Problem Set #4

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2270

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Mechanical Engineering

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Dec 6, 2023

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HK*2270 Problem Set #4 For this problem set, read the questions carefully. When setting up your equations of static equilibrium consider what you know/don’t know. Do you first need to calculate Net Joint Moment to help you solve for the muscle forces (F m )? Can you use information about Muscle forces to determine the Net Joint Moment about a particular joint? 1. The forearm (including the hand) segment has a mass of 3 kg and its center of mass is located 10 cm from the elbow as illustrated. The single muscle equivalent flexor force (F m ; assume no other muscles are acting active) acts 5 cm from the elbow joint. Assuming no other forces are acting, what muscular force (F m ) pulling at an angle of 35° would be required to keep the segment position 20° from the horizontal. The system is in Static Equilibrium. Remember, a complete free body diagram is always considered an integral part of a statics solution. 2. A certain amount of abdominal muscle force (F Abs-muscle ) is needed to hold the lifting posture illustrated. The force (F com ) is the effect of the mass of the barbell, arms and upper trunk combined (100 Kg). The perpendicular distance between the line of action of F com and the axis of rotation about the Hip (point C) is 25 cm; the moment arm for the abdominal muscle force to point C is 12.5 cm. Calculate the force exerted by the abdominal musculature; assume that no other tissues can create a moment of force about C. Remember the hint above! Principles of Biomechanics 1

F trice p F tricep 160° d1 elbow d2 d3 W 20° W F ½body 20° F body 3. An 80 kg man performs strengthening exercises with a 10 kg weight boot. Visualize that this man is facing you, standing in the frontal plane. While standing on his right foot, he externally (lateral) rotates his leg slightly and then moves his entire left leg (thigh, lower leg & foot; which is 65 cm in length) from the hip joint and holds it in an abducted position 20° from the global vertical axis . Given the following information (below), draw a free body diagram of the system. How much and what moment (abduction or adduction) must be provided by the hip to maintain this position? (a) the center of mass of the weight boot is located 60 cm below the hip joint (measured along the length of the leg) (b) the mass of the whole limb (thigh, lower leg & foot) is 14.55% of the total body weight (c) the center of mass of the limb is located at a point that is 43.3% of the distance of the leg, as measured from the proximal end (i.e. hip joint). 4. A 70 kg gymnast is exercising on parallel bars. He is supporting himself off the ground on both hands such that the forearms are 20° from the vertical and the elbow’s relative joint angle is 160°. The axis of the elbow joint (along the bone) is 30 cm from the hand’s contact point with the bar. The moment arm length of the triceps at the elbow is 2.5 cm and the triceps are pulling directly upwards. The mass of the forearm and hand, acting at 15 cm from the elbow is 1.4 kg. Assuming both arms carry equal loads: (A) Calculate the elbow joint net joint reaction forces at the right elbow, and (B) Calculate the muscle force exerted by the athlete’s right arm only that is needed to maintain this posture. Principles of Biomechanics 2

5 (a) & (b) 70° F 5 (c) 5. A below knee amputee is just beginning to walk with a prosthesis. The total below knee mass is 3.5 kg. The ground reaction force just after heel contact is 572 N acting as illustrated below. Assuming this is a static situation, and that there is minimal movement between the residual limb and the socket of the prosthetic, estimate: (a) the net joint reaction forces at the knee (b) the net joint moment about the knee joint. Use anatomical terms (knee flexion or extension) to describe the direction of this net joint moment. Which muscle group would be active? (c) This individual cannot use the prosthesis for very long because of skin tenderness and rapid fatigue. You are asked by a physiotherapist colleague working with this individual to help with this. Use the drawing of an able bodied individual doing a similar exercise (below) to help you draw your FBD for this amputee situation. A therapist places a weight on the foot (picture to left) and passively raises the leg in an extended position . The rehabilitation exercise for the patient is to maintain this static leg position as long as possible (situation to right). Use your advanced knowledge about biomechanics to assist him in determining the amount of weight the patient should exercise with ( with knee in extended position ) to create the same muscle demand as experienced at this point in the gait cycle. The perpendicular distance (dotted line) from the knee to a weight attached to the residual limb is 15 cm. 4 5 c m 2 7 c m

