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