moments of force d. gordon e. robertson, phd, fcsb biomechanics laboratory, school of human...

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Moments of Force D. Gordon E. Robertson, PhD, FCSB Biomechanics Laboratory, School of Human Kinetics, University of Ottawa, Ottawa, Canada

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Page 1: Moments of Force D. Gordon E. Robertson, PhD, FCSB Biomechanics Laboratory, School of Human Kinetics, University of Ottawa, Ottawa, Canada D. Gordon E

Moments of ForceMoments of Force

D. Gordon E. Robertson, PhD, FCSB

Biomechanics Laboratory,

School of Human Kinetics,

University of Ottawa, Ottawa, Canada

D. Gordon E. Robertson, PhD, FCSB

Biomechanics Laboratory,

School of Human Kinetics,

University of Ottawa, Ottawa, Canada

Page 2: Moments of Force D. Gordon E. Robertson, PhD, FCSB Biomechanics Laboratory, School of Human Kinetics, University of Ottawa, Ottawa, Canada D. Gordon E

Moment of a Force (when F is at 90º to d)

• turning effect of a force, also called torque

• product of force (F) and moment arm (d) of the force from the axis (A) of rotation

• moment arm is the perpendicular distance from the axis of rotation to the line of the force

M = F d

force (F)

axis (A)

line of action of force

momentarm (d)

Page 3: Moments of Force D. Gordon E. Robertson, PhD, FCSB Biomechanics Laboratory, School of Human Kinetics, University of Ottawa, Ottawa, Canada D. Gordon E

Moment of a Force (when F is at 90º to d)

M = F d• direction (+ / − sign) of

the moment of force depends on the right-hand rule

• i.e., counter-clockwise is positive

• units are newton metres or N.m

force (F)

axis (A)

line of action of force

momentarm (d)

Page 4: Moments of Force D. Gordon E. Robertson, PhD, FCSB Biomechanics Laboratory, School of Human Kinetics, University of Ottawa, Ottawa, Canada D. Gordon E

Moment of a Force (r, F, )

• if moment arm length is difficult to compute use:

M = r F sin • r is length of line from

axis to line of force

• theta is angle between line of force and line of r

force (F)

axis (A)

line of action of force

line from Ato force (r)

angle betweenr and F ()

Page 5: Moments of Force D. Gordon E. Robertson, PhD, FCSB Biomechanics Laboratory, School of Human Kinetics, University of Ottawa, Ottawa, Canada D. Gordon E

Moment of a Force (r, F, )

M = r F sin • direction (sign) of

moment follows right-hand rule

• i.e., if force “turns” line r counter-clockwise about axis at A then moment is positive

force (F)

axis (A)

line of action of force

line from Ato force (r)

angle betweenr and F ()

Page 6: Moments of Force D. Gordon E. Robertson, PhD, FCSB Biomechanics Laboratory, School of Human Kinetics, University of Ottawa, Ottawa, Canada D. Gordon E

Moment of a Force (r, F, )

• to simplify and clarify positive directions of moments and forces, add reference axes to each figure

force (F)

axis (A)

line of action of force

line from Ato force (r)

angle betweenr and F ()

+

• positive directions are defined by the arrows

• add axes labels (X, Y) for additional clarity

Y

X

Page 7: Moments of Force D. Gordon E. Robertson, PhD, FCSB Biomechanics Laboratory, School of Human Kinetics, University of Ottawa, Ottawa, Canada D. Gordon E

Moment of a Force (r, F, )

Factors that increase moment of force

• increase force (F)• increase lever arm

length (r)• increase angle ()

between lever and line of force to perpendicular

M = r F sin

force (F)

axis (A)

line of action of force

line from Ato force (r)

angle betweenr and F ()

Page 8: Moments of Force D. Gordon E. Robertson, PhD, FCSB Biomechanics Laboratory, School of Human Kinetics, University of Ottawa, Ottawa, Canada D. Gordon E

Moment of a Force (cross-product)

• if components of force and line connecting axis to line of force are known, use vector cross-product:

M = r x F

force (F)

axis (A)

r

Page 9: Moments of Force D. Gordon E. Robertson, PhD, FCSB Biomechanics Laboratory, School of Human Kinetics, University of Ottawa, Ottawa, Canada D. Gordon E

Moment of a Force (cross-product)

• first resolve vectors r and F into their rectangular coordinates

• then apply:

M = r x F

= ( rx Fy − ry Fx ) k

• k is the unit vector about the Z-axis

axis (A)

force (F)

axis (A)

r

Page 10: Moments of Force D. Gordon E. Robertson, PhD, FCSB Biomechanics Laboratory, School of Human Kinetics, University of Ottawa, Ottawa, Canada D. Gordon E

Moment of a Force (cross-product)

• first resolve vectors r and F into their rectangular coordinates

• then apply:

M = r x F

= ( rx Fy − ry Fx ) k

• k is the unit vector about the Z-axis

force (F)

axis (A)

r

Fy

Fx

ry

rx

Page 11: Moments of Force D. Gordon E. Robertson, PhD, FCSB Biomechanics Laboratory, School of Human Kinetics, University of Ottawa, Ottawa, Canada D. Gordon E

Moment of a Force (cross-product)

• put components at their original points of application

force (F)

axis (A)

Fy

Fx

ry

rx

Page 12: Moments of Force D. Gordon E. Robertson, PhD, FCSB Biomechanics Laboratory, School of Human Kinetics, University of Ottawa, Ottawa, Canada D. Gordon E

Moment of a Force (cross-product)

• put components at their original points of application

• next slide force vectors along their lines of action and multiply

force (F)

axis (A)

r

Fy

Fx

ry

rx

Page 13: Moments of Force D. Gordon E. Robertson, PhD, FCSB Biomechanics Laboratory, School of Human Kinetics, University of Ottawa, Ottawa, Canada D. Gordon E

Moment of a Force (cross-product)

• if only scalar part of the moment is wanted, use this notation:

M = [ r x F ]z

M = ( rx Fy − ry Fx )

force (F)

axis (A)

r

Fy

Fx

ry

rx

Page 14: Moments of Force D. Gordon E. Robertson, PhD, FCSB Biomechanics Laboratory, School of Human Kinetics, University of Ottawa, Ottawa, Canada D. Gordon E

Example

Given:

r = (20.0, −65.0) cm

F = (220, 150.0) N

M = [ rx Fy − ry Fx ]z

=[ (20.0 150.0) −

(-65.0 220) ]

= 3000 + 14 300

= 17 300 N.cm

= 173.0 N.m

force (F)

axis (A)

r

Fy

Fx

ry

rx