Slide 1

T = MR * F

What is the Torque (T) due to force F?F=100N; distance to F: rF = 1m, q=30oMRFrFq86.6 N100 Nm86.6 Nm50 Nm50 NC is correct1Elbow torque due to weight of forearm T=MR * FT = 0.16m * 11NT = 1.8 Nm

Direction?T = -1.8Nm0.16 m11 N

2A person holds their forearm so that it is 30 below the horizontal. Elbow torque due to forearm weight? Fw=11N; rF = 0.16mFwrF

1.76 Nm1.5 Nm0.88 Nm-1.5 Nm-0.88 Nm

qD is correct3T=MR*FF = 11 NMR=rFcos 30 MR = 0.16 cos 30 = 0.14 mT = 1.5 N mUse right-hand rule:T = -1.5 Nm11 NMR300.16 mA person holds their forearm so that it is 30 below the horizontal. Elbow torque due to forearm weight?Fw=11N; rF = 0.16m

4We 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 N11 N

5A 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 = Tarm+TbriefcaseT = (-Tarm) + TbriefcaseT = Tarm + (- Tbriefcase)It depends

0.4 m100 N0.1611 N

6A 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 Nm0.4 m100 N0.1611 N

7Torque about shoulder due to external forces when 5 kg briefcase is held with straight arm.

BriefcaseforceForearmforceUpperarmforce

8

49N0.65 m0.48 m20N0.16 m15N0.48 m15N

BriefcaseforceForearmforceUpperarmforce31 Nm20.6 Nm- 41Nm41 NmNone of the aboveT= (49N * 0.65m) + (15N * 0.48m) + (20N * 0.16m)T = (31 Nm) + (7.2 Nm) + (3.2 Nm)T = 41 Nm

D is correct9T= (49N * 0.65m) + (15N * 0.48m) + (20N * 0.16m)T = (31 Nm) + (7.2 Nm) + (3.2 Nm)T = 41 Nm

49N0.65 m0.48 m20N0.16 m15N0.48 m15N

10Muscles create torques about jointsElbowflexormuscleElbowUpperarmForearmBicepsforceT11Static equilibriumMelbow = 0 Fj creates no moment at elbow-(Tw) + (Tm) = 0 -(Fw * Rw) + (Fm * Rm) = 0 Fm = (Fw * Rw) / RmSubstitute: Fw = 11 N, Rw = 0.16 m, Rm = 0.03 m Fm = 59 NSolve for Muscle force, FmRwFwFmRmFj,y

12Ignore weight of forearm Information Rm = 0.03 m Rext = 0.4 mStep 1: Free body diagramIf a person holds a 100 N weight in their hand 0.4 m from the elbow, what is the elbow flexor muscle force?MuscleElbowUpperarmForearmFext13Solve forElbow flexor forceRm = 0.03 mFext = 100 NRext = 0.4 mRextFmRmFextFj,yFj,x

14Solve forElbow flexor forceRm = 0.03 mFext = 100 NRext = 0.4 mMelbow = 0(Tm) (Text) = 0(FmRm) - (FextRext) = 0Fm = Fext (Rext / Rm)Fm = 1333 NRextFmRmFextFj,yFj,x

15When Rm < Rext,muscle force > external forceBicepsbrachialisElbowUpperarmFextRextRmFm = Fext (Rext / Rm)Last example Fext = 100N Rext > Rm Fm = 1333 N

16If 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 equationsRextFmRmFj,yFextFj,x

17If 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 = 0Fj,y + Fm - Fext = 0Fm = 1333 N, Fext = 100 NFj,y = -1233 N

Fx = 0Fj,x = 0 NRextFmRmFj,yFextFj,x

18If 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 = 0Fm = 1333 N, Fext = 100 NFj,y = 1233 N

Fx = 0Fj,x = 0 NRextFmRmFj,yFextFj,x

19What is the muscle force when a 5 kg briefcase is held with straight arm?

BriefcaseforceForearmforceUpperarmforceFmFj

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

49N0.65 m0.48 m20N0.16 m15N0.48 m15N

21Rm = 0.025 m, Fm = ???

