1. Chapter 3 – Basic Kinetic Concepts Inertia – resistance to acceleration (reluctance of a body to change its state of motion) 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 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. Force – push or pull acting on a body. If strong enough, it may alter the body’s state of motion. Units - Newtons (N)

2. Free 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

3.  Weight – The force exerted on a body due to gravity. due to gravity. W = m x g Where m = mass (kg) and g = acceleration due to gravity (-9.81 m/s 2 ). W = Weight (N) Pressure – amount of force acting over a  Pressure – amount of force acting over a given area. given area. P = F/A Units - N/m 2 P = F/A Units - N/m 2

5. 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 cm 2, and the surface area of the court shoe is 175 cm 2. Known: F = 556 N; A 1 = 4 cm 2 ; A 2 = 175 cm 2 Required: Pressure P; use P = F/A For the spiked heel: P = 556 N / 4 cm 2 = 139 N/ cm 2 For the court shoe: P = 556 N / 175 cm 2 = 3.18 N/cm 2 P spiked / Pcourt = 139/3.18 = 43.75

6.  Volume - The amount of space a body occupies Units – liters. 1 liter = 1000cm 3  Density – mass per unit volume. ρ (“rho”) = m/v Units - kg/m 3  Torque – rotational effect created by an off-center force. Units – Nm

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

9. Mechanical loads on the body

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

14. Sample problem # 3 page 76 Sample problem # 3 page 76 How much compressive stress is present on the L1, L2 vertebral disk (20 cm 2 ) 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: Solution: a. Known: F = 625 N x 0.45 = 281.25 N ; A = 20 cm 2 a. Known: F = 625 N x 0.45 = 281.25 N ; A = 20 cm 2 Stress = F/A = 281.25 N / 20 cm 2 = 14.1 N/cm 2 Stress = F/A = 281.25 N / 20 cm 2 = 14.1 N/cm 2 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 cm 2 A = 20 cm 2 Stress = F/A = 503.25 N / 20 cm 2 = 25.2 N/cm 2 Stress = F/A = 503.25 N / 20 cm 2 = 25.2 N/cm 2