Physics of Fitness Fridays - Incline vs Decline Push-Ups

Welcome to Physics of Fitness Fridays! Here I’ll take a random fitness or exercise move and explain the physics behind it. The images are created for my Instagram feed, which you can find here. In this blog, I’ll get more into the math details for all of you nerds out there.

A common way to change the intensity level of pushups is to elevate the hands or the feet. But why are incline pushups easier and decline pushups more difficult? Let’s take a look…

(Note: I’m showing the plank position but I will refer to this as a pushup position to keep things simple.)

Let’s say that you are holding your pushup position as you see here on the right. (Look ma! I’m holding a plank forever!) The force of the move comes from my hands pushing into the ground.

So what forces are acting on my body? My hands are pushing into the ground and the ground is pushing back at my hands with an equal and opposite force, labeled here as Nhands. Similarly, the ground is pushing back at my feet with a force Nfeet. The “N” here stands for the normal force, or perpendicular force from a surface.

Also, gravity is pulling down on my body. I represent this as Wcom, the weight of my body at its center of mass (COM). There are also the forces of friction acting on my hands and feet, but for these purposes I’ll assume I’m on a frictionless surface.

Since I am in equilibrium, all of the forces on me cancel out. So the normal forces pushing on my hands and feet equal the weight of my body pulling down, or: Wcom = Nhands + Nfeet.

So we know that the forces balance out. But how much force is in our hands and how much is in our feet? We can figure this out with some help from the second rule of equilibrium that states that all torques acting on a body must also cancel out.

What is torque? In a nutshell, it’s a force provided at some distance from a point of rotation. In our pushup example, that is the pivot point at our ankles. Torque (T) is calculated as the cross product of the distance from the rotation point (r) and the force applied (F): T = r x F.

So in our diagram…the torque from Nfeet is zero since it’s being applied at our pivot point (that is, r = 0). That leaves the torque from gravity applied at the height of my center of mass (Hcom) and the force on my hands applied at the height of my shoulders (Hs).

Now imagine if you pulled on the little force arrows but my feet stayed in place. The force from gravity (Wcom) would make me spin clockwise around my feet, while Nhands would make me spin counterclockwise. That means that the torques are opposite. And remember, equilibrium tells us they are equal, so: Hcom x Wcom = Hs x Nhands.

So what is a cross product? Here, the cross product means we aren’t interested in the full length, just the component of length that is perpendicular to the force applied; in this case, since the forces applied are vertical, we want the horizontal components of Hcom and Hs. If we look at the angle theta made between the arms and the body, we see that the body makes a right triangle. Thus, the horizontal distance components are Hcom*sin(theta) and Hs*sin(theta).

Since the torques are equal, we have:

Wcom* Hcom*sin(theta) = Nhands*Hs*sin(theta).

Since we’re interested in the force on the hands, we solve for Nhands:

Nhands = Wcom * (Hcom/Hs)

We know that Wcom, or our weight, stays constant. The height to our shoulders (Hs) is also constant. But what about Hcom? Does that stay the same?

What—and where— is our center of mass? Simply put, the center of mass (COM, or Rcom) is the average location of all of the mass in your body; it’s where your body can be balanced. Imagine if you could cut up your body into a bunch of cubes, a la Resident Evil. (Ew. Maybe not.) Each chunk of body (ew!) has a mass and position from some set point in space. You multiply the two together, add them all up, then divide by that total mass and voila!

Rcom = sum(tiny body pieces * each piece’s distance from some point)/[total body mass]

Your body’s COM changes depending on how your body is “distributed” in space. Sometimes your COM is located in your body, sometimes it’s not.

So here’s a more concrete example: Your arms make up ~10% of your total body mass. As you raise your arms at some angle, your COM shifts upward (and to the right a little). When your arms are completely overhead, your COM no longer has a horizontal component, and is even higher than when you started.

So let’s put this all together…in an incline pushup, the angle between our arms and torso decreases, the COM shifts downward, and thus from our earlier equation for Nhands, the total force decreases. Thus, incline pushups are easier.

The opposite happens for a decline pushup: The angle between arms and torso increases, the COM moves higher, thus the force on the hands increases and they are harder to perform!

On the left shows different inclines and the percentage of your body weight supported by your hands. Of course, this can change based on your actual body shape, but it gives you a general idea. As you can see, in a normal pushup your hands support about 65% of your bodyweight (reference1, reference2). If you take it to the extremes, you’re either performing pushups against a wall (almost all of your bodyweight is supported by your feet), or you’re doing a handstand pushup (all of the weight is supported by your hands).

So if you’re building up to a full pushup, just start with your hands high and keep moving lower as you get better. And if you can do a full pushup, challenge yourself by getting your feet higher and higher!

So there you have it! You may have more questions, like, “how do you calculate the actual center of mass?”, or “how did they determine that your arms are 10% of your bodyweight?”, or “wait, what about wall pushups, or if my hands are on an incline?” I’d love to answer those, but this blog is already way too long, so I will save these for another time.

Questions? Comments? I wanna hear them? You can “weigh in” (ha! get it?) down below…