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What is the Torque (T) due to force F? F=100N; distance to F: r F = 1m, q =30 o

What is the Torque (T) due to force F? F=100N; distance to F: r F = 1m, q =30 o. T = MR * F. F. 86.6 N 100 Nm 86.6 Nm 50 Nm 50 N. MR. q. r F. Elbow torque due to weight of forearm . T =MR * F T = 0.16m * 11N T = 1.8 Nm Direction? T = -1.8Nm. 11 N. 0.16 m.

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What is the Torque (T) due to force F? F=100N; distance to F: r F = 1m, q =30 o

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  1. What is the Torque (T) due to force F?F=100N; distance to F: rF = 1m, q=30o T = MR * F F 86.6 N 100 Nm 86.6 Nm 50 Nm 50 N MR q rF

  2. Elbow torque due to weight of forearm T=MR * F T = 0.16m * 11N T = 1.8 Nm Direction? T = -1.8Nm 11 N 0.16 m

  3. A person holds their forearm so that it is 30° below the horizontal. Elbow torque due to forearm weight? Fw=11N; rF = 0.16m • 1.76 Nm • 1.5 Nm • 0.88 Nm • -1.5 Nm • -0.88 Nm Fw q rF

  4. A person holds their forearm so that it is 30° below the horizontal. Elbow torque due to forearm weight?Fw=11N; rF = 0.16m T=MR*F F = 11 N MR=rFcos30° MR = 0.16 cos30° = 0.14 m T = 1.5 N m Use right-hand rule: T = -1.5 Nm 11 N MR 30° 0.16 m

  5. 100 N 11 N A person is holding a 100N weight at a distance of 0.4 m from the elbow. What is the total elbow torque due to external forces? We must consider the effects of 2 forces: forearm (11N) weight being held (100N)

  6. A person is statically holding a 100N weight at a distance of 0.4 m from the elbow. What is the elbow torque due to external forces? • T= (-Tarm) + (-Tbriefcase) • T = Tarm+Tbriefcase • T = (-Tarm) + Tbriefcase • T = Tarm + (- Tbriefcase) • It depends 100 N 11 N 0.16 0.4 m

  7. A person is statically holding a 100N weight at a distance of 0.4 m from the elbow. What is the elbow torque due to external forces? T = (-Tarm) + (-Tbriefcase ) T = -(11 N * 0.16 m) - (100 N * 0.4 m) T = -42 Nm 100 N 11 N 0.16 0.4 m

  8. Torque about shoulder due to external forces when 5 kg briefcase is held with straight arm. Forearm force Upper arm force Briefcase force

  9. Upper arm force Forearm force Briefcase force 20N 0.16 m 15N 15N 0.48 m 49N 0.48 m 0.65 m 31 Nm 20.6 Nm - 41Nm 41 Nm None of the above

  10. 20N 0.16 m 15N T= (49N * 0.65m) + (15N * 0.48m) + (20N * 0.16m) T = (31 Nm) + (7.2 Nm) + (3.2 Nm) T = 41 Nm 15N 0.48 m 49N 0.48 m 0.65 m

  11. Muscles create torques about joints Upper arm Elbow flexor muscle Biceps force Elbow Forearm T

  12. Fj,y Fm Fw Solve for Muscle force, Fm Rm Static equilibrium Melbow = 0 Fj creates no moment at elbow -(Tw) + (Tm) = 0 -(Fw * Rw) + (Fm * Rm) = 0 Fm = (Fw * Rw) / Rm Substitute: Fw= 11 N, Rw = 0.16 m, Rm = 0.03 m Fm = 59 N Rw

  13. If a person holds a 100 N weight in their hand 0.4 m from the elbow, what is the elbow flexor muscle force? Ignore weight of forearm Information Rm = 0.03 m Rext = 0.4 m Step 1: Free body diagram Upper arm Muscle Fext Elbow Forearm

  14. Fext Fm Rm Rext Solve forElbow flexor force Rm = 0.03 m Fext= 100 N Rext= 0.4 m Fj,y Fj,x

  15. Fext Fm Rm Rext Solve forElbow flexor force Rm = 0.03 m Fext= 100 N Rext= 0.4 m Melbow = 0 (Tm) – (Text) = 0 (FmRm) - (FextRext) = 0 Fm= Fext(Rext / Rm) Fm = 1333 N Fj,y Fj,x

  16. When Rm < Rext,muscle force > external force Fm = Fext(Rext / Rm) Last example Fext = 100N Rext > Rm Fm = 1333 N Upper arm Biceps brachialis Fext Rm Elbow Rext

