1. What is the Torque (T) due to force F
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
C is correct 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
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
D is correct q rF

4. T=MR*F F = 11 N MR=rFcos 30° MR = 0.16 cos 30° = 0.14 m T = 1.5 N m
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=rFcos 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

5. We must consider the effects of 2 forces:
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)
100 N
11 N

6. T= (-Tarm) + (-Tbriefcase ) T = Tarm+Tbriefcase
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?
100 N
T= (-Tarm) + (-Tbriefcase )
T = Tarm+Tbriefcase
T = (-Tarm) + Tbriefcase
T = Tarm + (- Tbriefcase)
It depends
11 N
0.16
0.4 m

7. T = (-Tarm) + (-Tbriefcase ) T = -(11 N * 0.16 m) - (100 N * 0.4 m)
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?
100 N
T = (-Tarm) + (-Tbriefcase )
T = -(11 N * 0.16 m) (100 N * 0.4 m)
T = -42 Nm
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
T= (49N * 0.65m) + (15N * 0.48m) + (20N * 0.16m)
T = (31 Nm) + (7.2 Nm) + (3.2 Nm)
T = 41 Nm
D is correct
31 Nm
20.6 Nm
- 41Nm
41 Nm
None of the above

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

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

12. Solve for Muscle force, Fm
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. Ignore weight of forearm Information
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. Solve for Elbow flexor force
Rm = 0.03 m
Fext = 100 N
Rext = 0.4 m
Solve for Elbow flexor force
Rext Fm Rm
Fext
Fj,y
Fj,x

15. Solve for Elbow 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
Solve for Elbow flexor force
Rext Fm Rm
Fext
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. Free body diagram Apply equations Fj,y Fext Fm Fj,x Rm
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
Fj,x Rm
Rext

18. Free body diagram Apply equations: Fy = 0 Fj,y Fext Fm
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 = N
Fx = 0
Fj,x = 0 N
Fj,y
Fext Fm
Fj,x Rm
Rext

19. Free body diagram Apply equations Fy = 0 Fj,y Fext Fm
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
Fj,x Rm
Rext

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

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

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

23. Fm Fj 20N 0.16 m 15N 0.48 m 49N 0.65 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

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

25. What is the ankle torque due to Fg?
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. Step 1: Find moment arm of Fg,x (MRx) & Fg,y (MRy) about ankle.
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. 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
30°
350N
200N
MRx
0.2 m

28. 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?
30°
350N
200N
MRx
MRy
0.2 m
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. 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°
350N
200N
MRx
MRy
0.2 m

30. T = 79.5 Nm  Mankle = 0 0=79.5 Nm – Tmusc 0=79.5 Nm – MRmusc*Fmusc
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
30°
350N
200N
MRx
MRy
0.2 m

31. A person is holding a weight (100 N) in their hand
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 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
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 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 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)] = 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?
SMknee = 0 = -[(40)(0.2 cos 20)] + [(150)(0.35 cos 20)] • Fm,quad
Fm,quad = 2.27 kN

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

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

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

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

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

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

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

42. (Iprox = 0.06 kg * m2, Fw = 15 N, Rw = 0.2 m)
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? 
Elbow
Fflex Fj Fw
Fext

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

44. (Iprox = 0.06 kg * m2, Fw = 15 N, Rw = 0.2 m)
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) N
c) N
d) I’m lost
Fm*0.05=0.03*mg

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
Fm*0.05=0.03*mg

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) 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/s2. The moment of inertia of the head about the head-neck joint is 0.10 kg m2.
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 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
Td=60*0.6
Tw=30*0.25
Tm=Fm*0.05

51. 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 (muscle force)
570 Nm (muscle force and sign error)
43.5 Nm
-870 Nm