Chapter 3 – Basic Kinetic Concepts

Chapter 3 – Basic Kinetic ConceptsChapter 3 – Basic Kinetic Concepts

InertiaInertia – – resistance to accelerationresistance to acceleration (reluctance of (reluctance of a body to change its state of motion)a body to change its state of motion)Has no unitsHas no units

MassMass – – the quantity of matterthe quantity of matter (i.e. the amount of (i.e. the amount of stuff). A direct measure of inertia stuff). A direct measure of inertia Units - Units - kilogramskilograms (kg) (kg)

ForceForce – – push or pull acting on a bodypush or pull acting on a body. If strong . If strong enough, it may alter the body’s state of motion.enough, it may alter the body’s state of motion.Units - Units - NewtonsNewtons (N) (N)

Free body diagramFree body diagram

Diagram showing vector representations of all forces acting on a defined system

ball weight

Force applied by racquet

Air resistance

Ball being struck by a

racquet

Weight Weight – – The force exerted on a bodyThe force exerted on a body due to gravitydue to gravity. .

W = m x gW = m x g

Where m = mass (kg) and g = acceleration Where m = mass (kg) and g = acceleration due to gravity (-9.81 m/sdue to gravity (-9.81 m/s22). W = Weight (N)). W = Weight (N)

PressurePressure – – amount of force acting over aamount of force acting over a given areagiven area. .

P = F/A Units - P = F/A Units - N/mN/m22

Sample problem # 2 page 70Sample problem # 2 page 70

Is it better to be stepped on by a woman wearing a Is it better to be stepped on by a woman wearing a spike heel or by a woman wearing a smooth soled spike heel or by a woman wearing a smooth soled court shoe given the following? The woman weighs court shoe given the following? The woman weighs 556 N, the surface area of the heel is 4 cm556 N, the surface area of the heel is 4 cm22, and the , and the surface area of the court shoe is 175 cmsurface area of the court shoe is 175 cm22..

Known: F = 556 N; AKnown: F = 556 N; A11= 4 cm= 4 cm22; A; A2 2 = 175 cm= 175 cm2 2

Required: Pressure P; use P = F/ARequired: Pressure P; use P = F/A

For the spiked heel: P = 556 N / 4 cmFor the spiked heel: P = 556 N / 4 cm2 2 = = 139 N/ cm139 N/ cm22

For the court shoe: P = 556 N / 175 cmFor the court shoe: P = 556 N / 175 cm2 2 = = 3.18 N/cm3.18 N/cm22

PPspikedspiked//PcourtPcourt = 139/3.18 = = 139/3.18 = 43.7543.75

Volume - The amount of space a body occupies

Units – liters. 1 liter = 1000cm3

Density – mass per unit volume.

ρ (“rho”) = m/v

Units - kg/m3

Torque – rotational effect created by an off-center force.

Units – Nm

T = Fd (the product of force and the perpendicular distance from the force’s line of action to the axis of rotation – moment or lever arm

axisd = 2m

F = 10N

T = Fd

T = (10N)(2m)

T = 20 Nm

Mechanical loads on the bodyMechanical loads on the body

Shear force at knee joint produced by the axial force in the femur

Compression at the patellofemoral joint – vector sum of Fm and Ft

Sample problem # 3 page 76Sample problem # 3 page 76

How much compressive stress is present on the L1, L2 How much compressive stress is present on the L1, L2 vertebral disk (20 cmvertebral disk (20 cm22) of a 625 N women if 45% of her ) of a 625 N women if 45% of her body weight is supported by the disk a) when she body weight is supported by the disk a) when she stands in the anatomical position? b) when she is stands in the anatomical position? b) when she is holding a 222 N suitcase?holding a 222 N suitcase?

Solution: Solution:

a. Known: F = 625 N x 0.45 = 281.25 N ; A = 20 cma. Known: F = 625 N x 0.45 = 281.25 N ; A = 20 cm2 2

Stress = F/A = 281.25 N / 20 cmStress = F/A = 281.25 N / 20 cm22 = = 14.1 N/cm14.1 N/cm22

b. Known:F = (625 N x 0.45)+222 N = 503.25 N; b. Known:F = (625 N x 0.45)+222 N = 503.25 N;

A = 20 cmA = 20 cm2 2

Stress = F/A = 503.25 N / 20 cmStress = F/A = 503.25 N / 20 cm2 2 = = 25.2 N/cm25.2 N/cm22