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6. Calculate the joint reaction forces and net knee joint moment (assuming static equilibrium). Hint: Part of this question will be discussed in lecture in Week 5; be sure to review these notes if you get stuck. Given: Muscle moment arm is 4 cm and the muscle force (Fm) is acting upwards. Calculate : a) Bone on Bone x & y forces; b) Bone on Bone compression & shear forces. Y + F m M + 4 cm X + 45 cm 250N 20cm 30° 40N

ANSWERS * Note: More significant digits have been included in some of these solutions to make it easier for you to check your answers. 1. [Intermediate step: M elbow = 2.766 Nm]; Muscular force required= 96.4 N 2. [Intermediate step: M pointC = -245.3 Nm]; Force exerted by abdominal musculature= -1962 N 3. [Intermediate step: d perpendicular for COM = 0.09626; d perpendicular for Boot = 0.2052] Moment at Hip= 31.1 Nm, thus hip abduction is required to maintain this position 4. R elbow-x = 0 N; R elbow-y = -330 N (downwards); M elbow = -34.52 Nm; F muscle = + 1380 N 5. (a) R knee-x = 0 N; R knee-y = -537.66N (b) Moment at knee = -84.87 Nm (clockwise). Knee Flexion: Hamstrings would the muscle group active (c) Weight required for exercise is 566 N (or a mass of 57.7 kg) acting downward on the residual limb. 6. [Intermediate steps: F knee-x = 216.5 N; F knee-y = 165.0 N; M knee = 116.5 Nm; F muscle = 2912.5 N] BnBx = 217 N; BnBy= -2747.5 N Total-compression = 2488 N (of the femur on the tibia) Total-shear = -1186 N (Posterior shear)

1) 0= 35° 20° D1= 0.05 m = r D2 = 0.10 m Mass= 3kg Fcom= 3(9.81) = 29.43 N 4) F1/2Body= (Mtotalbody * g) 2 d ⁒ ⊥for com Felbow-y Me Fcom Felbow-x 2) Fhip-y Mass= 100kg Fcom= 100(9.81) Mhip = 981 N Mcom Fcom Fhip-x d for com (given) = 0.25m Momen arm for Fabdominal = 0.125m ∑M = 0 0= Mhip + Mcom Mhip= - (981 * 0.25) The net joint movement at the hip is -245 Nm Mhip = - 245.25 Nm Mhip = Fmuscle * d NOTE: Static equilibrium so the Hip Net Joint Moment you -245.25 = -[Fm * (0.125)] calculated from the External forces in the system must be= to Fm= 1962 N the internal muscle moment - [Mhip] = - [Fabs-muscle * Dperp] 3) d1= 0.65 m Θ = 20° The mass of leg = 14.55% of 80kg Mboot = 10kg = 11.64 kg Fboot= 10(9.81) Fcom = (11.64 * 9.81) = 98.1 N = 114.1884 N D2 = 0.60 m Com of hip is 43.3% of the leg length d ⊥ for com d 3sin Θ d ⊥ for boot d 2sin Θ = 43.3% of 0.65m = (0.28145)(sin20) = 0.60 sin 20 Me= Fmd Me= [Fm * (r sin Θ)] 2.766 = [(Fm) * (0.05 sin 35°)] 2.766= Fm (0.02868) Fm= 96.44 N The muscular force required to sustain the static would be 96.4 N ∑Me = 0 0= Me – Mcom Me = (Fcom * d for com) = (2943 * 0.09397) = 2.766 Nm A= H cos ∞ = d2 cos ∞ = (0.1) cos 20 = 0.09397 = A / H For com 3 ) ∑ M = 0 : 0 = M h i p – M b o o t – M c o m 0 = M h i p – ( F b o o t * d ⊥ f

