What is the Torque (T) due to force F?
F=100N; distance to F: rF = 1m, q=30o
T = MR * F
F
a)
b)
c)
d)
e)
86.6 N
100 Nm
86.6 Nm
50 Nm
50 N
MR
q
rF
Elbow torque due to weight of forearm
11 N
T=MR * F
T = 0.16m * 11N
T = 1.8 Nm
Direction?
T = -1.8Nm
0.16 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
Fw
q
rF
a)
b)
c)
d)
e)
1.76 Nm
1.5 Nm
0.88 Nm
-1.5 Nm
-0.88 Nm
A person holds their forearm so that it is 30° below the
horizontal. Elbow torque due to forearm weight?
Fw=11N; rF = 0.16m
11 N
MR
30°
0.16 m
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
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
We must consider the
effects of 2 forces:
forearm (11N)
weight being held (100N)
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
11 N
0.16
0.4 m
A)
B)
C)
D)
E)
T= (-Tarm) + (-Tbriefcase )
T = Tarm+Tbriefcase
T = (-Tarm) + Tbriefcase
T = Tarm + (- Tbriefcase)
It depends
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
11 N
0.16
0.4 m
T = (-Tarm) + (-Tbriefcase )
T = -(11 N * 0.16 m) (100 N * 0.4 m)
T = -42 Nm
Torque about shoulder due to external
forces when 5 kg briefcase is held with
straight arm.
Briefcase
force
Forearm
force
Upper
arm
force
Briefcase
force
Forearm
force
Upper
arm
force
20N
0.16 m
15N
15N
0.48 m
0.48 m
49N
0.65 m
a)
b)
c)
d)
e)
31 Nm
20.6 Nm
- 41Nm
41 Nm
None of the above
20N
0.16 m
15N
15N
0.48 m
0.48 m
49N
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
Muscles create torques about joints
Upper
arm
Elbow
Elbow
flexor
muscle
Biceps
force
Forearm
T
Solve for Muscle
force, Fm
Static equilibrium
Fj,y Fm
Rm
Rw
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
Fw
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
Upper
arm
Rm = 0.03 m
Rext = 0.4 m
Muscle
Fext
Step 1: Free body diagram
Elbow
Forearm
Solve for
Elbow flexor force
Fm
Fj,y
Fj,x
Rm
Rext
Rm = 0.03 m
Fext = 100 N
Rext = 0.4 m
Fext
Solve for
Elbow flexor force
Fm
Fj,y
Fj,x
Rm
Rext
Fext
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
When Rm < Rext,
muscle force > external force
Upper
arm
Biceps
brachialis
Fext
Rm
Elbow
Rext
Fm = Fext (Rext / Rm)
Last example
Fext = 100N
Rext > Rm
Fm = 1333 N
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).
Fj,y Fm
Fj,x
Rm
Rext
Fext
Free body diagram
Apply equations
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).
Fj,y Fm
Fj,x
Rm
Rext
Fext
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
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).
Fj,y Fm
Fj,x
Rm
Fext
• Free body diagram
• Apply equations
Fy = 0
-Fj,y + Fm - Fext = 0
Fm = 1333 N, Fext = 100 N
Fj,y = 1233 N
Rext
 Fx = 0
Fj,x = 0 N
What is the muscle force when a 5 kg
briefcase is held with straight arm?
Fm
Fj
Forearm
Briefcase force
force
Upper
arm
force
20N
0.16 m
15N
15N
0.48 m
0.48 m
49N
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
Fm
20N
15N
0.48 m
49N
0.65 m
Rm = 0.025 m, Fm = ???
a)
b)
c)
d)
-1640 Nm
1640 Nm
1640 N
None of the above
Fj
0.16 m
Fm
20N
15N
Fj
0.16 m
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
Fm
20N
15N
0.48 m
49N
0.65 m
Does Fjx = 0?
a) Yes
b) No
c) It depends
Fj
0.16 m
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?
0.2 m
30°
200N
350N
What is the ankle extensor muscle
force?
(MRmusc = 0.05m)
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
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
MRx
0.2 m
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
MRy
MRx
0.2 m
30°
200N
350N
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
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
MRy
MRx
0.2 m
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?
MRy
MRx
0.2 m
30°
200N
350N
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
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 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)?
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)?
Melbow = 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
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 0.025 m, what
is the muscle force needed to hold the shank at 20° below the
horizontal?
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?
Mknee = 0 = -[(40)(0.2 cos 20)] + [(150)(0.35 cos 20)] + 0.025 • Fm,quad
Fm,quad = 2.27 kN
What is the moment of inertia of the forearm
about the elbow?
Given: ICOM = 0.0065 kg * m2
m =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
Elbow
C.O.M.
0.2m
What is the moment of inertia of the forearm
about the elbow?
Given: IC.O.M. = 0.0065 kg * m2
m =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
Elbow
C.O.M.
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 elbow
Hand: ICOM = 0.0008 kg * m2, m =0.3kg, COM is 0.056m distal to wrist
Length 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
Elbow
COM
0.2m
COM
0.056m
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 elbow
Hand: ICOM = 0.0008 kg * m2, m =0.3kg, com is 0.056m distal to wrist
Length 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
Elbow
COM
0.2m
COM
0.056m
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)
Step 1: Draw free body diagram.
Step 2: :  M = I 
Fj
Fm
Rm
Elbow
Rw

Fw
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)
Step 1: Draw free body diagram.
Step 2: :  M = I 
a) Melbow = Iprox 
b) Melbow = Icom 
c) Mcom = Iprox 
d) I’m lost
Fj
Fm
Rm
Elbow
Rw

Fw
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)
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
Fj
Elbow
Fm
Rm
Rw

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?
Fext
Fj F
flex

Elbow
Fw
Net muscle moment: net moment due
to all active muscles
Mmus =- (Fm,ext*Rm,ext) + (Fm,flex*Rm,flex)
Fm,ext
Fj
Fm,flex

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)
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
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
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
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.
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
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
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.
A. 870 Nm
B. 570 Nm
C. 43.5 Nm
D. -870 Nm
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.
A. 870 Nm (muscle force)
B. 570 Nm (muscle force and sign error)
C. 43.5 Nm
D. -870 Nm