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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

Pull-Ups

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

Qmechanic
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Poomrokc The 3years
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    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
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    That guy is ripped. – Reinstate Monica -- notmaynard Apr 30 '14 at 14:30
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    @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

4 Answers4

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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:

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

Bar

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.

John Rennie
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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.

YungHummmma
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    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
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    @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
  • Hmm, and if someone can't apply adequate force to actually pull themselves up, but try to, aren't they still applying force to the bar? Shouldn't the force be greater in this case as well? – Cruncher Apr 30 '14 at 15:17
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    @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
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    @ChocoPouce That's an engineering problem, I'm a physicist :-p – Thriveth Apr 30 '14 at 20:24
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    @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
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    This answer is essentially correct but somewhat misleading given the way the question is framed. In the picture on the right, the person doing the pull-up may still be accelerating upwards, but since he's at the top of his pull-up, he's more likely accelerating downwards (or simply stopped and not accelerating), in which case, the answer is no, there is not more force on the bar. (Also, since pullups are performed starting from a resting position, there is always upward acceleration; this has nothing to do with it being "nearly impossible" for humans to rise at constant speed.) – Kyle Strand Apr 30 '14 at 22:17
  • @KyleStrand, the case of him accelerating downwards is actually also included in the equation, for $a<0$ you get $m(g-a)$ which physically makes sense. It also seems like he's asking two things now, because in his first sentence it seems like he's asking about the act of raising yourself, while at the end he asks about the second picture, where it seems like the shredded dude is probably at the end of the pullup and thus stationary. My answer basically covers all possibilities, though. – YungHummmma May 03 '14 at 00:44
  • @KyleStrand, It's not true that there's "always upward acceleration": In a simplified picture, if you could exceed the gravitational force $mg$ for even an instant and then decrease to exactly $mg$, you will not be accelerating, yet you would be going up. I mention the "human aspect" because IME beginning physics students are confused by things like this, and the classic "if $W=Fd$, then why does it take me energy to hold weights stationary ($d=0$)?" That question is one they directly experience in life but it (naively) goes against the work equation. – YungHummmma May 03 '14 at 00:48
  • I understand the physics in your answer and recognize that the equation covers all cases; I'm just saying that it's important to clearly distinguish between the effect of acceleration on the force of the bar versus the (negligible) effect of height on the force on the bar. By "always" I meant "during every pull-up," not "at every moment during a pull-up." We know that there is acceleration at the bottom of a pull-up and deceleration at the top. – Kyle Strand May 03 '14 at 09:05
  • As for the question about work, there is movement at the cellular level even when the arm is perfectly still, so even in the ideal case there would be work done; this is not an example of theory and observation conflicting due to human imperfection. (See http://physics.stackexchange.com/questions/1984/why-does-holding-something-up-cost-energy-while-no-work-is-being-done) – Kyle Strand May 03 '14 at 09:06
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If you are accelerating upward then the force on the bar will be greater. F=ma.

DavePhD
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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.

Taejang
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