# Physics Question

May 25, 2006 10:58 AM Subscribe

Simple physics question.

If I do a push-up with both feet and both hands on the ground each point in contact with the ground supports 1/4 of my weight. So if I do a push-up with two hand and only one foot on the ground is each of the points in contact with the ground then supporting 1/3 of my weight, and through that have I effectively increased the weight that my arms will be lifting?

If I do a push-up with both feet and both hands on the ground each point in contact with the ground supports 1/4 of my weight. So if I do a push-up with two hand and only one foot on the ground is each of the points in contact with the ground then supporting 1/3 of my weight, and through that have I effectively increased the weight that my arms will be lifting?

Best answer: Well, when you're doing a pushup, your arms and legs aren't supporting exactly the same amount of weight, because your body weight isn't distributed evenly. Your arms are supporting more because your upper body weighs more than your lower body. One way to figure out exactly how much is being supported by each limb is to do pushups with your hands and feet on bathroom scales.

posted by number9dream at 11:12 AM on May 25, 2006

posted by number9dream at 11:12 AM on May 25, 2006

You only support 1/4 the weight if you are symmetrical and your weight is distributed evenly. But it's probably close enough.

The way I would solve this is to use some Statics. Assume that you are at least symmetrical left to right and assume you are a beam with a support at each end (arms and legs) with your weight as a lumped force pointing down at your "centroid" see my awesome ASCII Freebody diagram below :)

........a............ b

X======O======X

|R1..........|W...........|R2

| is the force,

R1 is your leg contribution

R2 is your arm contribution

W is your weight.

posted by johnstein at 11:17 AM on May 25, 2006

The way I would solve this is to use some Statics. Assume that you are at least symmetrical left to right and assume you are a beam with a support at each end (arms and legs) with your weight as a lumped force pointing down at your "centroid" see my awesome ASCII Freebody diagram below :)

........a............ b

X======O======X

|R1..........|W...........|R2

| is the force,

R1 is your leg contribution

R2 is your arm contribution

W is your weight.

posted by johnstein at 11:17 AM on May 25, 2006

Best answer: If you do a push up with one foot on the ground. That one foot will support twice the weight it did when you had both feet on the ground and your arms will not be effected at all.

posted by zgott300 at 11:18 AM on May 25, 2006

posted by zgott300 at 11:18 AM on May 25, 2006

I don't think each of your contact points is supporting 1/4 of your weight. Your center of weight isn't directly in the center of your body vertically and horizontally. Like, if you made your body stiff and had to balance your body on a pole, you wouldn't put the end of the pole on your belly button, you'd put it higher probably.

So it depends on where your weight is centered, I think. If you can bend your leg around so your foot is on your head and shift your center of gravity toward your arms, then your arms will have more weight.

But it has been 10 years since I've had physics. Physics geniuses, correct me. Man, I'd love to take a physics class again.

posted by orangemiles at 11:18 AM on May 25, 2006

So it depends on where your weight is centered, I think. If you can bend your leg around so your foot is on your head and shift your center of gravity toward your arms, then your arms will have more weight.

But it has been 10 years since I've had physics. Physics geniuses, correct me. Man, I'd love to take a physics class again.

posted by orangemiles at 11:18 AM on May 25, 2006

Response by poster: I realize the weight isn't distributed evenly.

What I am wondering is if removing one support increases the load on the other supports and by that increases the resistance of the push-up. Making the arms, in effect, lift more.

posted by 517 at 11:18 AM on May 25, 2006

What I am wondering is if removing one support increases the load on the other supports and by that increases the resistance of the push-up. Making the arms, in effect, lift more.

posted by 517 at 11:18 AM on May 25, 2006

Yeah, what everybody else said. I took too long to formulate my answer. When I started there weren't any answers yet. Thanks for confirming my thought, folks.

posted by orangemiles at 11:19 AM on May 25, 2006

posted by orangemiles at 11:19 AM on May 25, 2006

Ever see a picture of a top fuel dragster? Four wheels, big car chassis. Two wheels are immense; two are really tiny. That's because the weight is distributed unevenly, and the two big tires carry most of it.

Because of the way things in your body are balanced, lifting one of your legs during pushups will increase the pressure on the other leg, but not really increase the pressure on the arms.

You want to do that, you use only one arm at a time. (Good luck!)

posted by Steven C. Den Beste at 11:28 AM on May 25, 2006

Because of the way things in your body are balanced, lifting one of your legs during pushups will increase the pressure on the other leg, but not really increase the pressure on the arms.

You want to do that, you use only one arm at a time. (Good luck!)

posted by Steven C. Den Beste at 11:28 AM on May 25, 2006

Best answer: umm. didn't mean to post so soon...

Anyways. a and b are the distance from your 'center of gravity' to your arms and legs.

Newton says the sum of the vertical forces have to be zero, so that means R1 + R2 - W = 0. (weight pulls down, your arms and legs push you up). He also says the sume of the moments have to be zero as well. Basically, the distance of the vertical forces from some stationary point. Let's choose your legs at R1.

so W is 'a' distance away and R2 is "a+b" away. so we can say that a*W = (a+b)*R2.

