Torque• The perpendicular distance from the line of action to the

pivot point is called the moment arm (r).

• Torque is the force multiplied by the moment arm. T = F*r, Units: N-m

• If the line of action of a force vector does not go through the pivot point of an object, it will tend to rotate the object.

F

r

F

r

Torque

• centric force– applied through axis (center) of rotation– creates no torque so causes no rotation

• eccentric force– applied some distance away from axis of

rotation– creates torque so causes rotation

Which door is easier to open?

Will the green person go up or

down?

What causes a limb to rotate -force or torque?

• angular motion occurs at a joint, so torque causes the limb to rotate

• torque is developed because a force acts at a distance from the axis of rotation

muscle force (Fm)

Perpendicular distance between pt of application and joint axis (dm)

muscle torque (Tm=Fm*dm)

Calculation of Muscle Torque

400 N

NOTE: The torque created by the muscle depends on1) the size of the muscle force2) the angle at which the muscle pulls3) the perpendicular distance from the muscle to the joint axis

Example

Fm= ?

Wfa=13.35 N Wbb=44.5 N

0.02 m

0.15 m

0.45 m

Give me a lever long enough and a fulcrum strong enough, and, single-handedly, I can move the world.

--Archimedes (287 - 212 B.C.)

Levers

• A lever consists of two forces (motive and resistance forces) acting around a pivot point (axis or fulcrum).

• The perpendicular distance from the line of action of the effort force to the fulcrum is called the motive arm.

• The perpendicular distance from the line of action of the resistance force to the fulcrum is called the resistance arm.

Elements of a Lever

motive force (effort force)

axis(fulcrum)

motivearm

resistance

arm

resistance force

Mechanical Advantage of a Lever

• MA =

• The ratio of the motive arm to the resistance arm is called the mechanical advantage (MA).

• If MA is approximately 1 the lever simply acts to redirect the applied force.

• If MA is > 1 the lever acts to amplify the force.• If MA is < 1 the lever acts to amplify the speed and

range of motion.

arm resistance

arm motive

MA = 1

motive force direction of the force vector is redirected

motive arm = resistance arm

MA > 1

motive force

motive arm > resistance armforce is amplified

MA < 1

motive arm < resistance armROM / speed is amplified

motive force

Muscles have MA <1

0.02 m(motive arm)

0.15 m(resistance arm)

0.45 m(resistance arm)

motive arm < resistance arms

Mechanical Advantage

• Muscles typically have a MA in ROM and speed.

Classes of Levers

• Classified according to the relative positions of the axis, motive force and resistive force.

How to remember the class of lever

1st class Axis is between resistance and motive force.

2nd class Resistance force is in between the axis and the motive force.

3rd class Motive force is in between the axis and the resistance force.

ARM

1st Class Lever

• axis in the middle

• e.g. see-saw

• most versatile lever because it can be used for any type of mechanical advantage

• e.g. in body– pushing down gas pedal– elbow/triceps extensions overhead

2nd Class Lever

• resistance in middle

• force advantage usually exists for motive force

• e.g. push-up– body is lever, feet are axis, resistance is weight

of body and motive is arms

3rd Class Lever

• motive in middle

• most musculoskeletal arrangements are 3rd class levers

• muscle is motive force

• advantage in ROM and speed but disadvantage in F

Gears

• A gear is similar in function to a lever.

• The torque on the wheel and the gear is the same.

• The moment arms are different and therefore the forces are different.

Equilibrium – Linear Components

W

Ry

if Ry = Wthen resultant force = 0if v = 0 and F = 0STATIC EQUILIBRIUM

W

Ry

FpFr

Fr = resistive forceFp = propulsive forceif v = 0 and F = 0DYNAMIC EQUILIBRIUM

V

Equilibrium – Angular Components

if T1 = T2then resultant torque = 0if = 0 and = 0STATIC EQUILIBRIUM

T1=T2DYNAMIC EQUILIBRIUMb/c = constant and = 0

T1 T2

If the object is rotating with a non-zero velocity and…

Stability

• Stability is the resistance to linear and angular acceleration.

• There are 3 major factors that influence the stability of an object.

1) The line of gravity with respect to the base of support.

2) The height of the center of mass.

unstable

stable

3) The mass.

