Static Torque.pdf
Transcript of Static Torque.pdf
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Slide 1 Torques ESS 3093 Biomechanics
Torques and Moments of Force
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Slide 2 Torques ESS 3093 Biomechanics
Class Objectives
Introduce Levers and Mechanical Advantage Define Torque and moment arm
Moment arm as a radius vector Torque as a vector cross product
Static Equilibrium Including Torque Define Center of Mass Estimate the Location of the Center of Mass
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Slide 3 Torques ESS 3093 Biomechanics
Levers and Leverage
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Slide 4 Torques ESS 3093 Biomechanics
Mechanical Advantage
armreistance
armeffortMA =
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Slide 5 Torques ESS 3093 Biomechanics
Levers Trade Force for Distance If Mechanical Advantage is > 1,
Effort force < Resistance force Effort movement > Resistance movement
If Mechanical Advantage is < 1,
Effort force > Resistance force Effort movement < Resistance movement
What does that tell you about range of motion? Speed?
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Slide 6 Torques ESS 3093 Biomechanics
Classes of Levers
1st Class: Direction reversed
2nd Class Direction maintained MA > 1
3rd Class: Direction maintained MA < 1
Effort
Effort
Effort
Resistance
Resistance
Resistance
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Slide 7 Torques ESS 3093 Biomechanics
Torque The turning effect caused by a force Moment of force
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Slide 8 Torques ESS 3093 Biomechanics
Mathematical Definition of Torque T = r F
T = torque (Nm) r = moment arm (m)
Distance from the axis of rotation to the line of action of the force
Perpendicular to the line of action of the force F = force (N) Units: Newton x meters (Nm)
Axis of rotation
Moment arm
Force
x
90
Moment arm
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Slide 9 Torques ESS 3093 Biomechanics
Torque Direction
Direction + Torque = counter clockwise - Torque = clockwise
Right hand rule
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Slide 10 Torques ESS 3093 Biomechanics
Internal and External Torque
Internal Torque External Torque
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Slide 11 Torques ESS 3093 Biomechanics
Moment Arm
Definition: Distance from the axis of rotation to the line of action or the force AND Perpendicular to the line of action of the force
Axis
Moment arm Force
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Slide 12 Torques ESS 3093 Biomechanics
Moment Arm: General Case 1. What is the moment arm to the muscle
force, OR 2. What is the component of muscle force
to the bone?
Moment arm?
Muscle Force
Axis Bone
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Slide 13 Torques ESS 3093 Biomechanics
Moment Arm: General Case Method 1 Use trig to find the length of the moment arm from the axis to the line of action of the force Moment arm = d sin
Torque = Fm d sin
Moment arm?
Muscle Force (Fm)
Axis, Joint center
Bone
d
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Slide 14 Torques ESS 3093 Biomechanics
Moment Arm: General Case Method 2 Use trig to find the component of muscle force which acts to the bone F = Fm sin
T = d Fm sin Muscle
Force
Axis Bone
Component of Muscle Force to the bone
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Slide 15 Torques ESS 3093 Biomechanics
Example 1
A force of 200 N is exerted 0.25 m from the axis of rotation. What is the resulting torque?
r
F
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Slide 16 Torques ESS 3093 Biomechanics
Example 2
What is the torque about the elbow produced by the following conditions: a 900 N force pulling on the forearm at an angle of 110 deg from the horizon at a point 3 cm to the right of the elbows
axis of rotation with the forearm horizontal
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Slide 17 Torques ESS 3093 Biomechanics
Fm = 900 N
110 deg
d = 0.03 m
rm = ?
T = rm Fm rm = d sin 110
T = Fm d sin 110
T = 900 N x 0.03 x sin 110
T = 25.4 Nm
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Slide 18 Torques ESS 3093 Biomechanics
The muscle tendon moment arms vary with joint angle.
Arthur Jones
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Slide 19 Torques ESS 3093 Biomechanics
Moment arm as a vector Torque as a vector cross product
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Slide 20 Torques ESS 3093 Biomechanics
Simple Torque Calculation
T = r F +rx +Fy = +T (CCW) T = rx Fy
+rx
+Fy
Convention
+Y
+X
+ rotation
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Slide 21 Torques ESS 3093 Biomechanics
Simple Torque calculations
T = r F +ry +Fx = -T (CW) T = -(ry Fx)
+ry
+Fx Convention +Y
+X
+ rotation
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Slide 22 Torques ESS 3093 Biomechanics
Torque as a Vector Cross Product
Tz = (rx Fy) (ry Fx)
Tz= (xpa-xar) Fy (ypa-yar) Fx
Fy, Fx
rx = xpa xar ry = ypa - yar
Axis of rotation (ar), x,y
Point of application (pa) x,y
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Slide 23 Torques ESS 3093 Biomechanics
Torque as a Vector Cross Product
Tz = (rx Fy) (ry Fx)
Tz= (xpa-xar) Fy (ypa-yar) Fx
Fy
rx = xpa xar ry = ypa - yar
Axis of rotation (ar), x,y
Point of application (pa) x,y Fx
rx
ry
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Slide 24 Torques ESS 3093 Biomechanics
Fm = 900 N
110 deg
d = 0.03 m
rm = ?
