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T = MR * F What is the Torque (T) due to force F? F=100N; distance to F: r F = 1m,  =30 o MR F rFrF  a)86.6 N b)100 Nm c)86.6 Nm d)50 Nm e)50 N.

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Presentation on theme: "T = MR * F What is the Torque (T) due to force F? F=100N; distance to F: r F = 1m,  =30 o MR F rFrF  a)86.6 N b)100 Nm c)86.6 Nm d)50 Nm e)50 N."— Presentation transcript:

1 T = MR * F What is the Torque (T) due to force F? F=100N; distance to F: r F = 1m,  =30 o MR F rFrF  a)86.6 N b)100 Nm c)86.6 Nm d)50 Nm e)50 N

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

3 A person holds their forearm so that it is 30° below the horizontal. Elbow torque due to forearm weight? F w =11N; r F = 0.16m FwFw rFrF a)1.76 Nm b)1.5 Nm c)0.88 Nm d)-1.5 Nm e)-0.88 Nm 

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

5 We must consider the effects of 2 forces: forearm (11N) weight being held (100N) 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? 100 N 11 N

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? A)T= (-T arm ) + (-T briefcase ) B)T = T arm +T briefcase C)T = (-T arm ) + T briefcase D)T = T arm + (- T briefcase ) E)It depends 0.4 m 100 N N

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 = (-T arm ) + (-T briefcase ) T = -(11 N * 0.16 m) - (100 N * 0.4 m) T = -42 Nm 0.4 m 100 N N

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

9 49N 0.65 m 0.48 m 20N 0.16 m 15N 0.48 m 15N Briefcase force Forearm force Upper arm force a)31 Nm b)20.6 Nm c)- 41Nm d)41 Nm e)None of the above

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

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

12 Static equilibrium  M elbow = 0 F j creates no moment at elbow -(T w ) + (T m ) = 0 -(F w * R w ) + (F m * R m ) = 0 F m = (F w * R w ) / R m Substitute: F w = 11 N, R w = 0.16 m, R m = 0.03 m F m = 59 N Solve for Muscle force, F m RwRw FwFw FmFm RmRm F j,y

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

14 Solve for Elbow flexor force R m = 0.03 m F ext = 100 N R ext = 0.4 m R ext FmFm RmRm F ext F j,y F j,x

15 Solve for Elbow flexor force R m = 0.03 m F ext = 100 N R ext = 0.4 m  M elbow = 0 (T m ) – (T ext ) = 0 (F m R m ) - (F ext R ext ) = 0 F m = F ext (R ext / R m ) F m = 1333 N R ext FmFm RmRm F ext F j,y F j,x

16 When R m external force Biceps brachialis Elbow Upper arm F ext R ext RmRm F m = F ext (R ext / R m ) Last example F ext = 100N R ext > R m F m = 1333 N

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 R ext FmFm RmRm F j,y F ext F j,x

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:  F y = 0 F j,y + F m - F ext = 0 F m = 1333 N, F ext = 100 N F j,y = N  F x = 0 F j,x = 0 N R ext FmFm RmRm F j,y F ext F j,x

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  F y = 0 -F j,y + F m - F ext = 0 F m = 1333 N, F ext = 100 N F j,y = 1233 N  F x = 0 F j,x = 0 N R ext FmFm RmRm F j,y F ext F j,x

20 What is the muscle force when a 5 kg briefcase is held with straight arm? Briefcase force Forearm force Upper arm force FmFm FjFj

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

22 R m = m, F m = ??? 49N 0.65 m 0.48 m 20N 0.16 m 15N FmFm FjFj a)-1640 Nm b)1640 Nm c)1640 N d)None of the above

23 R m = m  M shoulder = 0 0 = T briefcase + = T lowerarm + T upperarm – ( T muscle ) 0 = (49N * 0.65m) + (15N * 0.48m) + (20N * 0.16m) - (F m * 0.025m) 0 = 41 Nm – F m *0.025 F m = 1,640 N 49N 0.65 m 0.48 m 20N 0.16 m 15N FmFm FjFj

24 Does F jx = 0? 49N 0.65 m 0.48 m 20N 0.16 m 15N FmFm FjFj a)Yes b)No c)It depends

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

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

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

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

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

30 T = 79.5 Nm  M ankle = 0 0=79.5 Nm – T musc 0=79.5 Nm – MR musc *F musc If MR musc = 0.05m F musc = 79.5/0.05 = 1590 N 30° 350N 200N MR x MR y 0.2 m At the end of stance phase while running, F g,x under the right foot is 200N and F g,y is 350N. F g is applied to the foot 0.2 m from the ankle. What is the ankle extensor muscle force?

