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 ...
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T = MR * F
What is the Torque (T) due to force F?F=100N; distance to F: rF = 1m, q=30o
MR
F
rF
q
a) 86.6 Nb) 100 Nmc) 86.6 Nmd) 50 Nme) 50 N
Elbow torque due to weight of forearm
T=MR * FT = 0.16m * 11NT = 1.8 Nm
Direction?T = -1.8Nm
0.16 m
11 N
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
rF
a) 1.76 Nmb) 1.5 Nmc) 0.88 Nmd) -1.5 Nme) -0.88 Nm
q
T=MR*FF = 11 NMR=rFcos 30°
MR = 0.16 cos 30° = 0.14 mT = 1.5 N mUse right-hand rule:T = -1.5 Nm
11 NMR30°
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
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
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= (-Tarm) + (-Tbriefcase )B) T = Tarm+Tbriefcase
C) T = (-Tarm) + Tbriefcase
D) T = Tarm + (- Tbriefcase)E) It depends
0.4 m
100 N
0.16
11 N
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
0.4 m
100 N
0.16
11 N
Torque about shoulder due to external forces when 5 kg briefcase is held with straight arm.
Briefcaseforce
Forearmforce
Upperarmforce
49N
0.65 m
0.48 m
20N0.16 m15N
0.48 m
15N
Briefcaseforce
Forearmforce
Upperarmforce
a) 31 Nmb) 20.6 Nmc) - 41Nmd) 41 Nme) None of the above
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
20N0.16 m15N
0.48 m
15N
Muscles create torques about joints
Elbowflexormuscle
Elbow
Upperarm
Forearm
Bicepsforce
T
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
Solve for Muscle force, Fm
Rw
FwFm
Rm
Fj,y
Ignore weight of forearm Information
Rm = 0.03 m
Rext = 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
Upperarm
Forearm
Fext
Solve forElbow flexor force
Rm = 0.03 m
Fext = 100 N
Rext = 0.4 m
Rext
Fm
Rm
FextFj,y
Fj,x
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
Rext
Fm
Rm
FextFj,y
Fj,x
When Rm < Rext,muscle force > external force
Bicepsbrachialis
Elbow
Upperarm
Fext
Rext
Rm
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).
Free body diagramApply equations
Rext
Fm
Rm
Fj,y Fext
Fj,x
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 diagramApply equations: Fy = 0
Fj,y + Fm - Fext = 0
Fm = 1333 N, Fext = 100 N
Fj,y = -1233 N
Fx = 0
Fj,x = 0 N
Rext
Fm
Rm
Fj,y Fext
Fj,x
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
Rext
Fm
Rm
Fj,y Fext
Fj,x
What is the muscle force when a 5 kg briefcase is held with straight arm?
Briefcaseforce
Forearmforce
Upperarmforce
Fm
Fj
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
20N0.16 m15N
0.48 m
15N
Rm = 0.025 m, Fm = ???
49N
0.65 m
0.48 m
20N0.16 m
15N
Fm
Fj
a) -1640 Nmb) 1640 Nmc) 1640 Nd) None of the above
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
49N
0.65 m
0.48 m
20N0.16 m
15N
Fm
Fj
Does Fjx = 0?
49N
0.65 m
0.48 m
20N0.16 m
15N
Fm
Fj
a) Yesb) Noc) It depends
What is the ankle torque due to Fg?
What is the ankle extensor muscle force? (MRmusc = 0.05m)
30°
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?
200N
0.2 m
Step 1: Find moment arm of Fg,x (MRx) & Fg,y (MRy) about ankle.
30°
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?
200N
0.2 m
Step1MRx = 0.2 sin 30° = 0.10 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?
30°
350N
200N
MRx0.2 m
Step 1
MRx = 0.2 sin 30° = 0.10 m
MRy = 0.2 cos 30° = 0.17 mStep 2T = (Tx) + (Ty)
T = (Fg,x *MRx) + (Fg,y * MRy)
T = (200 * 0.10) + (350 * 0.17)T = 79.5 Nm
30°
350N
200N
MRx
MRy
0.2 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?
30°
350N
200N
MRx
MRy
0.2 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 extensor muscle force?
MRmusc = 0.05m
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
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?
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 * RR = 0.3cos 30° = 0.26 metersTelbow = 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 meters100 * 0.19 – Fm * 0.03 = 0Fm = 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?
SMknee = 0 = -[(40)(0.2 cos 20)] + [(150)(0.35 cos 20)] + 0.025 • Fm,quad
Fm,quad = 2.27 kN
Iprox = ?a) 0.0415 kg m2
b) 0.0545 kg m2
c) 0.2465d) I have no idea
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
ElbowC.O.M.
0.2m
• 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
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
ElbowC.O.M.
0.2m
• Iprox = ?a) 0.0917 kg m2
b) 0.0546 kg m2
c) 0.0933 kg m2
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: 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 wristLength of forearm: 0.3m
ElbowCOM
0.2m
COM
0.056mW
rist
• 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
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 wristLength of forearm: 0.3m
ElbowCOM
0.2m
COM
0.056mW
rist
Step 1: Draw free body diagram.Step 2: : M = I
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)
Elbow
FmFj
Fw
Rm
Rw
Step 1: Draw free body diagram.Step 2: : M = I
a) Melbow = Iprox b) Melbow = Icom c) Mcom = Iprox d) I’m lost
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)
Elbow
FmFj
Fw
Rm
Rw
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
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)
Elbow
FmFj
Fw
Rm
Rw
Net muscle moment?
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)
Elbow
FflexFj
Fw
Fext
Net muscle moment: net moment due to all active muscles
Mmus =- (Fm,ext*Rm,ext) + (Fm,flex*Rm,flex)
Fm,flexFm,ext
Elbow
Fj
Fw
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
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)
Fj
FwRw
Mmus
Another sleepy day in 4540Keeping 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 4540Keeping 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 NightmareWhile 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 NightmareCalculate 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 NightmareCalculate 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 NmB. 570 NmC. 43.5 NmD. -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 NmD. -870 Nm
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