A 75 kg person falls front he second floor of a building and lands directly on o

Question

A 75 kg person falls front he second floor of a building and lands directly on one knee with his body otherwise vertical.
(a) If the fall is from a height of 10 m, find the velocity on impact with the ground.
(b) If it takes 5 ms for the person to come to rest, find the average force acting during the collision.
(c) Using the data of Example 3.9, will the femur break?

Example 3.9:
Estimate Young's modulus for compression of bone from the following data. The femur of an 85 kg person has an effective cross-sectional area of about 6 cm^2 and a length of about 0.5 m. When the person lifts a 100 kg mass, careful measurements show that the femur compresses by about 0.04 mm. Also, if the ultimate compressive strength of the femur is 1.7 x 10^8 Pa, find the maximum weight that the femur can support.

The 100 kg mass is assumed to be carried equally by both legs, so that the load on each leg is a force of (50kg)(9.8) = 490 N. The added stress is then found to be F/A = 8.2 x 10^5 Pa, which results in a strain of (0.04 mm)(0.5 m) = 8 x 10^-5. Young's modulus is then found as the ratio of the stress to strain, or Y = (8.2 x 10^5)/(8 x 10^-5) = 10^10 Pa.
From the ultimate compressive strength, we find that the maximum weight that the femur can support is F = (ultimate strength)(area) = (1.7 x 10^8)(6 x 10^-4) = 10^5 N. This enormous weight implies that normally the femur will not fracture under compressive forces.

Explanation / Answer

a) v^2 = v0^2 + 2ax v^2= 0^2 + 2*9.81*10 v=14 m/s b) f = m dv/dt = 75*14/5E-3=2.1E5 N c) this is bigger than the 10^5 max force mentioned in the example so it will break

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