CP BIO Stress on the Shin Bone. The compressive strength of our bones is important in everyday life. Young’s modulus for bone is about 1.4 X 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.

Solution 944P Step 1 : Given 10 Youngs modulus of bone, y = 1.4×10 Cross sectional area of bone, A =3cm = 3×10 -4 Let l = the length of the bone 0 And F max Maximum force that can applied to the bone before it fractured. The bone can take 1% change in its length before fracturing. l= 1%of l 0 = 0.01 l0 Step 2: a) Youngs modulus is the measure of elasticity, equal to the ratio of the stress acting on a substance to the strain produced. Y = Fmax/A l/l0 Fmax× l0 Y = l× A Step 3: Put the value l= 0.01 l in0he above equation Y = Fmax× l0 (0.01) l0×A The l0 t canceled out and the equation become Y = Fmax 0.01 A From the above equation we write down the equation for F max Fmax Y ×0.01×A F = (1.4×10 ) ×0.01×( 3×10 ) -4 max 4 Fmax .2×10 This is the maximum force that can apply to the bone.