 17:20
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    $\begingroup$ I think the answer is two-fold, and you can test this in the gym with a machine or bench. The angle increases your body weight, yes, but it also changes the primary muscles under load. As the angle increases, you recruit more of your shoulder and less of your pectoral muscles (well, you start using the upper parts more). The shoulder is weaker than the chest in most people. $\endgroup$
    – Phil
    Dec 5, 2016 at 3:15
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Now this could be naive, but I'm inclined to think that any of the shifting around as a result of changing the angle of declination for a pushup shouldn't matter in terms of the net external force the hands have to produce. I would wager any change of measurement you find on a scale has more to do with fluids shifting in the body, and thus the center of mass changing.

For example, if one considers a very large simplification of the body as a rigid body(not a bad approximation if the athlete has good core strength) with two torques (F_g = weight, F_n = normal) acting on an axis of rotation, you can see why. Suppose there are two situations, one in which the 'rigid body' is parallel with the ground (normal pushup), and the second in which the 'rigid body' is tilted with respect to the ground:

enter image description here

In both situations the gravitation force/weight acts through the center of mass, and the force of 'pushing' is felt via the normal force in the rigid body. In the second situation where the rod has been rotated, the torque about the axis has changed, but it changes the same for both the weight and normal vector.

This means that relatively, nothing has changed between the forces, you've made gravitational torque less 'strong' but also the normal torque less 'strong' to the same degree. Your arms will have to push with the same force as they were in the parallel case. However, this does mean there is now a compressive component to the forces, but the internal forces don't matter for a rigid body, nor would effect the torque the arms in a pushup would have to produce. The mechanics on the arms change I'm sure, and probably change the moments on the elbow and shoulder, but the overall net force required to rotate the rod/body has not changed. Therefore, I think any change in weight measure on a scale would be from a non-mechanical influence.

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  • $\begingroup$ Please just improve your answer without stating what was wrong. Furthermore, please consider decreasing the size of your images, meta.stackexchange.com/questions/165795/… $\endgroup$
    – Semoi
    Apr 2, 2020 at 21:29
  • $\begingroup$ @Semoi Thanks for the tips. I'm new to stackexchange! $\endgroup$
    – John Buggeln
    Apr 3, 2020 at 1:20
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I'm obviously late to this. But I think the original posts calculations are probably correct. The angle of the body doesn't change the force required to rotate the body about the feet, assuming the arms are vertical in all instances.

Typically an incline push up is done against a fence or chair, in both instances the person likely holds their arms closer to perpendicular to the body rather than vertically. The angle of the arms relative to gravity makes a big difference to the maths here.

I tested this on my scales at home and if I made a concerted effort to hold my arms vertically for an incline and standard push up, I got very similar results.

So I think the change in effort required at different body angles comes more from the different muscles required ( as already highlighted here) and the inconsistencies of the angle of the arms relative to gravity.

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