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Chapter 3 – Basic Kinetic Concepts

Chapter 3 – Basic Kinetic Concepts. Inertia – resistance to acceleration (reluctance of a body to change its state of motion) Has no units Mass – the quantity of matter (i.e. the amount of stuff). A direct measure of inertia Units - kilograms (kg)

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Chapter 3 – Basic Kinetic Concepts

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  1. Chapter 3 – Basic Kinetic Concepts • Inertia – resistance to acceleration (reluctance of a body to change its state of motion) Has no units • Mass – the quantity of matter (i.e. the amount of stuff). A direct measure of inertia Units - kilograms (kg) • Force – push or pull acting on a body. If strong enough, it may alter the body’s state of motion. Units - Newtons (N)

  2. Force applied by racquet Ball being struck by a racquet Air resistance ball weight Free body diagram Diagram showing vector representations of all forces acting on a defined system

  3. Weight – The force exerted on a body due to gravity. W = m x g • Where m = mass (kg) and g = acceleration due to gravity (-9.81 m/s2). W = Weight (N) • Pressure – amount of force acting over a given area. P = F/A Units - N/m2

  4. Sample problem # 2 page 70 Is it better to be stepped on by a woman wearing a spike heel or by a woman wearing a smooth soled court shoe given the following? The woman weighs 556 N, the surface area of the heel is 4 cm2, and the surface area of the court shoe is 175 cm2. Known: F = 556 N; A1= 4 cm2; A2 = 175 cm2 Required: Pressure P; use P = F/A For the spiked heel: P = 556 N / 4 cm2 = 139 N/ cm2 For the court shoe: P = 556 N / 175 cm2 = 3.18 N/cm2 Pspiked/Pcourt = 139/3.18 = 43.75

  5. 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

  6. F = 10N d = 2m axis 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 T = Fd T = (10N)(2m) T = 20 Nm

  7. Mechanical loads on the body

  8. Shear force at knee joint produced by the axial force in the femur Compression at the patellofemoral joint – vector sum of Fm and Ft

  9. Sample problem # 3 page 76 How much compressive stress is present on the L1, L2 vertebral disk (20 cm2) of a 625 N women if 45% of her body weight is supported by the disk a) when she stands in the anatomical position? b) when she is holding a 222 N suitcase? • Solution: a. Known: F = 625 N x 0.45 = 281.25 N ; A = 20 cm2 Stress = F/A = 281.25 N / 20 cm2 = 14.1 N/cm2 b. Known:F = (625 N x 0.45)+222 N = 503.25 N; A = 20 cm2 Stress = F/A = 503.25 N / 20 cm2 = 25.2 N/cm2

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