Question

While unrealistic, we will examine the forces on a leg when one falls from a height by approximating the leg as a uniform cylinder of bone with a diameter of 2.3 cm and ignoring any shear forces. Human bone can be compressed with approximately 1.7 × 108 N/m2 before breaking. A man with a mass of 80 kg falls from a height of 3 m. Assume his acceleration once he hits the ground is constant. For these calculations, g = 10 m/s2.
Part A: What is his speed just before he hits the ground?
Part B: With how much force can the "leg" be compressed before breaking?
Part C: If he lands "stiff legged" and his shoes only compress 1 cm, what is the magnitude of the average force he experiences as he slows to a rest?
Part D: If he bends his legs as he lands, he can increase the distance over which he slows down to 50 cm. What would be the average force he experiences in this scenario?
Part E: Dyne is also a unit of force and 1 Dyn= 10−5 N. What is the maximum a bone can be compressed in Dyn/cm2?
Part F: Which of the following is the reason that we would recommend that the man bend his legs while landing from such a fall?
a. Bending his legs allows him to push back up on the ground and negate some of the effects of the force applied by the ground.
b. Bending his legs decreases the speed at which he hits the ground, thus decreasing the force applied by the ground.
c. Bending his legs decreases his overall change in momentum, thus decreasing the force applied by the ground.
d. Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground.

Answer 1

Answer:

Part A: 7.75 m/s

Part B: 2330.8 kN

Part C: 24.03 kN

Part D: 4.8 kN

Part E: $1.7\times 10^{9} Dyn/cm^{2}$

Part F: Option D

Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground.

Explanation:

Part A

From the fundamental kinematic equation

$v^{2}=u^{2}+2gh$ where v is the velocity of the man just before hitting the ground, g is acceleration due to gravity, u is initial velocity, h is the height.

Since the initial velocity is zero hence

$v^{2}=2gh$

$v=\sqrt 2gh$

Substituting 10 m/s2 for g and 3 m for h we obtain

$v=\sqrt 2\times 10\times 3 =\sqrt 60= 7.745967\approx 7.75 m/s$

Part B

Force exerted by the leg is given by

F=PA where P is pressure, F is force, A is the cross-section of the bone

$A=\frac {\pi d^{2}}{4}$

Substituting 2.3 cm which is equivalent to 0.023m for d and $1.7\times10^{8} N/m2$ for P we obtain the force as

$F=PA=1.7\times10^{8}*\frac {\pi (0.023)^{2}}{4}= 2330818.276\approx 2330.8 kN$

Part C

The fundamental kinematic equation is part (a) can also be written as

$v^{2}=u^{2}+2a\triangle x$ and making a the subject then

$a=\frac {v^{2}-u^{2}}{2\triangle x}$ where a is acceleration and $\triangle x$ is the change in length

Substituting the value obtained in part a, 7.75 m/s for v, u is zero and 1cm which is equivalent to 0.01 m for $\triangle x$ then  

$a=\frac {7.75^{2}-0^{2}}{2\times 0.01}= 3003.125 m/s^{2}$

Force exerted on the man is given by

$F=ma=80\times 3003.125= 240250 N\approx 24.03 kN$

Part D

The fundamental kinematic equation is part (a) can also be written as

$v^{2}=u^{2}+2a\triangle h$ and making a the subject then

$a=\frac {v^{2}-u^{2}}{2\triangle h}$ where a is acceleration and $\triangle h$ is the change in height

Also, force exerted on the man is given by $F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}$

Substituting 80 Kg for m, 50 cm which is equivalent to 0.5m for \triangle h and other values as used in part c

$F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}=80\times \frac {7.75^{2}-0^{2}}{2\times 0.5}= 4805 N\approx 4.8 kN$

Part E

$P=1.7\times 10^{8}=1.7\times 10^{8}\times (\frac {10^{5} Dyn}{10^{4} cm^{2}}=1.7\times 10^{9} Dyn/cm^{2}$

Part F

Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground