The compressive strength of our bones is important in everyday life. Young’s modulus for bone is about 1.4 * 1010 Pa. Bone can take only about a 1.0% change in its length before fracturing. (a) What is the maximum force that can be applied to a bone whose minimum cross-sectional area is 3.0 cm2 ? (This is approximately the cross-sectional area of a tibia, or shin bone, at its narrowest point.) (b) Estimate the maximum height from which a 70 kg man could jump and not fracture his tibia. Take the time between when he first touches the floor and when he has stopped to be 0.030 s, and assume that the stress on his two legs is distributed equally.

Explanation:

Young modulus ε = 1.4 × 10¹⁰ Pa

ΔL = 1% of the original length = 0.01 x where x is the original length

cross sectional area = 3.0 cm² =( 3 .0 / 10000) m²= 0.0003 m²

ε = Stress / strain

stress = ε × strain

stress = F /A

F force = ε × A × ( ΔL / L) =  1.4 × 10¹⁰ Pa × 0.0003 m² × 0.01 = 4.2 × 10⁴ N

b) F net = F max - mg ( weight) = 84000 - ( 70 × 9.8 m/s² ) ( F is double since the stress on the two leg is equally distributed)

f net = ma = 84000 - ( 70 × 9.8 m/s² )

a =  (84000 - ( 70 × 9.8 m/s² )) = 1190.2 m/s²

v = u + at

where final velocity equal zero

- u = -at since it coming downwards

u = at = 1190.2 m/s² × 0.03s = 35.706 m/s

using conservation of energy

1/2 mv² = mgh

1/2v²/ g = h

h = 0.5 × (35.706 m/s )² / 9.8 = 65.04 m

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.