2 If you jump from a desktop and land stiff-legged on a concrete floor,
3 you run a significant rist that you will break a leg. To see how that
4 happens, consider the average force stopping your body when you drop
5 from rest from a height of $h = 1.00\U{m}$ and stop in a much shorter
6 distance $d$. Your leg is likely to break at the point where the
7 cross-sectional area of the tibia is smallest. This point is just
8 above the anke, where the cross sectional area of one bone is about $A
9 = 1.60\U{cm}^2$. A bone will fracture when the compressive stress on
10 it exceeds about $\sigma_b = 1.60\E{8}\U{N/m}^2$. If you land on both
11 legs, the maximum force $F_{max}$ that your ankles can safely exert on
12 the rest of your body is then about
14 F_{max} = 2 F_b = 2 \sigma_b A = 5.12\E{4}\U{N}
16 Calculate the minimum stopping distance $d$ that will not result in a
17 broken leg if your mass is $m = 60.0\U{kg}$.
18 \end{problem*} % problem 4.51
21 The problem breaks down into two constant-acceleration problems.
22 Call the top of the desk dropping-off-point $P_0$, the point of maximum
23 velocity when you are just starting to contact the floor $P_1$, and the
24 point where your shoe soles have compressed a distance $d$ and brought
25 you back to rest $P_2$.
27 First consider the constant acceleration portion from $P_0$ to $P_1$.
28 Your final velocity $v_1$ is given by
30 v_1^2 = v_0^2 + 2 a_{01} \Delta y_{01}
31 = 2 a_{01} \Delta y_{01}
33 Now applying the same equation to
34 the second constant acceleration portion from $P_1$ to $P_2$.
36 v_2^2 &= v_1^2 + 2 a_{12} \Delta y_{12} = 0 \\
37 v_1^2 &= -2 a_{12} \Delta y_{12} = 2 a_{01} \Delta y_{01} \\
38 d &= \Delta y_{12} = -\frac{a_{01}}{a_{12}} \Delta y_{01}
40 The acceleration $a_{12}$ is given by Newton's second law
41 (picking down as the $+\vect{x}$ direction)
43 \sum F_x &= m a_{12x} \\
44 a_{12x} &= (\sum F_x)/m
45 = \frac{mg - F_{max}}{m}
46 = g - \frac{F_{max}}{m}
48 (I forgot to include $mg$ in the sum of the forces when I was doing
49 the problem, so I didn't take off points if you forgot it as well.)
52 d &= -\frac{a_{01}}{a_{12}} \Delta y_{01}
53 = -\frac{g}{g - \frac{F_{max}}{m}} h
54 = -\frac{1}{1 - \frac{F_{max}}{mg}} h \\
55 &= -\frac{1}{1 - \frac{5.12\E{4}}{60 \cdot 9.8}} \cdot 1.00\U{m}
59 (Ignoring gravity in your sum of forces, you would have gotten
60 $d = \frac{mg}{F_{max}} h = 1.15\U{cm}$.
61 The correction is very small because $F_{max} \gg mg$.)