49N0.65 m0.48 m20N0.16 m15NFmFj

-1640 Nm1640 Nm1640 NNone of the above 0 = Tbriefcase + = Tlowerarm + Tupperarm ( Tmuscle ) 0 = (49N * 0.65m) + (15N * 0.48m) + (20N *0.16m) - (Fm * 0.025m) 0 = 41 Nm Fm*0.025Fm = 1,640 N

1640 N-1640 N (add Tmuscle)

22Rm = 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*0.025Fm = 1,640 N

49N0.65 m0.48 m20N0.16 m15NFmFj

23Does Fjx = 0?

49N0.65 m0.48 m20N0.16 m15NFmFj

YesNoIt dependsC is correct24What is the ankle torque due to Fg?

What is the ankle extensor muscle force? (MRmusc = 0.05m)30350NAt 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?200N0.2 m

25Step 1: Find moment arm of Fg,x (MRx) & Fg,y (MRy) about ankle.30350NAt 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?200N0.2 m

26Step1MRx = 0.2 sin 30 = 0.10 mAt 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?30350N200NMRx0.2 m

27Step 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 Nm30350N200NMRxMRy0.2 mAt 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?

2830350N200NMRxMRy0.2 mAt 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

29T = 79.5 Nm Mankle = 00=79.5 Nm Tmusc0=79.5 Nm MRmusc*Fmusc

If MRmusc = 0.05mFmusc = 79.5/0.05 = 1590 N30350N200NMRxMRy0.2 mAt 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?

30A 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 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)?

31A 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.

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 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,quadFm,quad = 2.27 kN

Iprox = ?a) 0.0415 kg m2b) 0.0545 kg m2c) 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.0.2mIprox = ?a) 0.0415 kg m2b) 0.0545 kg m2c) 0.2465 Parallel axis theorem Iprox = IC.O.M. + mr2 Iprox = 0.0065 + (1.2)(0.22) = 0.0545 kg * m2What 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 elbowElbowC.O.M.0.2mIprox = ?a) 0.0917 kg m2b) 0.0546 kg m2c) 0.0933 kg m2 d) I have no ideaLets 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 elbowHand: ICOM = 0.0008 kg * m2, m =0.3kg, COM is 0.056m distal to wristLength of forearm: 0.3mElbowCOM0.2mCOM0.056mWristIprox = ?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 m2Lets 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 elbowHand: ICOM = 0.0008 kg * m2, m =0.3kg, com is 0.056m distal to wristLength of forearm: 0.3mElbowCOM0.2mCOM0.056mWristStep 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) ElbowFmFjFwRmRw

Step 1: Draw free body diagram.Step 2: : M = I

a) Melbow = Iprox b) Melbow = Icom c) Mcom = Iprox d) Im 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) ElbowFmFjFwRmRw

Step 1: Draw free body diagram.Step 2: : M = I Melbow = Iprox Melbow = 0.054 * 20 = 1.1 NmStep 3: Find moments due to each force on forearm(Fm * Rm) - (Fw * Rw) = 1.1 N mFw = 11 N, Rw = 0.16 m, Rm = 0.03 mFm = 95 NWhat is Fmusc needed to accelerate the forearm at 20 rad/s2 about the elbow? ( ICOM = 0.0065 kg * m2 ; Iprox = 0.054 kg * m2) ElbowFmFjFwRmRw

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) ElbowFflexFjFwFext

Net muscle moment: net moment due to all active musclesMmus =- (Fm,ext*Rm,ext) + (Fm,flex*Rm,flex)Fm,flexFm,ext ElbowFjFw

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 mWhat 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)FjFwRwMmus

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) Im lost

Fm*0.05=0.03*mg45Another 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) Im lost

Fm*0.05=0.03*mg

46I-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) Im 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) Im 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.870 Nm570 Nm43.5 Nm-870 Nm

Td=60*0.6Tw=30*0.25Tm=Fm*0.05

50How 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