We can estimate W and 'a' and 'b'. Let's just assume that we have a perfectly symmetrical being and a=b. this reduces our equation to:

R2 = 1/2 W, which leads to R1 = 1/2 W as well. which if you assume each arm and leg do equal work, that will be 1/4 W for each limb.

But if you remove one leg, what will happen? Well, it depends on your assumption if you want to do a quick check (and someone smarter than me can give you a better answer probly).

but instead of assuming only R1 and R2. let's say that we have R1 (one leg) and 2*R2 (2 arms). Solve the above equations and you get R2 = 1/4 W and R1 = 1/2 W. meaning, each arm is still doing 1/4 the weight, while your poor leg is now doing twice the work.

so maybe doing one handed pushups would be better (2*R1 and R2). or change your center of gravity (extending legs to the side or moving one arm closer to your head and the other toward your feet).

um. sorry for the long post. bored at work. someone check my numbers since I know I probably missed something in my assumptions.

posted by johnstein at 11:29 AM on May 25, 2006

Anyways. a and b are the distance from your 'center of gravity' to your arms and legs.

Newton says the sum of the vertical forces have to be zero, so that means R1 + R2 - W = 0. (weight pulls down, your arms and legs push you up). He also says the sume of the moments have to be zero as well. Basically, the distance of the vertical forces from some stationary point. Let's choose your legs at R1.

so W is 'a' distance away and R2 is "a+b" away. so we can say that a*W = (a+b)*R2.

We can estimate W and 'a' and 'b'. Let's just assume that we have a perfectly symmetrical being and a=b. this reduces our equation to:

R2 = 1/2 W, which leads to R1 = 1/2 W as well. which if you assume each arm and leg do equal work, that will be 1/4 W for each limb.

But if you remove one leg, what will happen? Well, it depends on your assumption if you want to do a quick check (and someone smarter than me can give you a better answer probly).

but instead of assuming only R1 and R2. let's say that we have R1 (one leg) and 2*R2 (2 arms). Solve the above equations and you get R2 = 1/4 W and R1 = 1/2 W. meaning, each arm is still doing 1/4 the weight, while your poor leg is now doing twice the work.

so maybe doing one handed pushups would be better (2*R1 and R2). or change your center of gravity (extending legs to the side or moving one arm closer to your head and the other toward your feet).

um. sorry for the long post. bored at work. someone check my numbers since I know I probably missed something in my assumptions.

posted by johnstein at 11:29 AM on May 25, 2006

Response by poster: So I pulled out the scale and it looks like zgott300 has it.

posted by 517 at 11:29 AM on May 25, 2006

posted by 517 at 11:29 AM on May 25, 2006

What

If the single foot isn't in line with your center of mass, the force an individual arm carries will change (one goes up, one goes down) to keep you from rotating longitudinally. But the sum of the force on your two arms won't change.

posted by Opposite George at 11:34 AM on May 25, 2006

**zgott300**said, with a minor caveat.If the single foot isn't in line with your center of mass, the force an individual arm carries will change (one goes up, one goes down) to keep you from rotating longitudinally. But the sum of the force on your two arms won't change.

posted by Opposite George at 11:34 AM on May 25, 2006

If you lift one foot, the weight that your arms has to lift WILL increase, but not very much. It depends on where you place that raised foot. If you bend the knee, and move your foot closer to your torso, the weight distribution will move farther from the fulrum (your toes), and it will take more strength to lift. Not really enough to matter, probably.

posted by tadellin at 11:48 AM on May 25, 2006

posted by tadellin at 11:48 AM on May 25, 2006

One more approach.... a first order approximation that is also based on physics but no calculations...

If you compare the weight supported by two legs versus one, you'll roughly double the leg load on a single leg. Ditto arms. The weight distribution would not significantly change due to removing a support, so the load at the legs would roughly double with a leg lifted, and the load at the arm would roughly double with an arm lifted. Conceptually, the leg on the arm-lifted side would see slightly more and the arm on the leg-lifted side likewise, but probably not enough to warrant calculating.

Close enough for government work.

posted by FauxScot at 12:33 PM on May 25, 2006

If you compare the weight supported by two legs versus one, you'll roughly double the leg load on a single leg. Ditto arms. The weight distribution would not significantly change due to removing a support, so the load at the legs would roughly double with a leg lifted, and the load at the arm would roughly double with an arm lifted. Conceptually, the leg on the arm-lifted side would see slightly more and the arm on the leg-lifted side likewise, but probably not enough to warrant calculating.

Close enough for government work.

posted by FauxScot at 12:33 PM on May 25, 2006

May I ask how you were able to read the scale under your foot while doing a pushup? Or did you have a helper?

posted by randomstriker at 3:45 PM on May 25, 2006

posted by randomstriker at 3:45 PM on May 25, 2006

This thread is closed to new comments.

If I do a push-up with both feet and both hands on the ground each point in contact with the ground supports 1/4 of my weight?No.

Take, for example, a car. Most cars have more weight on the front tires than the rear tires, since the engine is typically in the front end of the car.

So if I do a push-up with two hand and only one foot on the ground is each of the points in contact with the ground then supporting 1/3 of my weight?No

posted by malp at 11:10 AM on May 25, 2006