BME 5210 Problem Solutions
		Chapter 2
	Chapter 2 Problem 1 Problem 2 Problem 3 Figures: Problem 1 Problem 2 Problem 3
	1. A person at the gym is going to strengthen his/her quadriceps muscles by doing a series of leg extension exercises. Case 1: Assume that the femur is at a right angle to the tibia and that the weight exerts an external force, F, which acts perpendicular to the tibia through the ankle joint at a distance, d1, to the center of the tibial plateau. Calculate the external moment generated at the knee due to the force, F, at the ankle and the mass of the shank, m. The distance from the ankle to the center of the knee joint is d1 = 0.4 m, the weight lifted is F = 400 N, and the mass of the shank is m = 2.89 kg. Case 2: Assume that the leg is fully extended. Use the same values given for Case 1. The distance from the mass center to the tibial plateau is d2 = 0.1 m. Solution: Case 1: Knowns: d1 = 0.4 m F = -400 N M = 2.89 kg Unknowns: Mk Fk Assumptions: acceleration is 0 Reaction force at ankle is 0 Force balance: Sum F = 0 Fkv - mg = 0 Fkv - 2.89 kg * 9.8 m/s2 = Fkv - 28.3 N = 0 Fkv = 28.3 N Fkh + F = 0 Fkh - 400 N = 0 Fkh = 400 N Resultant force Fk = 401 N angle = 4.04° Moment balance Sum M = 0 F will act to move leg clockwise - creates negative moment Mk - F*d1 = 0 Mk - 400 N*0.4 m = Mk - 160 Nm = 0 Mk = 160 Nm Will act counter-clockwise to extend leg Case 2: Known: F = 400 N d1 = 0.4 m d2 = 0.1 m m = 2.89 kg Unknown: Fk Mk Assume: see case 1 Force balance: Sum F = 0 Fkv - mg + F = 0 Fkv - 2.89 kg * 9.8 m/s2 - 400N = Fkv - 28.3 N - 400 N = 0 Fkv = 428.3 N Fkh = 0 Resultant force Fk = 428.3 N angle = 0° Moment balance Sum M = 0 F and mg will act to move leg clockwise - creates negative moments Mk - F*d1 -mg*d2= 0 Mk - 400 N*0.4 m -28.3N*0.1 m = Mk - 160 Nm - 2.83 Nm = 0 Mk = 162.83 Nm Will act counter-clockwise to extend leg 2. Using the result for the external reaction moment at the knee, obtained from Problem 1, calculate the internal force in the quadriceps muscle necessary to balance the external reaction moment when the knee joint is flexed at 90° and when it is at full extension. Assume that the quadriceps is the only muscle acting and the distance from the quadriceps muscle force and the tibiofemoral contact point is d = 0.02 m. Calculate the force in the quadriceps muscle if the distance from the quadriceps force, Fqv to the tibiofemoral contact point, 0, is d = 0.04 m. Solution General assumptions: external moment and force are equal and opposite to the internal reaction moment and force calculated in problem 1. Case 1: Knowns: Fextv = -400 N Fexth = -28.3 N d = 0.02 or 0.04 m Unknowns: Fq Fc Assume: Quadriceps act purely vertically Moment balance: Sum M = 0 Mext + Fq*d = 0 -160 Nm + Fq*0.02 m = 0 Fq = 8000 N If d = 0.04 m, Fq = 4000 N Force balance: Sum F = 0 Fch + Fexth = 0 Fch - 400 N = 0 Fch = 400 N Fcv + Fextv + Fq = 0 Fcv - 28.3 N + 8000 N = 0 Fcv = 7971.7 N (or 3971.7 N if d = 0.04 m) Case 2: Known: Fextv = - 428.3 N Fexth = 0 N Mext = - 162.83 Nm d = 0.02 or 0.04 m Unknown: Fq Fc Assume: Quadriceps pull with only horizontal component Moment balance: Sum M = 0 Mext + Fq*d = 0 -162.83 Nm + Fq*0.02 m = 0 Fq = 8141 N If d = 0.04 m, Fq = 4070 N Force balance: Sum F = 0 Fch + Fexth + Fq= 0 Fch + 0 N - 8141 N = 0 Fch = 8141 N (or 4070 N if d = 0.04 m) Fcv + Fextv = 0 Fcv - 428.3 N = 0 Fcv = 428.3 N   3. A person steps across a force platform in a gait analysis laboratory. At heel-strike, the floor exerts a vertical reaction force on the foot measured by the force platform to be Fgv = 700 N with a horizontal component, Fgh = 100 N. Neglecting the inertial forces, determine the external reaction forces and moments at the hip joint. The configuration of the lower limb at heel strike is shown in the figure. All distances are measured from either the vertical or horizontal components of the ground reaction force and are labeled. The weight of the thigh (Fwt = 70 N) and leg (Fws = 29 N) are located at the mass center of the thigh and leg, respectively. Solution Known: Weight of leg (Fws): 29 N Weight of thigh (Fwt): 70 N Geometry (see figure) Ground reaction force: Fgv = 700 N; Fgh = 100 N Unknown: Reaction forces and moment at hip Assumptions: Mass of foot is negligible Inertial forces negligible (ignored) Ankle reaction forces (on foot) are equal and opposite to ground reaction forces Leg: Sum F = 0 Vertical: Fgv - Fws + Rkv = 0 Rkv = Fws - Fgv Rkv = -700 N + 29 N Rkv = -671 N Horizontal: Fgh + Rkh = 0 Rkh = -Fgh Rkh = -100N Sum M = 0 Moments taken about knee Clockwise: negative Counterclockwise: positive Mk - Fws*(d5 - d6) + Fgv*(d5) + Fgh*(d2) = 0 Mk = 29 N * 0.7 m - 700 N * 0.21 m - 100 N * 0.5 m Mk = -176.7 Nm (clockwise - flexing leg) Thigh: Reaction forces at knee are equal and opposite to those calculated above Rkv = 671 N Rkh = 100 N Mk = 176.7 Nm Sum F = 0 Vertical: Rkv - Fwt + Rhv = 0 Rhv = Fwt - Rkv Rhv = 70 N - 671 N Rhv = -601 N Horizontal: Rkh + Rhh = 0 Rhh = -Rkh Rhh = -100 N Sum M = 0 About hip Mh - Fwt*(d3 - d4) + Rkv*(d3 - d5) + Rkh*(d1 - d2) + Mk = 0 Mh = 70 N * 0.18 m - 671 N * 0.29 m - 100 N * 0.2 m - 176.7 Nm Mh = -378.69 Nm (clockwise - extending thigh)
		Problem 1 Figures
		Problem 2 Figures
		Problem 3 Figures