10 kg

unstable

100kg

stable

Center of Mass• The center of mass is the point about which the body's

mass is evenly distributed.

• The line of gravity is the line that defines the center of mass in the transverse plane.

• The sum of the torques about an axis caused by the weights of multiple particles is equal to the distance from the axis to the center of mass multiplied by the sum of the weights.

Center of Mass

• The center of mass is the point about which the body's mass is evenly distributed.

Symmetricdistribution

CM in the middle

balance point

Asymmetricdistribution

CM closer to larger weight

The sum of the torques about an axis caused by the weights of multiple particles is equal to the distance from the axis to the center of mass multiplied by the sum of the weights.

Y

X

2 kg(1,3)

1 kg(1,1)

3 kg(3,3)

COM(2, 2.67)

1

2

3

line of gravity(if Y is vertical)

m1x1 + m2x2 + m3x3 = Mxcm

2kg(1m) + 1kg(1m) + 3kg(3m) = 6kgxcm

xcm = = 2 m

Y

X

2 kg(1, 3)

1 kg(1,1)

3 kg(3, 3)

COM(2, 2.67)

1

2

3

6kg

m12kg

COM Location in the x direction

m1y1 + m2y2 + m3y3 = Mycm

2kg(3m) + 1kg(1m) + 3kg(3m) = 6ycm

ycm = = 2.67 m

Y

X

2 kg(1, 3)

1 kg(1,1)

3 kg(3, 3)

COM(2, 2.67)

1

2

3

6kg

m16kg

COM Location in the y direction

General Formulas:

x = m x

m cm

i ii=1

n

ii=1

n

y =

m y

m cm

i ii=1

n

ii=1

n

where,

xi is the distance from the y-axis to the ith mass

yi is the distance from the x-axis to the ith mass

mi is the mass of the ith element (segment)

An alternative approach is to use the proportion of each mass (pi) instead of the actual masses. For the human body the proportion can be found in many text books for each body segment.

x = p x cm i ii=1

n

y = p y cm i ii=1

n

X

Y

Relative Mass (% Total Body Mass)

Head 7.3 46.4% from vertexTrunk 50.7 43.8% from suprasternaleL Upper Arm 2.6 49.1% from shoulderL Forearm 1.6 41.8% from elbowL Hand 0.7 82.0% from wristR Upper Arm 2.6 49.1% from shoulderR Forearm 1.6 41.8% from elbowR Hand 0.7 82.0% from wristL Thigh 10.3 40.0% from hipL Shank 4.3 41.8% from kneeL Foot 1.5 44.9% from heelR Thigh 10.3 40.0% from hipR Shank 4.3 41.8% from kneeR Foot 1.5 44.9% from heel

Segment CM Location (% Length) Body SegmentParameters

Derived fromdirect cadaver measurements

Elderly, male,Caucasiancadavers

From these data it is apparent that to determine the center of massof a segment it is necessary to locate the segment endpoints

X

Y

X

Y

X

Y

6.8 cm

To locate the segment CM1st measure the length of the segment

X

Y

3.0 cm

CM of the trunk is 43.8% of the length of the trunk away from the suprasternale43.8% of 6.8 cm = 3.0 cm

X

Yto find the whole body CM you need to express the segmental CM locations with respect to a common reference point - we’ll use the origin

yi

xi

xp = xn

1=iiicm

y = p y cm i ii=1

n

X

YDo this for every segment

yi

xi

xp = xn

1=iiicm

y = p y cm i ii=1

n

Use these distances and the segment masses to computethe whole body CM location

X

Y

Y-distance = 92 mm

X-distance = 120 mm

Plot the final coordinates of the CM

Straddle Jump

Fosbury Flop

A basketball player can appear to remain at a constant height for brief periods of time by manipulating the body segments about their center of mass. The COM will always follow the path of a parabola while the body is in the air. (Michael Jordan is very good at this.)

Resistance to angular motion (like linear motion) is dependenton mass.

The more closely mass is distributed to the axis of rotation, the easier it is to rotate.

therefore: resistance to angular motion is dependent on the distribution of mass

This resistance is called the Moment of Inertia.

Moment of Inertia

Moment of Inertia

• ANGULAR FORM OF INERTIA– resistance to changes in the state of angular

motion

• I = mr2

– for a single particle I is proportional to the distance squared

• SI unit = kg-m2

Each block is .5 m by 1.5 m and has a mass of 2 kg. The mass in each block is uniformly distributed. What is the moment of inertia about the x axis?

