Week 5 Forces And Centre Of Mass

Question 1

Mass (m)

Answer 1

Objects tend to stay at rest (or move with constant velocity) unless they are acted upon. This property is called “inertia”.
- inertia is the resistance to change a state
Mass is the measure of the inertia of an object, i.e., the measure of the resistance to the change in the state of motion.

-Mass is also the measure of the gravitational pull exerted by an object.
More mass the more inertia

-Mass is a base scalar physical quantity with unit of “kilogram” (kg).

Anthropometric tables give the information of the mass of body segments expressed as fractions of the total body mass.
Segment where mass is distributed

Question 2

Centre of mass

Answer 2

Centre of mass of an object (e.g. a body segment) is the point in space whose coordinates are calculated as the mass-weighted average position of all the particles that constitute that object. Mathematically this is expressed as:
Xc\Yc\Zc = Sum of miXi/ sum of mi

Sum of mass x postion/ mass

The centre of mass is a very useful reference point in mechanics. Representing the location of an object by the position of its centre of mass simplifies the analysis of many problems in mechanics

Question 3

Centre of mass of body segments

Answer 3

The locations of the centres of mass of body segments are determined experimentally and reported in the anthropometric tables. Centres of mass of several segments put together, or of the whole body can be calculated by treating each segment as a point mass located at its centre of mass.
Can manipulate the CoM TO AN ADVANTAGE SUCH AS A HIGH JUMP WHERE THE CENTRE OF MASS IS UNDERNEATH THE BAR = parabolic trajectory

Question 4

Force (F)

Answer 4

Force is the measure of interaction between physical objects. That interaction results in the change of state of motion (acceleration) and in deformation.

Force is a vector physical quantity and it can be represented in component form resolved along the axes of the Cartesian coordinate system
Fx Fy Fz
The unit for force is Newton (N).

Question 5

Weight (W) and acceleration due to gravitational pull of the earth (g)

Answer 5

Weight (W) is the force exerted on an object by gravitational pull.
Near the surface of the earth the weight of an object of mass m equals:
W= mg (g = acceleration due to gravity)

Here, g=9.81 m/s2 is the acceleration due to gravitational pull, i.e., the acceleration of a free-falling object in the absence of any other forces.

The weight is about ten times greater than the mass, e.g., a person with a mass of 80 kg will weigh 9.81*80=784.8 N

Although the weight and acceleration due to gravity are vectors, they are usually denoted without an arrow because their direction is known.

Question 6

The net force acing on an object

Answer 6

The net force acting on an object is the vector sum of all the forces acting on that object.

Net force that dictates the corresponding movement

Question 7

Mechanical system

Answer 7

A mechanical system comprises of two or more objects that are related in some sense. We can define mechanical systems of any shape and form that suits the purpose of the analysis that we want to conduct.

In biomechanics, the most common mechanical system is the human body which is represented as a collection of objects (body segments) connected by joints. The segments are usually: the head, torso, upper arm, forearm, hand, thigh, shank, and foot.

We can also define a portion of the body as a separate system, for example a leg comprising of the thigh, shank, and foot.

Question 8

Internal and external forces

Answer 8

In the context of a mechanical system, forces can be classified as internal and external.

The internal forces are the forces that the components of a system exert on each other. For the human body, the internal forces are the muscle forces and the forces between the bones forming the joints.

The external forces are the forces that act on the components of a system from the outside. The most commonly encountered forces acting on the human body are the weight and the ground reaction force.
Can change whole bodys Com

Question 9

Newton’s first law

Answer 9

A body originally at rest will remain at rest if the net force acting upon it is zero
A body moving at constant velocity will continue to move at constant velocity if the net force acting upon it is zero

Equilibrium:
—Rest
— Constant velocity

A person is standing still.
The body weight is counterbalanced
by the ground reaction force,
so that the net force is zero. ^=v

A sky diver reaches “terminal velocity”
when the net aerodynamic drag force (FD)
equals the body weight. The sky diver then
moves with a constant velocity.

LAW OF INERTIA

Question 10

Newton’s second law

Answer 10

A body accelerates in the direction of the net force acting upon it with the magnitude proportional to the force:

Ma= F

From Newton’s second law we can relate Newton [N] with the base physical units:

(N). Kg m / s2

Law of acceleration

Force = proportional to magnitude and direction

F
M A

Question 11

Newton’s second law in component form

Answer 11

Newton’s second law is a vector equation relating the acceleration and the net force.

By resolving these two vectors into components we obtain three equations relating each acceleration component with the corresponding force component.

ma x. Fx
ma y. Fy
ma

Question 12

Newton third law

Answer 12

The forces of action and reaction between interacting bodies are of the:
Same magnitude and orientation
Opposite direction

Question 13

Particle

Answer 13

Strictly speaking, a “body” mentioned in Newton’s first and second laws is a “particle”, i.e., a “point mass” without dimensions. In that case, a net zero force is sufficient to assure equilibrium since all forces act at a single point. For an object with finite dimensions two opposite forces which are not acting at a single point will tend to spin the object.

Question 14

Newton’s second law applied to the centre of mass

Answer 14

For a “mechanical system” such as the human body, the net vector sum of all internal forces is zero as a consequence of Newton’s third law.
This does not imply that body parts cannot accelerate, as action and reaction forces act on different objects.
However, this does mean that the centre of mass of the system cannot be accelerated by internal forces. We can therefore define the centre of mass as a point which accelerates only due to the action of external forces.

As a consequence, we can approximate the human body to a point-mass equalling the net mass of the system and located at the centre of mass. That point-mass (particle) will accelerate due to external forces alone.

Question 15

Use of the CoM concept in a biomechanical analysis

Answer 15

In many situations such as walking and running the only external forces are the body weight and the ground reaction force.
By focusing on the movement of the centre of mass alone, we greatly simplify biomechanical analysis at the expense of tracking the movement of a single point (centre of mass).

Question 16

Measurement of the ground reaction force =force plates

Answer 16

Force plates are the devices used for the measurement of ground reaction forces

Force plates are sophisticated scales capable of measuring all three components of the ground reaction force, as well as the location at which the force acts (centre of pressure)

Force plates are usually built into the floor of the laboratory

Question 17

Position of the centre of mass from force plate data

Answer 17

Velocity and position of the centre of mass can be estimated from the force plate readings in the following way:
Find the net force by subtracting the weight from the force plate reading for the vertical component of the ground reaction force (add the weight vector)
Divide all three components of the net force by the body mass to get the three components of acceleration:
Ax = fGRX/M
Ay= FGRY/M
AZ=FGRZ- MG/M

Integrate the acceleration components using the approximate methods discussed in Lecture 3 to get the velocity components; you need to know the velocity at initial time
Integrate the velocity components using approximate methods discussed in Lecture 2 to get the coordinates; you need to know the coordinates at initial time

Question 18

Centre of mass trajectories in walking and running

Answer 18

Walking

For each stride the trajectory of the centre of mass is approximately an arc of constant radius. The elevation increases from initial contact to mid-stance, and decreases from mid-stance to toe-off. At mid-stance the elevation is at maximum and vertical ground reaction force is at minimum.

Running
In the aerial phase the trajectory is parabolic. In the contact phase the elevation decreases from initial contact to mid-stance, and increases from mid -stance to toe-off. At mid-stance the elevation is at minimum and vertical ground reaction force at maximum.