Finding the Error in Solving a Chin-up Force Problem

[Honoratus]

Homework Statement

When you do a chin-up, you raise your chin just over a bar (the chinning bar), supporting yourself with only your arms. Typically, the body below the arms is raised by about 30 cm in a time of 1.0 s, starting from rest. Assume that the entire body of a 700 N person doing chin-ups is raised by 30 cm, and that half the 1.0 s is spent accelerating upward and the other half accelerating downward, uniformly in both cases. Find the force his arms must exert on him during the accelerating part of the chin-up.

Relevant Equations

$$F_{net} = m*a$$
$$\frac{d}{dt}y(t) = v(t)$$
$$\frac{d}{dt}v(t) = a{t}$$

This is a fairly straightforward problem. I'll just post the way that I'd solved it: for some reason I'm getting the wrong answer.

$$m\times a_{0} = F - w$$
$$a_{0} = \frac{F}{m} - g$$
$$v_{0} = (\frac{F}{m} - g) \times t$$
$$y_{0} = (\frac{F}{m} - g) \times \frac{t^{2}}{2}$$
This gives us the initial values for position and velocity during the first part of the chin-up. Now, in the second part, we have $a = -g$, so:

$$a = -g$$
$$v = -gt + v_{0}$$
$$y = -\frac{gt^{2}}{2} + v_{0}t + y_{0}$$

Notice that we reach $y = 0.3$ m at $t = 0.5$ s in the second part of the chin-up, so $t = t_{1} = 0.5$ s.

$$0.3 = -\frac{gt^{2}}{2} + (\frac{F}{m} - g) \times t^{2} + (\frac{F}{m} - g) \times \frac{t^{2}}{2}$$
$$0.3 = -\frac{0.5^{2}g}{2} + (\frac{F}{m} - g)\times(0.5^{2} + \frac{0.5^{2}}{2})$$
$$0.3 + \frac{0.5^{2}g}{2} = (\frac{F}{m} - g)\times(0.5^{2} + \frac{0.5^{2}}{2})$$
$$\frac{F}{m} - g = \frac{(0.3 + \frac{0.5^{2}g}{2})}{(0.5^{2} + \frac{0.5^{2}}{2})}$$
$$F = (\frac{(0.3 + \frac{0.5^{2}g}{2})}{(0.5^{2} + \frac{0.5^{2}}{2})} + g)\times m$$
With $g = 9.8$ and $m = \frac{700}{9.8}$, this evaluates to 990 N, and this answer (unless my calculator is failing me) is wrong: the answer given in the textbook is 786 N. In the solutions I'd searched through online, an assumption is made that, after the first part of the chin-up, $y = 0.15 m$. I am not quite sure why that would be true; and, in any case, it wouldn't matter if the answer I'd gotten using calculus was correct. Alas, it is not. Can you please point out the mistake I'd made here? Thank you so much.

 

[PeroK]

In the solutions I'd searched through online, an assumption is made that, after the first part of the chin-up, $y = 0.15 m$. I am not quite sure why that would be true;

Just on this point. If you accelerate from rest (at constant acceleration), then decelerate to rest (again at constant deceleration), then your average speed over both phases is the same (half the maximum speed). If these two phases take the same time, then you must travel the same distance during both phases.

I haven't looked through your solution in detail to find the mistake.

 

[TSny]

Now, in the second part, we have $a = -g$, so:
$$a = -g$$

Is it correct to assume $a = -g$ during the second part?

 

[Honoratus]

Just on this point. If you accelerate from rest (at constant acceleration), then decelerate to rest (again at constant deceleration), then your average speed over both phases is the same (half the maximum speed). If these two phases take the same time, then you must travel the same distance during both phases.

I haven't looked through your solution in detail to find the mistake.

Oh, thank you so much! I didn't quite make the connection between the "uniform acceleration" and the "taking the same time" parts. It makes sense now. I can replicate the successful solution now without a hitch, (in fact, if $(\frac{F}{m} - g)\times\frac{t_{1}^{2}}{2} = 0.15$, I get the correct answer.)

It seems my mistake is confined solely to the second part.

 

[Honoratus]

Is it correct to assume $a = -g$ during the second part?

It seemed reasonable to me when I started solving it. It is uniform, and I'd assumed that all the force the person doing the chin-up exerts is solely confined to the first half of the chin-up.

Although now that you've pointed it out, the athlete in question doesn't really let go of the bar, so I'm not sure the assumption of free-fall holds here. There could be any constant value of force applied in the second part of the pull-up and the athlete would decelerate anyways.

I think I got it. The acceleration in the second part isn't $-g$. Since both of these take the same time, are uniform and both end at $v = 0$, they should have the same magnitude, and in my solution they obviously don't. I'm going to try and play around with the equations here some more to see if I can make it work.

Edit: yes, that is the issue, thank you so much for pointing this out to me, you've really helped me get this!

For the record, this is the correct way to solve this:

Everything holds up to the assumption that $a = -g$. If we instead set $a = g - F$, we get
$$v = (g-\frac{F}{m})t + (\frac{F}{m}-g)t_{1}$$
This already gives $v = 0$ at $t = t_{1}$, which is promising. Further on,
$$y = (g-\frac{F}{m})\frac{t^{2}}{2}+(\frac{F}{m}-g)t_{1}t+(\frac{F}{m}-g)\frac{t_{1}^{2}}{2}$$
At $t = t_{1}$, the first and third parts of this equation cancel out, and we're left with
$$y = (\frac{F}{m}-g)t^{2}$$
Which gives us the correct answer. We can even verify that by looking at the first equation for $y_{0}$: if we set it equal to $\frac{0.3}{2}$, we get the same equation, save for a minor algebraic difference.

Thank you all so much for the help, I really appreciate it! Physics never fails to challenge.

 

[Orodruin]

Typically, the body below the arms is raised by about 30 cm in a time of 1.0 s, starting from rest.

For the sake of completeness, I feel that this is a person with very short arms doing relatively slow chin-ups ...