- #1
Honoratus
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- 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.
$$m\times a_{0} = F - w$$
$$a_{0} = \frac{F}{m} - g$$
$$v_{0} = (\frac{F}{m} - g) \times t$$
$$y_{0} = (\frac{F}{m} - g) \times \frac{t^{2}}{2}$$
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.