Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When I do pull-ups, I feel I push down to the bar. But does the bar really take more weight than just hang down?

For people who don't know pull-ups and hang down, here is an illustration.

Left: Hang Down-----------------------Right: Pull ups


So, does in right picture the bar take more weight than the left one?

Thanks in advance.

share|cite|improve this question
If the picture shows a stationary person: no, the bars are taking the same weight. (Yung's answer deals with the acceleration of moving up, which may be what you're really asking) – Tim S. Apr 30 '14 at 14:30
That guy is ripped. – iamnotmaynard Apr 30 '14 at 14:30
@iamnotmaynard actually he's just been flayed. It turns out someone recently removed his skin and fat layers. That's also why he has no face. – Racheet Apr 30 '14 at 15:07
@YungHummmma's answer is essentially correct, but fails to address the "feeling" of pushing down on the bar that you mention. You must apply a downward force on the bar in order to keep yourself suspended; however, this is just to counteract the downward force of gravity, and the bar feels (essentially) the same total amount of force regardless of whether you're at the top or bottom of a pull-up. (Gravity does diminish with distance, though; see the first couple of comments on YungHummmma's answer.) – Kyle Strand Apr 30 '14 at 22:20

Yes, you do put more weight on the bar.

Your mass here is $m$.

To hang, you simply put force $F=mg$ on the bar (and equivalently, the bar puts that same force on you, so the forces cancel and you don't move anywhere).

For you to move upwards at some acceleration $a$, now you need the net force on you to equal $ma$: $\sum F = ma = F_{bar}-mg$. So, $F_{bar} = m(g+a)$.

This is the force the bar must put on you to accelerate up, so it's the force you put on the bar.

Note that in the case of hanging ($a=0$), it reduces to the first case, $F_{bar} = mg$.

Edit: I'll add a tiny caveat because it can possibly be confusing. You'll notice that in this scenario, there's only more force on the bar if you're accelerating up. If you manage to go up at a totally constant speed, the force should be $mg$ still. I guess the reason this doesn't happen is because it's nearly impossible for a human to do.

share|cite|improve this answer
But once you're up, the slight height difference means the force is ever-so-slightly smalller than it was before ;-) – Thriveth Apr 30 '14 at 13:50
@Thriveth I dare you to design a device able to measure the difference between the force one apply to the bar when hanged down and the force applied to the bar when one is up :) – ChocoPouce Apr 30 '14 at 14:01
@Cruncher if there is no acceleration then, no, there is no increase in force. For example: an object suspended from the bar by a string does not weigh less than an object suspended from a spring of the same mass, no more or less force is exerted on the bar. – David Wilkins Apr 30 '14 at 16:04
@ChocoPouce That's an engineering problem, I'm a physicist :-p – Thriveth Apr 30 '14 at 20:24
@Cruncher: if they can't move then they're applying no additional force to the bar. They're redistributing some forces within their arms, maybe taking some weight off ligaments and onto muscles or maybe just creating additional balancing forces by the static force they exert. But it's all to no overall effect on the bar. In practice of course it doesn't take much to at least bounce up and down a little bit, which will change the force on the bar a little bit. In fact just kicking your legs around will probably move your centre of mass up and down, so the force on the bar will change. – Steve Jessop Apr 30 '14 at 20:24

Let's suppose you are pulling yourself up and down in approximately simple harmonic motion so your height above the ground will be give by:

$$ h = h_o + h' sin(\omega t) $$


Your acceleration is just $d^2h/dt^2$, and the force is just your mass times the acceleration, so the force due to your motion will be:

$$ F = - m h' \omega^2 sin(\omega t) $$

and the total force on the bar is:

$$ F = - m \left( g + h' \omega^2 sin(\omega t)\right) $$

So in this model the force is greatest at the bottom of your cycle as you are slowing your decent and accelerating yourself back up. It is lowest at the top where your ascent is slowing and you're allowing gravity to pull you back down.

share|cite|improve this answer

If you are accelerating upward then the force on the bar will be greater. F=ma.

share|cite|improve this answer

The act of pulling oneself up (or holding oneself up) will require a change in force applied to your arms, or at least a redistribution of it when holding oneself up. This will change the force applied to a portion of the bar- specifically, where the hands are, as they will almost assuredly grip the bar harder than when hanging limply. The overall weight applied to the bar, as a whole, won't have changed, but by gripping tighter, additional force has been applied.

This, of course, in addition to the increase in force required to actively accelerate upwards.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.