  17. If a person holds a 100 N weight in their hand 0.4 m from the elbow, what is the joint reaction force? (ignore weight of forearm). Free body diagram Apply equations Fj,y Fext Fm Rm Fj,x Rext

  18. If a person holds a 100 N weight in their hand 0.4 m from the elbow, what is the joint reaction force? (ignore weight of forearm). Free body diagram Apply equations: Fy = 0 Fj,y + Fm - Fext = 0 Fm = 1333 N, Fext = 100 N Fj,y = -1233 N Fx = 0 Fj,x = 0 N Fj,y Fext Fm Rm Fj,x Rext

  19. If a person holds a 100 N weight in their hand 0.4 m from the elbow, what is the joint reaction force? (ignore weight of forearm). • Free body diagram • Apply equations Fy = 0 -Fj,y + Fm - Fext = 0 Fm = 1333 N, Fext = 100 N Fj,y = 1233 N Fx = 0 Fj,x = 0 N Fj,y Fext Fm Rm Fj,x Rext

  20. What is the muscle force when a 5 kg briefcase is held with straight arm? Fm Fj Upper arm force Forearm force Briefcase force

  21. 20N 0.16 m 15N T = (49N * 0.65m) + (15N * 0.48m) + (20N * 0.16m) T = (31 Nm) + (7.2 Nm) + (3.2 Nm) T = 41 Nm 15N 0.48 m 49N 0.48 m 0.65 m

  22. Fm Fj 20N 0.16 m Rm = 0.025 m, Fm = ??? 15N 0.48 m 49N 0.65 m -1640 Nm 1640 Nm 1640 N None of the above

  23. Fm Fj 20N 0.16 m Rm = 0.025 m Mshoulder = 0 0 = Tbriefcase+ = Tlowerarm + Tupperarm – (Tmuscle) 0 = (49N * 0.65m) + (15N * 0.48m) + (20N * 0.16m) - (Fm * 0.025m) 0 = 41 Nm – Fm*0.025 Fm = 1,640 N 15N 0.48 m 49N 0.65 m

  24. Fm Fj 20N 0.16 m Does Fjx = 0? 15N 0.48 m 49N 0.65 m Yes No It depends

  25. At the end of stance phase while running, Fg,x under the right foot is 200N and Fg,y is 350N. Fg is applied to the foot 0.2 m from the ankle. What is the ankle torque due to Fg? What is the ankle torque due to Fg? What is the ankle extensor muscle force? (MRmusc= 0.05m) 0.2 m 30° 200N 350N

  26. At the end of stance phase while running, Fg,x under the right foot is 200N and Fg,y is 350N. Fg is applied to the foot 0.2 m from the ankle. What is the ankle torque due to Fg? Step 1: Find moment arm of Fg,x(MRx) & Fg,y(MRy) about ankle. 0.2 m 30° 200N 350N

  27. 0.2 m MRx 30° 200N 350N At the end of stance phase while running, Fg,x under the right foot is 200N and Fg,y is 350N. Fg is applied to the foot 0.2 m from the ankle. What is the ankle torque due to Fg? Step1 MRx= 0.2 sin 30° = 0.10 m

  28. MRy 0.2 m MRx 30° 200N 350N At the end of stance phase while running, Fg,x under the right foot is 200N and Fg,y is 350N. Fg is applied to the foot 0.2 m from the ankle. What is the ankle torque due to Fg? Step 1 MRx= 0.2 sin 30° = 0.10 m MRy= 0.2 cos 30° = 0.17 m Step 2 T = (Tx) + (Ty) T= (Fg,x*MRx) + (Fg,y * MRy) T = (200 * 0.10) + (350 * 0.17) T = 79.5 Nm

  29. MRy 0.2 m MRx 30° 200N 350N At the end of stance phase while running, Fg,x under the right foot is 200N and Fg,y is 350N. Fg is applied to the foot 0.2 m from the ankle. What is the ankle extensor muscle force?MRmusc = 0.05m

  30. MRy 0.2 m MRx 30° 200N 350N At the end of stance phase while running, Fg,x under the right foot is 200N and Fg,y is 350N. Fg is applied to the foot 0.2 m from the ankle. What is the ankle extensor muscle force? T = 79.5 Nm Mankle = 0 0=79.5 Nm – Tmusc 0=79.5 Nm – MRmusc*Fmusc If MRmusc = 0.05m Fmusc = 79.5/0.05 = 1590 N

  31. A person is holding a weight (100 N) in their hand. The weight is a distance of 0.3 meters from the center of rotation of the elbow. • If the forearm is parallel to the ground, what is the torque about the elbow due to the weight? • If the forearm has an angle of 30° above the horizontal (hand is higher than elbow), what is the torque about the elbow due to the weight? 3. If an elbow flexor muscle has a moment arm of 0.030 m about the elbow, what force must it generate to hold the forearm at an angle of 50° above the horizontal with the weight in the hand (ignoring the weight of the forearm)?