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= (70 * 9.81) 2 = d2 sin20° ⁒ = 343.35 N = (0.15sin20) *Force supported by each hand* = 0.0513m M f+h= 1.4kg d1= 0.025m Fcom= (1.4 * 9.81) d2= 0.15m d ⊥ for F1/2body = 13.734 N d3= 0.30m = d3 sin20° = (0.3 sin20) = 0.1026 ∑Fx= 0 Felbow-x = 0 ∑Fy= 0 0= F1/2body – Fcom + Felbow-y 0= 343.35 – 13.734 + Felbow-y 0= 329.616 + Felbow-y Felbow-y = -330 N ∑ M=0 0= Me – Mcom + Mbody 0= Me –(Fcom*d ⊥ for com)+(F1/2body * d⊥ for body) 0=Me- (13.734 * 0.0513) + (343.35 * 0.1026) 0= Me – 0.7046 + 35.22771 0= Me + 34.523 Me= -34.5 Nm b. Me = Fm * d1 -34.5 = 1 (Fmuscle * 0.025) Fmuscle= + 1380 N The muscle exerted on the athletes right arm is +1380N 5) Θ= 70° d1= 0.27m Mass of leg= 3.5kg d2= 0.45m Fcom= 3.5*9.81 Fgnd= 572N = 34.34 N d ⊥ for com d⊥fro gnd d1 cos Θ d2 cos Θ = 0.27 cos70 = 0.45 cos70 = 0.09235 = 0.15391 ∑Fy= 0 0= Fknee-y – Fcom + Fgnd 0= Fknee-y – 34.34 + 572 Fknee-y – 537.66 N b. ∑M=0 0= Mknee – Mcom + Mgnd

0= Mknee – (Fcom * d ⊥ for com) + (Fgnd * d⊥ gnd) 0= Mknee – (34.34 * 0.09235)+ (572 * 0.15391) 0= Mknee – (3.171299) + (88.03652) 0= Knee + 84.865 Mknee= - 87.9 Nm Mknee clockwise, flexion of the knee, hamstring must be active c. To exercise the quadriceps in an equivalent manner, assume that weights can be applied at the end of the residual limb and the weight of this limb is negligible (less than 1-2kg) d3= 0.15m ∑Mk= 0 0= -Mknee – Mload 0= -Mknee – (Fload * d3) 0= -84.9 – (Fload * 0.15) 84.9= - 0.15 F-load F-load = -566 N The weight for the required exercise is 566 N (or a mass of 57.7kg) acting downward at the end of the residual limb 6) d1= 0.45m Fkick= 250N D2= 0.20 m d ⊥muscle = 0.04m Θ= 30° Fcom= 40N ∑Fx= 0 0= Fknee-x – Fkick-x 0= Fknee-x – (250 cos30) Fknee-x= 216.5 N ∑Fy= 0 0= Fknee-y – Fcom – Fkick-y 0= Fknee-y – (40) – (250 sin30) Fknee-y = 165.0 N ∑M= Mknee – Mcom – Mkick 0= Mknee – (Wcom * d ⊥ for com)- (Fkick * d⊥) 0= Mknee – (40 * 0.100) – (250 * 0.45) 0= Mknee – (4) – (112.5) Mknee= + 116.5 Nm Mk= Msme 116.5 = Fsme * d 116.5 0.04 = Fsme 2912.5 N= Fsme (only in y)

Jrx= BnBx + Fsmex 216.5N = BnBx Jry= BnBy + Fsmey 165= BnBy + 2912.5 -2745.5 N = BnBy BnBy compression BnBy shear = +2747.5 cos30 = - 2747 sin30 = 2379.4 N = - 1373.75 N BnBx compression BnBx shear = + 216.5 sin 30 = 216.5 cos30 = 108.25 N = 187.49 N Total compression = BnBycomp + BnBxcomp = 2379.4 N + 108.25 N = 2488 N Total shear= BnByshear + BnBxshear = - 1373.75 N + 187.49 N = - 1186 N

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