F = 900 cos 110, 900 sin 110 r = 0.03, 0
T = (rx Fy) (ry Fx)
T = 0.03 x 900 sin 110 - 0
T = 25.4 Nm
FMx = 900N x cos 110 = -307.8 FMy = 900N x sin 110 = 845.7 Rx = 0.03 Ry = 0 Tz = (rx Fy) (ry Fx) Tz = (0.03 x 845.7) (0 x -307.8) Tz = 25.4 0 T = 25.4
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Slide 25 Torques ESS 3093 Biomechanics
Static Equilibrium
F = 0 T = 0
Combination allows solutions of indeterminate problems
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Slide 26 Torques ESS 3093 Biomechanics
A person is holding a 420N (94lb) dumbbell 0.4m from the elbow in a static position as shown.
The arm weighs 45N and the center of mass of the arm is 0.15m from the elbow.
What is the torque generated by the elbow flexors if the muscle tendon moment arm is 0.05m?
What is the force generated by the elbow flexors to maintain this static position (assume Fm is vertical)?
Tm = ?
Fm = ?
Fj = ?
F j
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Slide 27 Torques ESS 3093 Biomechanics
Tdumb = r x F Tdumb = 0.4 x -420 N Tdumb= -168 Nm Tarm = r x F Tarm = 0.15 x -45 Tarm = -6.75 Nm Sum of Torques = Tm + -168Nm + -6.75Nm = 0 Tm -174.75 = 0 Tm = 174.75 Nm Tm = r x F 174.75 Nm = 0.05 m x Fm Fm = 3495 N Ftotal=-420 -45+3495 + Fj = 0 Fj = -3030 N
F j
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Slide 28 Torques ESS 3093 Biomechanics
Food for thought: The torques produced by free weights vary as the moment arms of these weights change during the movement
Wdb
Wdb
Dumbbells dont get heavier, but the torque gets larger as the elbow is flexed
T = r x F
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Slide 29 Torques ESS 3093 Biomechanics
Application: Low Back Pain Lower back pain (LBP) is
the most costly musculoskeletal disorder in industrial nations
80% will suffer LBP
LBP can be completely debilitating to temporarily annoying
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Slide 30 Torques ESS 3093 Biomechanics
Biomechanics of Lifting The disk between L5/S1 incurs the greatest moment in lifting and is one of the most vulnerable tissues to force-induced injuries.
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Slide 31 Torques ESS 3093 Biomechanics
Lifting Calculate the forward
bending torque about L5/S1 axis: Lw = 0.25 m Lp = 0.4 m
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Slide 32 Torques ESS 3093 Biomechanics
Lifting T = 0 T = Tb + Tw + Tp = 0 Tb = -Tw - Tp Tb = -(Fw Lw) - (Fp Lp) Tb = -(-450*.25) - (-200*.4) Tb = 192.5 Nm
If the moment arm of the erector spinae is .05 m then what is the magnitude of the Fm?
T = F x moment arm F= T/moment arm = 192.5Nm/.05m = 3850N This generally represents the force on the disc
as well as muscle and connective tissue
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Slide 33 Torques ESS 3093 Biomechanics
Center of Mass Mean position of the mass of a body A point about which a bodys mass is
evenly distributed The CoM need not be within body
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Slide 34 Torques ESS 3093 Biomechanics
Example
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Slide 35 Torques ESS 3093 Biomechanics
Center of Mass: Balance Point
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Slide 36 Torques ESS 3093 Biomechanics
CoM by Mass Moments Mass Moment = Mass Displacement
m1
x1
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Slide 37 Torques ESS 3093 Biomechanics
CoM by Mass Moments MMs of all components = MMCoM
m1 m2
x1 x2
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Slide 38 Torques ESS 3093 Biomechanics
CoM by Mass Moments MMs = m1x1 + m2x2 = MtotalxCoM MtotalxCoM = m1x1 + m2x2 = (m1 + m2)xCoM xCoM = (m1x1 + m2x2)/(m1 + m2) Works for any number of masses (not just 2)
m1 m2
x1 x2 xCoM
Total Mass
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Slide 39 Torques ESS 3093 Biomechanics
Segmental Center of Mass
Thigh CoM: 43% of the length from the hip to the knee Mass: 14.6% of body mass
Leg CoM: 43% of the length from the knee to the ankle Mass: 4.3% of body mass
Foot CoM: 50% of the length from the ankle to the 2nd metarsal head Mass: 1.4% of body mass
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Slide 40 Torques ESS 3093 Biomechanics
CoM by Mass Moments
43% LThigh
14.6% BM
Thigh CoM: 43% of the length from the hip to the knee Mass: 14.6% of body mass
Leg CoM: 43% of the length from the knee to the ankle Mass: 4.3% of body mass
Foot CoM: 50% of the length from the ankle to the 2nd metarsal head Mass: 1.4% of body mass
4.3% BM 1.4% BM
43% LLeg 50% Lfoot
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Slide 41 Torques ESS 3093 Biomechanics
2D components of CoM
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Slide 42 Torques ESS 3093 Biomechanics
CoM in Two Dimensions?
The center of mass of the thigh is located 43% of the segment length from the hip joint.
What is the location of the CoM of the thigh segment with the hip located at (0.5, 1.0) and the knee located at (0.275, 0.61)?
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Slide 43 Torques ESS 3093 Biomechanics
(0.275, 0.61)
(0.5, 1.0) X direction
CoMx = Xprox - 0.43(Xprox-Xdist)
= 0.5 0.43(0.5-0.275)
= 0.403
Y direction CoMy = Yprox - 0.43(Yprox-Ydist) = 1.0 0.43(1.0-0.61)
= 0.832
CoM will be at (0.403, 0.832)