1. Forces Don 't forget mass in kg distance in meters time in seconds

2. Equilibrium (F R = 0) all the x components have to add up to 0 all the y components add up to 0 x components F3 Totalx y components F3 Totaly The object is in equilibrium when Total x and Total y b oth add up to 0. Angle will be 180 o - 22.5 0 = 157.5 o Force will be 153 N

3. What happens to F N ? Mass of the box = 10 kg F g = F N = F T = 30N at 45 o -98 +98 76.8 FgFg

5. Newton's Laws of Physics

6. Newton's first law A body at rest will stay at rest, A body in motion will stay in motion in the absence of an external force....

7. Example: When we are in the car at a fast speed and the driver applies the brakes quickly... Without a seat belt we are likely to hit the dash board. Our bodies want to keep moving forward. this is called inertia - kick snow off our boots. - salad spinner -when a magician pulls a tablecloth out from under dishes and the dishes stay put.

8. Why do objects in motion stop? Friction.

9. F f(static) ≤ μ s F N m eans that F f will be equal to the opposing force up until its maximum value at which point the object will move F f(kinetic) = μ k F N t he moving object will experience this force of friction Force of Friction

10. Newton's Second Law F = ma Force = mass (kg) x acceleration (m/s 2 ) "the fundamental principle of dynamics" The force exerted on a body is equal to the product of the mass (m) and its acceleration (a)

11. The acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass. The direction of the acceleration is in the direction of the applied net force F = ma

12. Example: Estimate the net force needed to accelerate a 1500 kg race car at 5.0 m/s 2 F = ma = 1500kg x 5.0m/s 2 = 7500N = 7.5 x 10 3 N (sig figs)

13. What net force is required to bring a 1500 kg car to rest from a speed of 100 km/h within a distance of 55m? First determine the acceleration a. The car is acting on the horizontal. Vi = 100 km/h = 28m/s Vf = 0 m/s Vf 2 = Vi 2 + 2 a(Δx) 0 = 28 2 + 2a(55m) a = (-28m/s) 2 /2(55m) = -7.1m/s 2 Th e net force required is then F = ma = (1500kg) (-7.1m/s 2 ) F = -1.1 x 10 4 N

14. The elevator problem.... If a 75 kg person rides the elevator which accelerates at 1.5 m/s 2 What is the total force acting on him or her? F = ma = -(75kg (-9.8m/s 2 )) + 75kg (1.5m/s 2 ) = 847.5N

15. The elevator problem.... If a 75 kg person rides the elevator DOWN which accelerates down at 1.5 m/s 2 What is the Normal force acting on him or her? Total F = FN + other forces = 75kg (9.8m/s 2 ) - 75kg (1.5m/s 2 ) = 622.5N Going down... F N = - F g = (75 (-9.8) )= 735

16. The spider problem When the spider is in the elevator there is tension on the thread. p. 314 # 3 A 10.2 g spider is suspended from an elevator ceiling by a thread.

17. The spider problem 0.0102 kg a) What is the tension in the thread if the elevator rises at a constant velocity? It is in equilibrium therefor the only force on the string if Fg Fg = 0.0102kg x -9.8 m/s 2 = -0.100N

18. b) The elevator accelerates at a rate of 2.0 m/s 2 going up. What is the tension? F T = F g + ma = 0.100N + (0.0102kg x 2.0 m/s 2 ) = 0. 120N c) decelerates at the same rate: F T = F g + ma = 0.100N + (0.0102kg x -2.0 m/s 2 ) = 0.080N

19. Newton's third law Fo r every action there is an equal and opposite reaction F A ⇒ B = -F B ⇒ A

20. action - reaction the foot exerts a force on the ball and at the same time the ball exerts a force on the foot

21. In other words, things are never one sided... a hammer on a nail exerts a force and the reaction of the nail on the hammer is an equivalent force in the opposite direction.

22. When we walk, our foot pushes on the ground, The ground exerts an equal and opposite force back on the person. This force moves him or her forward.

23. At the skating rink Jessica pushes Jordan and while Jordan accelerates away from Jessica, Jessica also accelerates away from Jordan Action - Reaction