Question

One model for the length of a person's ACL (L_ACL, in millimeters) relates it to the person's height (h, in centimeters) with the linear function

L_ACL = 0.4606h ? 41.29.

Age, gender, and weight did not significantly influence the relationship.

(a) If a basketball player has a height of 2.37 m, how long is his ACL?
1    cm

(b) If a pressure of

8 ? 10⁶ N/m²

is applied to his ligament, how much will it stretch?
2    cm

Answer 1

ACL means that "Anterior Cruciate Ligament".

given that :

L_ACL = 0.4606 h - 41.29                { eq. 1 }

where, L_ACL = length of the person's ACL in mm

h = height of the person in cm

Age, gender, and weight did not significantly influence the relationship.

(a) If a basketball player has a height of 2.37 m, then his ACL length is given as :

h = 2.37 m = 237 cm

inserting the value of h in above eq.

L_ACL = 0.4606 (237 cm) - (41.29)   

L_ACL = 67.87 cm

or L_ACL = 678.7 mm

(b) using an eq.    Y = F L / A (L)                     { eq. 2 }

or   L = F L / A Y

where, Y = young's modulus of human bone = 0.1 x 10⁹ N/m²

F = force = p x A

A = area of cross-sectional

L = original length of ACL= 678.7 mm

inserting the values in eq.2,

L = (p x A ) L / A Y

where, p = applied pressure to ligament = 8 x 10⁶ N/m²

L = (8 x 10⁶ N/m²) (678.7 x 10^-3 m) / (0.1 x 10⁹ N/m²)

L = (5429.6 x 10³ N.m) / (0.1 x 10⁹ N/m²)

L = 54296 x 10^-6 m

L = 54.2 mm