(W= weight!) (W= weight!) W = m W = m g g

The main force acting on the body is the gravitational force! The main force acting on the body is the gravitational force!

Gravitational force Gravitational force WW applies at the center of gravity applies at the center of gravity CGCG of the body!of the body!

Stability of the body against the Stability of the body against the gravitational force is maintained by the bone gravitational force is maintained by the bone structure of the skeleton!structure of the skeleton!

CGCG depends on body mass distribution! to maintain depends on body mass distribution! to maintain stability stability CGCG must be located between feet, if feet are far apart must be located between feet, if feet are far apart forces in horizontal direction forces in horizontal direction FFxx have to be considered have to be considered

To maintain stability To maintain stability the vector sum of all forces the vector sum of all forces applying at theapplying at the CG CG must be must be zero!zero!

The torque r causes a rotational movement around a pivot point! The torque r causes a rotational movement around a pivot point!

Torque is defined by the Torque is defined by the force force FF applied at the distance applied at the distance r r from the pivot point.from the pivot point.

In rotational equilibrium (no rotation, In rotational equilibrium (no rotation, constant rotation) to maintain stability constant rotation) to maintain stability for a person standing on one leg the for a person standing on one leg the torque requires to shift torque requires to shift CGCG of body of body so, that:so, that:

New CG:New CG:

EXAMPLE: EXAMPLE: FORCES ON THE HIPFORCES ON THE HIP

(a person standing on one leg only) (a person standing on one leg only)

EXAMPLE:EXAMPLE: FORCES ON THE SPINAL COLUMNFORCES ON THE SPINAL COLUMN

Body movements are controlled by muscle forces, Body movements are controlled by muscle forces, initiated by contraction or extension of the muscles. Skeletal initiated by contraction or extension of the muscles. Skeletal muscles control the movements of the body limbs. muscles control the movements of the body limbs.

Most of the muscle forces involve levers!Most of the muscle forces involve levers!

Three examples for lever systems, Three examples for lever systems, WW is the is the applied weight, applied weight, FF is the force supporting the pivot is the force supporting the pivot point of the lever system, and point of the lever system, and MM is the muscles is the muscles force. force.

EXAMPLE:EXAMPLE: THE FOREARM AS LEVER SYSTEMTHE FOREARM AS LEVER SYSTEM

The biceps muscle pulls the arm upwards The biceps muscle pulls the arm upwards by muscle contraction with a force by muscle contraction with a force MM the opposing force is the the opposing force is the weight of the arm weight of the arm HH at its center of gravity at its center of gravity (CG)(CG)! !

Biceps can be strengthened by weight W lifting this adds Biceps can be strengthened by weight W lifting this adds another force which has to be compensated by the muscle force. another force which has to be compensated by the muscle force.

The lower arm can be hold by the biceps muscle at different angles The lower arm can be hold by the biceps muscle at different angles What muscle forces are required for the different arm positions? What muscle forces are required for the different arm positions?

EXAMPLE:EXAMPLE: THE ARM AS LEVER SYSTEM THE ARM AS LEVER SYSTEM

The deltoid muscle pulls the arm upwards by muscle contraction with a The deltoid muscle pulls the arm upwards by muscle contraction with a force force TT at a fixed angle at a fixed angle a a with respect to the arm the opposing force is with respect to the arm the opposing force is the weight of the arm the weight of the arm HH at its center of gravity at its center of gravity (CG)(CG) and the (possible) and the (possible) weight weight WW hold in the hand! hold in the hand!

Consider arm at an angle Consider arm at an angle hold by the deltoid muscle (hold by the deltoid muscle ( 15°) 15°)

If a body of mass If a body of mass mm is in constant motion no is in constant motion no acceleration or deceleration occurs !acceleration or deceleration occurs !

Acceleration Acceleration aa can be caused by leg muscle force can be caused by leg muscle force FF ! !

Deceleration can be caused by friction, muscle force or Deceleration can be caused by friction, muscle force or external forces (by running into a wall for example). external forces (by running into a wall for example).

Friction occurs between a moving surface and a surface at rest: Friction occurs between a moving surface and a surface at rest:

Friction force: Friction force: FFff 640 N 640 N

deceleration: deceleration: a a 7.8 m/s7.8 m/s22

N is the normal force!N is the normal force!

kk is coefficient for kinetic friction: is coefficient for kinetic friction: for rubber-concrete: for rubber-concrete: kk 0.8 0.8joints between bones: joints between bones: kk 0.003 0.003

As smaller the coefficient as less resistance by frictional forces!As smaller the coefficient as less resistance by frictional forces!

For a walker of N For a walker of N W W 800 N (m=82kg): 800 N (m=82kg):

Accelerating muscle forces maintain a constant walking speed!Accelerating muscle forces maintain a constant walking speed!

When the body bumps into a solid object (like a wall) rapid When the body bumps into a solid object (like a wall) rapid deceleration deceleration aa occurs: occurs:

The decelerating force The decelerating force FFdd applied by the wall to the body (or to applied by the wall to the body (or to whatever body part which hits first) causes pressure whatever body part which hits first) causes pressure PPdd which causes which causes deformation:deformation:

AA is the surface area of the body or body part exposed to the force. is the surface area of the body or body part exposed to the force.

Force is only applied over the time period Force is only applied over the time period tt until complete stop. until complete stop.

Therefore: Therefore:

To calculate the impact force the time structure of the To calculate the impact force the time structure of the deceleration process needs to be known. deceleration process needs to be known.

Approximation:Approximation: treatment of force as a square pulse actual time treatment of force as a square pulse actual time structure may depend on particular impact structure may depend on particular impact

Falls from great heightFalls from great height

The above equation has to be generalized The above equation has to be generalized because of energy transfer arguments!because of energy transfer arguments!

The average force acting on the part of the body which hits the ground is The average force acting on the part of the body which hits the ground is

Body decelerates with average deceleration Body decelerates with average deceleration aa from from impact velocity impact velocity vv to zero while the center of mass of to zero while the center of mass of body moves over a distance body moves over a distance a/a/CMCM during the collision during the collision

EXAMPLE: Stiffed leg jump on hard groundEXAMPLE: Stiffed leg jump on hard ground

Tolerance levels for whole body impactTolerance levels for whole body impact

Threshold for survival: Threshold for survival: 20 mi/h = 36 km/h = 8.9 m/s 20 mi/h = 36 km/h = 8.9 m/s

Effects of impact can be reduced by increasing Effects of impact can be reduced by increasing tt ( (hhCMCM)) or by or by distributing force distributing force FF over large area to reduce compressing stress. over large area to reduce compressing stress.

proper landing techniques for parachutes proper landing techniques for parachutes

To survive a fall the impact pressure should be: To survive a fall the impact pressure should be: 40 lbs/in 40 lbs/in22 = 27.6 N/cm = 27.6 N/cm22

For an impact pressure of 35 N/cmFor an impact pressure of 35 N/cm22 50 % survival chance! 50 % survival chance!

EXAMPLE: Free fall from large heightsEXAMPLE: Free fall from large heights

hc h

Vf2

Vt2

V2

free fall

actual fall

Solution of force equation yields final velocity Solution of force equation yields final velocity vv! !

h/hh/hcc ee-h/h-h/hcc V/VV/Vtt

11 0.370.37 0.790.7922 0.140.14 0.9750.97533 0.0500.050 0.9750.97544 0.0180.018 0.9910.991

Exponential approach of speed of fall towards the terminal velocity!Exponential approach of speed of fall towards the terminal velocity!

Terminal velocity represents the state where the forces are in equilibrium! Terminal velocity represents the state where the forces are in equilibrium!