Question

If you jump from a desktop and land stiff-legged on a concrete floor, you run a significant risk that you will break a leg. To see how that happens, consider the average force stopping your body when you drop from rest from a height of 1.10 m and stop in a much shorter distance d. Your leg is likely to break at the point where the cross-sectional area of the bone (the tibia) is smallest. This point is just above the ankle, where the cross-sectional area of one bone is about 1.60 cm². A bone will fracture when the compressive stress on it exceeds about 1.60 ✕ 10⁸ N/m². If you land on both legs, the maximum force that your ankles can safely exert on the rest of your body is then about the following.

2(1.60 ✕ 10⁸ N/m²)(1.60 ✕ 10^-4 m²) = 5.12 ✕ 10⁴ N

Calculate the minium stopping distance d that will not result in a broken leg if your mass is 55.0 kg. Don't try it! Bend your knees!

_____m

Answer 1

The problem breaks down into two constant-acceleration problems. Call the top of the desk dropping-off-point P0, the point of maximum velocity when you are just starting to contact the floor P1, and the point where your shoe soles have compressed a distance d and brought you back to rest P2.

First consider the constant acceleration portion from P0 to P1. Your final velocity v1 is given by

v1^2 = u1^2 + 2*a1*s1

u1 = 0

v1^2 = 2*a1*s1

Now applying the same equation to the second constant acceleration portion from P1 to P2

v2^2 = v1^2 + 2*a2*s2

now v2 = 0

v1^2 = -2*a2*s2

2*a1*s1 = -2*a2*s2

s2 = -(a1/a2)*s1

The acceleration a2 is given by Newton’s second law

F_T = m*a2

a2 = F_T/m = (mg - F_max)/m

a2 = [g - ( F_max/m)]

a1 = g

s2 = -(g/[g - ( F_max/m)])*s1

s2 = -(1/[1 - (F_max/mg)])*s1

s2 = -(1/[1 - 5.12*10^4/(55*9.81)])*1.10

s2 = 1.17 cm