Ix = miri2 = [2kg*(.25m)2] + [2kg*.75m)2] + [2*(.25m)2]

+ [2kg*(.75m)2] = 2.5 kg-m2

x

If the mass of the above object were concentrated at a single point (the center of mass) how far from the axis would it have to be located to have the same moment of inertia?Ix = 2.5kg-m2 = mk2 = 8kg*k2

k = = .559 m

x

/8kgm2.5kg 2

This value is called the radius of gyration:distance from axis of rotation to a point where the body’s mass could be concentrated without altering its rotational characteristics

for a system of particles

I = mk2 where k = ‘radius of gyration’

It is often expressed as a proportion of the segment length in biomechanics. Thus,

I = m(l)2

where I is the moment of inertia is the radius of gyration as a proportion of the segment length (l)

Different Axes

• recognize that rotation can occur about different axes– each axis has its

own moment of inertia associated with it

Whole Body Moment of Inertia• consider human movement to occur about 3 principal axes

• each principal axis has a principal moment of inertia associated with it

• when mass is distributed closer to the axis, the moment of inertia is lower

Angular Analog Newton’s Laws

1) A rotating body will continue to turn about its axis of rotation with constant angular momentum, unless an external couple or eccentric force is exerted upon it.

•linear momentump = m*v

•angular momentumH = I*

Angular Momentum

•linear momentump = m*v

•angular momentumH = I*

•In the linear case mass does not change but the moment of inertia can be manipulated by reorienting body segments.

- ice skaters- divers

Angular Analog Newton’s Laws

2) The rate of change of angular momentum of a body is proportional to the torque causing it and the change takes place in the direction in which the torque acts.

T = If - i

t

T = Ior

Angular Analog Newton’s Laws

3) For every torque that is exerted by one body on another there is an equal and opposite torque exerted by the second body on the first.

TRANSFER OFANGULAR

MOMENTUM

enter pike - Hlegs because legs slow down

Htrunk+arms to maintaina constant Htotal

the opposite occurs at entry - Htrunk + arms

to give a clean entry

Hlegs to maintain Htotal

Angular Momentum in the Long Jump

Htotal = Htrunk+head + Harms + Hlegs = constant CW

To prevent trunk+head from rotating forward (CW)rotate arms and legs CW to account for Htotal

Iarms and Ilegs are smaller than Itotal so

arms and legs must be larger to produce

H’s large enough to accommodate Htotal

Initiation of Rotation in Air

Newton’s Laws specificallystate that you can NOT initiate rotation (e.g. in theair) without an external torquebeing applied to you

So -- can you initiate rotationwhile airborne?

A cat does! (seemingly)

explanation -consider relationship between I’s of body parts that interact when rotation is initiated

1) As the cat begins to fall it bends in middle, brings its front legs in close to its head and rotates the upper body through 180 degrees.

1a) In reaction to the upper body the lower body will rotate in the opposite direction.

However -- since the body ispiked Ilower body is very largecompared to upper body sothe corresponding rotation is small (about 5 degrees).

2) To complete the 180 degreerotation the cat brings its hind legs and tail into line with its lower trunk such that its longitudinal axis runs through its hindquarters.

2a) The reaction of upper body is again small since Iupper body (about this axis of rotation) islarge, so there is little rotation of upper body

3) Minor adjustments are made by rotating tail in direction opposite to the desired rotation.

Centripetal vs. Centrifugal Force

• Centripetal force (center seeking force) = mass xcentripetal acceleration

• Centrifugal force (center fleeing force) -- reaction to the centripetal force; applied to the other body

rmr

vmmaF t

cc2

2

                                               

Consider Newton’s second law of motion:F = ma

Now substitute centripetal acceleration. In centripetal motion the centripetal acceleration is linked to a centripetal force. You can think of this force as being responsible for holding the object in a circular path.

Exampleyou make a right turn in your caryou feel the driver door push on you to the right(toward the center of the curvature of your

curved path)

the door applies a centripetal force to you,you apply a centrifugal force which is

equal and opposite to the centripetal force

door

youFcpFcf

Forces occurring Forces occurring along a curved pathalong a curved path