Assignment 1

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University of Guelph *

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2270

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Mechanical Engineering

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Dec 6, 2023

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Assignment #1 Goal: The goal of Assignment #1 is to complete Biomechanical analyses of static equilibrium . The Task: Solve for Net Joint Forces and Moments; Solve for Muscle Forces & Bone on Bone Forces; and finally solve for Compression & Shear forces. Given the questions provided: a) solve the questions asked in the space provided b) scan in your answers and submit these as a .pdf document. NOTE: Only a .pdf copy will be accepted. All other versions uploaded will NOT be graded and will receive a ZERO. There are several free software available that can compile multiple .jpg files into a single .pdf; these are very easy to use (e.g. https://jpg2pdf.com/ or Adobe Scanner) Due on Wednesday October 11 at 9:00 AM by CourseLink Drop box This assignment is worth 8% of your overall grade in course Penalty of -5% off of the total assignment for each day this assignment is late . For example: You deposit your assignment at 9:30 am Oct 11, 2023 Even if you get 35/35 (100%) on the assignment è you will be graded 33.25/35 and receive 7.6 out of 8% for this component of the course Assignments handed in ³ 7 days late will not be graded. A grade of zero will be recorded.
Student Name: ___________________________ Student ID#: ______________________ 1 HK*2270 Principles of Human Biomechanics Assignment #1 PLEASE SOLVE THE QUESTIONS BELOW IN THE SPACE PROVIDED. Question 1. (20 marks) _____________ Question 2. (6 marks) _____________ Question 3. (9 marks) _____________ Total (35 marks): ________ 1. (20 marks) You are a sport biomechanist who has been asked to perform a biomechanical analysis of a paralympic skier as they turn during a race. Note: his right leg is pushing off the snow to change direction. Draw a free body diagram of the skier’s right knee (red dot) and calculate the Net Joint Forces and Moment at his right knee. Assume Static Equilibrium. His body mass is 80 kg The lower leg is 6% of his body mass The foot is 2% of his body mass His lower leg/foot is oriented 55 degrees (from right horizontal) The ground reaction force acting under the right foot acts in XY space and is located 20 cm to the left and 32 cm below the knee joint (from his perspective) From the knee: Lower leg COM is located 17 cm along the extended leg Prosthetic foot COM is located 28 cm along the extended leg 150⁰ F ground = 640N Knee 55⁰ +y +x M + Space for calculations: (a) Free Body Diagram morgan brewster 1229518 - i .. 3 Y X Yi , x SOH CAH TOA A I 1300 x 1880 > 380 3 - 158 : - e stee ing H x 7gx weight of leg : weight of foot We = 0 02 x80 x 9 81 W , 0 06 x 80 x 9 81 We= 15 . 696 N W , 47 088N firm-foot from-leg
Student Name: ___________________________ Student ID#: ______________________ 2 (e) The right knee is being held in a static position by a __________ ( Choose one: flexion; extension; hyperextension; abduction; adduction; circumduction ) moment about the knee. (f) What is the plane of motion for the action identified in 1(e)? _______________________ (g) What is the axis of motion identified in 1(e)? ___________________________________ (h) Provide a definition of the movement identified in 1(e) in the space below. (b) F x = 0: (c) F y = 0: (d) M = 0: morgan brewster 1229518 Fknee-x-Fgx 0 Finee-y + fgy-from-leg - fcom-foot = 0 Finee -x = fgx Ficnee-y =- fgy+fcom-leg + fom-foot finee x 6 40N x C0S38 · Ficnee - y = -640NxS1n300 + 47 088N+ 15 696N Finee - x= 554 N f knee -y = -320N + 47 088N + 15 696N the force on the knee In the x direction Is SS4N ht finee-y =- 23TN rig : the force on the knee In the y direction Is ISTN down Mic - Mcom-leg - Mcom-foot + Megy - Megx Mic = + Mcom-leg + Mcom-foot - Megy + Megx Mis = Farm-leg x d1 + from-foot xd1-Fgy > d1 "fgx &1 M , x = 47 88840 17m + 15 696 + 0 28m -- 257 x0 . 32m) - (554 = 0 28m) Mx = 8 00496N - m + 4 39488N-m-(-82 . 24N m) + /10 . 8N M Mic = 20SNM the momentum in the knee is 205 N m counter clockwise adduction frontal plane anterior - posterior axis adduction is the movement of a body part towards the bodies midline
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Student Name: ___________________________ Student ID#: ______________________ 3 2. (6 marks) One of your classmates decided to analyze the role of the steering arm during a turning motion within a ski race. Given the information below, calculate the (a) muscle force (Fsme) and (b) bone-on-bone forces acting at the elbow. Your classmate calculated the F elbow-x = 178 N and F elbow-y = 273 and the M elbow = 84.1 Nm. Note that the single muscle equivalent acts at a distance (along dashed line) of 1.8 cm from the elbow joint at 141 degrees from right horizontal; Arrow heads indicate direction of forces applied. (a) Muscle Force (Fsme) (b) Bone-on-Bone Forces at the Elbow: morgan brewster 1229518 I a M S t 1 - Jux BNBx-fsmex Jry BNBy Sme-y IT8N BNB x -5400C0s39 · IT3N = BNBx+ S4WNSI 39: 178N = BNBx (4194 593 IT3N = BNBy+ 3398 33N BNBx= 43 TON BNBY=-3130N Me Fsme -y x d1 H -84 /Nm Fsme-y x 0 018m 0 390 1211 Esme-y = 4672 IN : the bone on bone forces in the X direction is 4370N right A Isme= Esme -y and in the y direction 3130N down Sin 39: -sme = 4672 2 N Sin39 Fsme = 5400N : Esme= S400N at 39 · Clockwise
Student Name: ___________________________ Student ID#: ______________________ 4 3. (9 marks) Your classmate has been given a task from the skier’s coach and has asked you for help. They have asked you to calculate the total compression and shear forces at the skier’s elbow. Be sure to use an appropriate, anatomically relevant axis system for these calculations. Please use the angle and Bone-on-Bone forces provided in the Diagram below; please note the direction of arrows for your calculations. Diagram: Compression Forces: Shear Forces: morgan brewster 1229518 compression - - Ve shear O A BrBy Shear 4000N 42 ampression B Y BrBx 55UON a 18420 espear Bett Compx= Buxx SinG COMPy - BuByxCOSt Shearx=BNBx x COSO - Sheary= BuByx SinG - -COMPX= 5500NxSin42 -COmpy= 4000NxCOS42 - Shearx=SS00NxCOS42 - Sheary 4000NxS1n42 · -COMPX=-504/N Compx= 5041N COMpy=-2973N - Shearx=-1200N -Sheary= 2677N Shear = 2200N Sheary=-2677N Comptotal = COMDx + COMPy Shear total = Shearx+ Sheary COMPtotal =5041N +12973N) Sheartal = 2200N +1 - 2677N) COMProtal = 20G8N Shearttal = - 47TN : total compression Is 2068N : total Shear 1S-477N