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I have been thinking and searching about this a lot and have still got all my doubts.

My main hypothesis is that, since our body is actually rotating (and not moving) during push ups, torque must be the main factor. When we push up, we have to produce a torque which should cancel the torque due to gravity. When we do flat push up we are rotating directly against the direction which gravitational torque is acting. However, when we do inclined push ups, the direction in which we are rotating is not same as the gravitational torque, a smaller component of gravitational torque is along the direction of our rotation. So, we have to produce lesser torque to rotate at an incline.

In other words, our body is being pulled down by same force still(at an incline), but we aren't moving directly against that pull. Its like standing against a wall and doing push ups on the wall. There is no Torque against the direction we rotate (i.e perpendicular to the wall).

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  • $\begingroup$ What is an inclined push-up? A push-up on an incline? Or a push-up where your body moves at a non-vertical angle? $\endgroup$
    – Steeven
    Commented Jul 12, 2022 at 16:39
  • $\begingroup$ A push-up on an incline $\endgroup$
    – Rohit Shekhawat
    Commented Jul 15, 2022 at 12:04

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When you are doing an inclined push-up, your center of mass moves less in the vertical direction than in the normal variant. So you have to spend less energy since you elevate your center of mass only to a smaller height at every rep.

Edit: btw, your explenation I think is also correct. Torque is calculated as $\vec{r} \times \vec{F}$, where $\vec{r}$ is the vector from the pivot point to the center of mass and $\vec{F}$ is the gravitational force. In other words, the only component of the lever that is relevant is the one that is perpendicular to $\vec{F}$. In an inclined push-up, you reduce the relevant projection of $\vec{r}$ (the one that is perpendicular to $\vec{F}$, i.e. parallel to the floor), so less torque is necessary to stay in equilibrium, i.e. not fall (or "rotate") towards the ground.

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    $\begingroup$ So why are declined pushups harder? Based on your answer, it seems that inclined and declined pushups should be the same difficulty. $\endgroup$
    – DanDan面
    Commented Jul 12, 2022 at 16:11
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    $\begingroup$ (1) Please keep in mind that assessing what makes an exercise hard has also a lot to do with human physiology. I don't know the answer to your question, but it might have something to do with which muscles are involved. I could also ask the question why are one-hand push-ups harder than regular push-ups, by my explenation they should also by exactly equally hard. $\endgroup$
    – Samuel
    Commented Jul 12, 2022 at 16:16
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    $\begingroup$ (2) To add to that: what I assume is that your center of mass is moving along a perfect circle. So I think, only by intuition, a declined push-up is harder because you move more up-down than along a circle, compared to a regular pushup. In fact, it might be also wrong to assume circular motion for a regular push-up in the first place. $\endgroup$
    – Samuel
    Commented Jul 12, 2022 at 16:19
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    $\begingroup$ (3) After googling a bit, I found an explanation by someone on quora: declined push-ups are harder, because your shoulder muscles have to do more work, while in regular push-ups your chest is doing most of the work (which is a larger muscle, hence it feels easier) quora.com/… $\endgroup$
    – Samuel
    Commented Jul 12, 2022 at 16:22
  • $\begingroup$ I had my doubts about regular push-ups not being circular motion, but now I'm quite convinced that it is circular motion. Its not a perfect circular arc, but even if its straight line motion, it can be approximated to circular motion since the arc length is very small. $\endgroup$
    – Rohit Shekhawat
    Commented Jul 14, 2022 at 12:18

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