  32. A person is holding a weight (100 N) in their hand. The weight is a distance of 0.3 meters from the center of rotation of the elbow. 1. If the forearm is parallel to the ground, what is the torque about the elbow due to the weight? Telbow = F * R = 100 N * 0.3 meters = 30 Nm 2. If the forearm has an angle of 30° above the horizontal (hand is higher than elbow), what is the torque about the elbow due to the weight? Telbow = F * R R = 0.3cos 30° = 0.26 meters Telbow = 100 N * 0.26 m = 26 Nm 3. If an elbow flexor muscle has a moment arm of 0.030 m about the elbow, what force must it generate to hold the forearm at an angle of 50° above the horizontal with the weight in the hand (ignoring the weight of the forearm)? SMelbow = 0: FwRw – FmRm = 0 where Fm & Rm are the muscle force and its moment arm respectively, and Fw & Rw are the 100N weight and its moment arm respectively. Rw = 0.3 * cos 50° = 0.19 meters 100 * 0.19 – Fm * 0.03 = 0 Fm = 642 N

  33. A person is wearing a weight boot (150N) and doing exercises to strengthen the knee extensor muscles. The center of mass of the weight boot is 0.35 meters from the center of rotation of the knee. The person's foot + shank has a weight of 40N and its center of mass is 0.2 meters from the center of rotation of the knee. These distances are along the length of the shank. • When the shank is perpendicular to the ground, what is the total torque about the knee? (Be sure to use a free-body diagram.) • When the shank is held at a position where it is 20° below the horizontal (foot is lower than knee), what is the torque about the knee? (Be sure to use a free-body diagram. Note that the moment arm of the weight boot about the knee changes with knee angle.) 3. If the quadriceps muscle group has a moment arm of 0.025 m, what is the muscle force needed to hold the shank at 20° below the horizontal?

  34. A person is wearing a weight boot (150N) and doing exercises to strengthen the knee extensor muscles. The center of mass of the weight boot is 0.35 meters from the center of rotation of the knee. The person's foot + shank has a weight of 40N and its center of mass is 0.2 meters from the center of rotation of the knee. These distances are along the length of the shank. 1. When the shank is perpendicular to the ground, what is the total torque about the knee? (Be sure to use a free-body diagram.) All of the moment arms are zero because the force vectors go directly through the center of rotation of the knee. As a result, the total torque is 0. 2. When the shank is held at a position where it is 20° below the horizontal (foot is lower than knee), what is the torque about the knee? (Be sure to use a free-body diagram. Note that the moment arm of the weight boot about the knee changes with knee angle.) T = - (T due to shank weight + T due to weight boot) = -[(40)(0.2 cos 20)] + [(150)(0.35 cos 20)] = -56.9 N • m 3. If the quadriceps muscle group has a moment arm of 0.025 m, what is the muscle force needed to hold the shank at 20° below the horizontal? SMknee = 0 = -[(40)(0.2 cos 20)] + [(150)(0.35 cos 20)] + 0.025 • Fm,quad Fm,quad = 2.27 kN

  35. C.O.M. Elbow 0.2m What is the moment of inertia of the forearm about the elbow?Given: ICOM= 0.0065 kg * m2m =1.2kg & COM is 0.2m distal to elbow Iprox = ? a) 0.0415 kg m2 b) 0.0545 kg m2 c) 0.2465 d) I have no idea

  36. C.O.M. Elbow 0.2m What is the moment of inertia of the forearm about the elbow?Given: IC.O.M. = 0.0065 kg * m2m =1.2kg & com 0.2m distal to elbow • Iprox = ? a) 0.0415 kg m2 b) 0.0545 kg m2 c) 0.2465 • Parallel axis theorem Iprox = IC.O.M. + mr2 Iprox = 0.0065 + (1.2)(0.22) = 0.0545 kg * m2

  37. COM Elbow 0.2m Lets include the hand:What is the moment of inertia of the hand and forearm about the elbow when fully extended?Given: Forearm: ICOM = 0.0065 kg * m2, m =1.2kg, COM is 0.2m distal to elbowHand: ICOM = 0.0008 kg * m2, m =0.3kg, COM is 0.056m distal to wristLength of forearm: 0.3m • Iprox = ? a) 0.0917 kg m2 b) 0.0546 kg m2 c) 0.0933 kg m2 d) I have no idea Wrist COM 0.056m

