While unrealistic, we will examine the forces on a leg when one falls from a height by approximating the leg as a uniform cylind

Question

4 days ago

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While unrealistic, we will examine the forces on a leg when one falls from a height by approximating the leg as a uniform cylind

er 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.

Physics

1 answer:

Yuliya22 [1.1K] · Answer 1 · 2026-03-01T04:50:21+03:00

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 one's legs lengthens the duration of force application from the ground, resulting in a reduction of the applied force.

Explanation:

Part A

Using the fundamental kinematic equations

$v^{2}=u^{2}+2gh$ where v represents the velocity just before ground impact, g denotes gravitational acceleration, u signifies initial velocity, and h is the fall height.

With the initial velocity at zero, thus:

$v^{2}=2gh$

$v=\sqrt 2gh$

Plugging in 10 m/s² for g and 3 m for h gives:

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

Part B

The force exercised by the leg can be expressed as

F = PA where P is pressure, F indicates force, and A denotes the cross-sectional area of the bone.

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

With a substitution of 2.3 cm or 0.023m for d and $1.7\times10^{8} N/m2$ for P, we derive 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 equations from part (a) can also be rearranged to show:

$v^{2}=u^{2}+2a\triangle x$ and solving for a yields

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

Using the previously derived value from part a, 7.75 m/s for v, and 0.01 m for $\triangle x$ gives us:

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

The force felt by the man is given by:

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

Part D

A similar approach with the fundamental kinematic equations shows:

$v^{2}=u^{2}+2a\triangle h$ and solving for a indicates:

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

The force experienced can thus be defined as $F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}$ .

For substitution, we use m = 80 Kg, and 0.5m for \triangle h along with other values calculated 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 one's legs extends the period over which the force acts, thus lessening the overall force exerted by the ground.