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. 1.If the forearm is parallel to the ground, what is the torque about the elbow due to the weight? 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? 3. If an elbow flexor muscle has a moment arm of 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? T elbow = 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? T elbow = F * R R = 0.3cos 30° = 0.26 meters T elbow = 100 N * 0.26 m = 26 Nm 3. If an elbow flexor muscle has a moment arm of 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)?  M elbow = 0: F w R w – F m R m = 0 where F m & R m are the muscle force and its moment arm respectively, and F w & R w are the 100N weight and its moment arm respectively. R w = 0.3 * cos 50° = 0.19 meters 100 * 0.19 – F m * 0.03 = 0 F m = 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. 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.) 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.) 3. If the quadriceps muscle group has a moment arm of 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 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)] = N m 3. If the quadriceps muscle group has a moment arm of m, what is the muscle force needed to hold the shank at 20° below the horizontal?  M knee = 0 = -[(40)(0.2 cos 20)] + [(150)(0.35 cos 20)] F m,quad F m,quad = 2.27 kN

35 I prox = ? a) kg m 2 b) kg m 2 c) d) I have no idea What is the moment of inertia of the forearm about the elbow? Given: I COM = kg * m 2 m =1.2kg & COM is 0.2m distal to elbow Elbow C.O.M. 0.2m

36 I prox = ? a) kg m 2 b) kg m 2 c) Parallel axis theorem I prox = I C.O.M. + mr 2 I prox = (1.2)(0.2 2 ) = kg * m 2 What is the moment of inertia of the forearm about the elbow? Given: I C.O.M. = kg * m 2 m =1.2kg & com 0.2m distal to elbow Elbow C.O.M. 0.2m

37 I prox = ? a) kg m 2 b) kg m 2 c) kg m 2 d) I have no idea Lets include the hand: What is the moment of inertia of the hand and forearm about the elbow when fully extended? Given: Forearm: I COM = kg * m 2, m =1.2kg, COM is 0.2m distal to elbow Hand: I COM = kg * m 2, m =0.3kg, COM is 0.056m distal to wrist Length of forearm: 0.3m Elbow COM 0.2m COM 0.056m Wrist

38 I prox = ? a) kg m 2 (subtracted) b) kg m 2 (forgot forearm length) c) kg m 2 Parallel axis theorem I prox = (I COM + mr 2 ) forearm +(I COM + mr 2 ) hand I prox = (1.2)(0.2 2 ) (0.3)( ) 2 I prox = I prox = kg m 2 Lets include the hand: What is the moment of inertia of the hand and forearm about the elbow when fully extended? Given: Forearm: I COM = kg * m 2, m =1.2kg, com is 0.2m distal to elbow Hand: I COM = kg * m 2, m =0.3kg, com is 0.056m distal to wrist Length of forearm: 0.3m Elbow COM 0.2m COM 0.056m Wrist

39 Step 1: Draw free body diagram. Step 2: :  M = I  What is F musc needed to accelerate the forearm at 20 rad/s 2 about the elbow? ( I COM = kg * m 2 ; I prox = kg * m 2 )  Elbow FmFm FjFj FwFw RmRm RwRw

40 Step 1: Draw free body diagram. Step 2: :  M = I  a  M elbow = I prox   b  M elbow = I com   c  M com = I prox   d) I’m lost  What is F musc needed to accelerate the forearm at 20 rad/s 2 about the elbow? ( I COM = kg * m 2 ; I prox = kg * m 2 )  Elbow FmFm FjFj FwFw RmRm RwRw

41 Step 1: Draw free body diagram. Step 2: :  M = I   M elbow = I prox   M elbow = * 20 = 1.1 Nm Step 3: Find moments due to each force on forearm (F m * R m ) - (F w * R w ) = 1.1 N m F w = 11 N, R w = 0.16 m, R m = 0.03 m F m = 95 N What is F musc needed to accelerate the forearm at 20 rad/s 2 about the elbow? ( I COM = kg * m 2 ; I prox = kg * m 2 )  Elbow FmFm FjFj FwFw RmRm RwRw

42 Net muscle moment? What is net muscle moment needed to accelerate the forearm at 20 rad/s 2 about the elbow? (I prox = 0.06 kg * m 2, F w = 15 N, R w = 0.2 m)  Elbow F flex FjFj FwFw F ext

43 Net muscle moment: net moment due to all active muscles M mus =- (F m,ext *R m,ext ) + (F m,flex *R m,flex ) F m,flex F m,ext  Elbow FjFj FwFw

44 Step 1: Free body diagram Step 2:  M elbow = I prox   M elbow = (0.06)(20) = 1.2 N m Step 3: Find sum of the moments about the elbow  M elbow = 1.2 = M mus - (F w * R w ) M mus = 4.2 N m What is net muscle moment needed to accelerate the forearm at 20 rad/s 2 about the elbow? (I prox = 0.06 kg * m 2, F w = 15 N, R w = 0.2 m) FjFj FwFw RwRw M mus

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) N c) 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) N c) 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/s 2. The moment of inertia of the head about the head-neck joint is 0.10 kg m 2. a) N b) -73 N c) N d) I’m lost

49 I-70 Nightmare Calculate the muscle force needed to cause a neck extension acceleration of 10 rad/s 2. The moment of inertia of the head about the head-neck joint is 0.10 kg m 2. a) N b) -73 N (sign mistake) c) 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 90 o 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. A.870 Nm B.570 Nm C.43.5 Nm D.-870 Nm

51 How much torque must be generated by the deltoid muscle to hold a 60N dumbbell straight out at a 90 o 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. A.870 Nm (muscle force) B.570 Nm (muscle force and sign error) C.43.5 Nm D.-870 Nm


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