  38. COM Elbow 0.2m Lets include the hand:What is the moment of inertia of the hand and forearm about the elbow when fully extended?Given: Forearm: ICOM = 0.0065 kg * m2, m =1.2kg, com is 0.2m distal to elbowHand: ICOM = 0.0008 kg * m2, m =0.3kg, com is 0.056m distal to wristLength of forearm: 0.3m • Iprox = ? a) 0.0917 kg m2 (subtracted) b) 0.0546 kg m2 (forgot forearm length) c) 0.0933 kg m2 • Parallel axis theorem Iprox = (ICOM + mr2)forearm+(ICOM + mr2)hand Iprox = 0.0065 + (1.2)(0.22) + 0.0008+(0.3)(0.3+0.056)2 Iprox= 0.0545 + 0.0388 Iprox = 0.0933 kg m2 Wrist COM 0.056m

  39. What is Fmusc needed to accelerate the forearm at 20 rad/s2 about the elbow? ( ICOM = 0.0065 kg * m2 ; Iprox = 0.054 kg * m2) Fj Fm Rm  Elbow Rw Fw Step 1: Draw free body diagram. Step 2: :  M = I 

  40. What is Fmusc needed to accelerate the forearm at 20 rad/s2 about the elbow? ( ICOM = 0.0065 kg * m2 ; Iprox = 0.054 kg * m2) Fj Fm Rm  Elbow Rw Fw Step 1: Draw free body diagram. Step 2: :  M = I  a)  Melbow = Iprox • b)  Melbow = Icom • c)  Mcom = Iprox • d) I’m lost

  41. What is Fmusc needed to accelerate the forearm at 20 rad/s2 about the elbow? ( ICOM = 0.0065 kg * m2 ; Iprox = 0.054 kg * m2) Fj Fm Rm  Elbow Rw Fw Step 1: Draw free body diagram. Step 2: :  M = I   Melbow = Iprox  Melbow = 0.054 * 20 = 1.1 Nm Step 3: Find moments due to each force on forearm (Fm * Rm) - (Fw * Rw) = 1.1 N m Fw = 11 N, Rw = 0.16 m, Rm = 0.03 m Fm = 95 N

  42. Fj Fflex Fext  Elbow Fw What is net muscle moment needed to accelerate the forearm at 20 rad/s2 about the elbow? (Iprox = 0.06 kg * m2, Fw = 15 N, Rw = 0.2 m) Net muscle moment?

  43. Fj Fm,flex Fm,ext  Elbow Fw Net muscle moment: net moment due to all active muscles Mmus =- (Fm,ext*Rm,ext) + (Fm,flex*Rm,flex)

  44. What is net muscle moment needed to accelerate the forearm at 20 rad/s2 about the elbow? (Iprox = 0.06 kg * m2, Fw = 15 N, Rw = 0.2 m) Step 1: Free body diagram Step 2:  Melbow = Iprox  Melbow = (0.06)(20) = 1.2 N m Step 3: Find sum of the moments about the elbow  Melbow = 1.2 = Mmus - (Fw * Rw) Mmus = 4.2 N • m Fj Mmus Rw Fw

  45. Another sleepy day in 4540 Keeping your head upright requires alertness but not much muscle force. Given this diagram/information, calculate the muscle force. Head mass = 4 kg. a) 2.4 N b) 23.544N c) 0.0589 N d) I’m lost

  46. Another sleepy day in 4540 Keeping your head upright requires alertness but not much muscle force. Given this diagram/information, calculate the muscle force. Head mass = 4 kg. a) 2.4 N (forgot 9.81) b) 23.544N c) 0.0589 N (multiplied moment arm) d) I’m lost

  47. I-70 Nightmare While driving, you start to nod off asleep, and a very protective reaction kicks in, activating your neck muscles and jerking your head up in the nick of time to avoid an accident.

  48. I-70 Nightmare Calculate the muscle force needed to cause a neck extension acceleration of 10 rad/s2. The moment of inertia of the head about the head-neck joint is 0.10 kg m2. a) 123.1 N b) -73 N c) -0.117 N d) I’m lost

  49. I-70 Nightmare Calculate the muscle force needed to cause a neck extension acceleration of 10 rad/s2. The moment of inertia of the head about the head-neck joint is 0.10 kg m2. a) 123.1 N b) -73 N (sign mistake) c) -0.117 N (mulitplied moment arm) d) I’m lost

  50. How much torque must be generated by the deltoid muscle to hold a 60N dumbbell straight out at a 90o arm position? The dumbbell is 0.6m from the shoulder joint. The center of mass of the arm, weighing 30N, is 0.25 m from the shoulder joint. The moment arm for the deltoid muscle is 0.05m. • 870 Nm • 570 Nm • 43.5 Nm • -870 Nm

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