1. When released, the lift provides a net force of 6 N on a 2 kg balloon. If it started at rest, how fast will it be moving in 4.0 s? F net = 6 Nm = 2 kg v o = 0t = 4 s v f = ? v f = v o + at, so we need “a”. a = F net / m a = 6 N / 2 kg = 3 m/s 2 v f = 0 + 3 m/s 2 x 4 s v f = 12 m/s F net = ma

2. Newton’s Three Laws 1.An object at rest remains at rest and an object in motion continues that motion unless acted upon by a net external force. 2.F net = mass x acceleration 3.For every force, there is an equal but opposite force.

3. A 1500 kg. car starts from rest and is moving at 10 m/s after 5.0 s. What was the net force on the car? m = 1500 kg V o = 0 V f = 10 m/s t = 5.0 s F net = ? F net = ma, so we need to find the acceleration. v f = v o + at a = ( v f – v o ) / t a = (10 m/s – 0) / 5 s = 2 m/s 2 F net = ma F net = 1500 kg x 2 m/s 2 F net = 3000 N F net = ma

4. Weight Weight, Wt. is the gravitational force acting on an object Your weight is determined by both your mass and the strength of the gravitational field (the acceleration due to gravity, “g”) Weight = mass x acceleration due to gravity Wt. = mg Since weight is a force, it is measured in Newtons, N Remember, “g” on Earth is 9.8 m/s 2

5. What is the weight of a 42 kg child on Earth? Wt. = mgm = 42 kg Wt. = 42 kg x 9.8 m/s 2 Wt. = 411.6 N Weight = mg

6. What is the mass of a 15000 N car on Earth? Wt. = mg m = Wt. / g m = 15000 N / 9.8 m/s 2 m = 1530.6 kg Weight = mg

8. Remember, one Newton is not a very big force (about the same as a ¼ pound). So, your weight in Newtons is MUCH bigger than your weight in pounds! In fact, you would have to multiply your weight in pounds by 4.45 to get your weight in Newtons. How much do you weigh in Newton’s?

9. Even if you weigh 550 Newtons, You still wouldn’t be much of a Sumo Wrestler! (that’s only around 120 lbs)

10. “Net” Force F net = ma

11. More than one force can act on an object at the same time. For example, two people could push on a book at the same time. One person could push toward the left and the other could push toward the right. In this case the two forces would act against each other. Since Newton’s Law requires NET force, what is the NET force? 2 N towards the left If the book had a mass of ½ kg, what would be its acceleration? a = F net / m a = 2 N /.5 kg a = 4 m/s 2 (left) F net = ma

12. What if the opposing forces were equal? What is the net Force? What is the acceleration? If the forces are “balanced” forces, the net Force is zero and there will be NO acceleration! F net = ma

13. What if the forces were in the same direction? What is the net force? If the book had a mass of 2 kg, what is its acceleration? a = F net / m a = 14 N / 2 kg = 7 m/s 2 F net = ma

14. Free Body Diagrams Free-body diagrams are pictures used to show the relative magnitude and direction of all forces acting upon an object in a given situation. These diagrams are often used in physics. The length of the arrow in a free-body diagram sometimes is used to represent the magnitude (size) of the force. The direction of the arrow shows the direction that the force is acting. Each force arrow in the diagram is labeled to indicate the exact type of force. The object itself is either drawn as a box or squeezed down to a dot. The force arrows are always drawn pointing away from the center of the box. 3 kg 18N6N 15N

15. Examples of Free-Body Diagrams What are the unknown forces for the given net force?

16. Forces on an Airplane: if the forces are not “balanced”, there will be an acceleration! When Lift is larger than Weight, the plane will…. When Weight is larger than Lift, the plane will… When Thrust is larger than Drag, the plan will…. When Drag is larger than Thrust, the plane will…. When Thrust = Drag, the plane will… When Lift = Weight, the plane will….

17. We usually label forces as negative or positive. Forces upward are positive. Forces downward are negative. Forces to the right are positive. Forces to the left are negative + - - +

18. What is the net Force? - 6 N + 18 N = + 12 N What is the acceleration? a = F net / m 12 N / 3 kg = 4 m/s 2 3 kg 18N6N F net = ma

19. What is the net Force? - 6 N – 15 N + 18 N = - 3 N What is the acceleration? a = F net / m a = - 3 N / 3 kg = - 1 m/s 2 3 kg 18N6N 15N F net = ma

20. A parachute provides a lift force of 400 N on a parachutist that weighs 500 N. What is the net force on him? Use g = 10 m/s 2 - 100 N What is his acceleration? a = F net / m What is his mass? Weight = mg a = -100 N / 50 kg a = - 2 m/s 2 Weight Parachute F net = ma

21. Tension Force Tension, T, is the force that cables, ropes, and strings pull with.

22. A child pulls up on a string that is holding 2 fish of total mass 5 kg. If he is providing a tension of 60 N, what is the net force on the fish? F net = Tension – Weight F net = 60 N – 50 N F net = 10 N What is the acceleration of the fish? a = F net / m a = 10 N / 5 kg a = 2 m/s 2 F net = ma

23. One child pulls up on a box with a force of 19 N. Another child pulls down on the box with a force of 5 N. What is the net Force? Hold on, there’s another force not drawn! The gravitational force of weight is also pulling down! Wt = mg, (g = 10 m/s 2 ) Wt = 2 kg x 10 m/s 2 = 20 N Draw the weight vector also! Now, what is the net Force? Net force = +19 N – 5 N – 20 N = Net Force = - 6 N What is the acceleration? a = Fnet / m = a = - 6 N / 2 kg a = -3 m/s 2 It will accelerate downward. 2 kg 19 N 5 N mg = 20 N F net = ma

24. , “sigma” is a Greek letter that is used to signify “the sum of” Quite often, in Newton’s 2 nd Law, we write  F = ma instead of F net = ma  F = ma

25. A child pulls a 5 kg bucket out of well with a rope. If the bucket accelerates upward at 1.2 m/s 2, what is the tension in the rope? m = 5 kg a = 1.2 m/s 2 T = ?  F = ma T – mg = ma T = ma + mg T = (5 kg x 1.2 m/s 2 ) + (5 kg x 9.8 m/s 2 ) T = 55 N T mg  F = ma

26. m = 2 kg a = 4 m/s 2 T = ?  F = ma T – mg = ma T = ma + mg T = (2 kg x 4 m/s 2 ) + (2 kg x 9.8 m/s 2 ) T = 27.6 N T mg A 2 kg cat grabs hold of a rope dangling from an alien spacecraft that is blasting off from Earth. What is the tension in the rope if the spacecraft accelerates straight upward at 4 m/s 2 ?  F = ma

27. Pre-AP: Now for some challenging problems…

28. A hockey player strikes the 0.5 kg puck with his hockey stick at an angle of 70 degrees. If he exerts a force of 60 N, with what acceleration will the puck slide across the frictionless ice?  F = ma F sin  = ma a = 

29. A boy pushing on a 15 kg lawnmower wants to produce an acceleration along the horizontal ground of 3 m/s 2. What force must he exert if his force is directed at an angle of 40 degrees measured from the horizontal? Required Force = mass x acceleration Force = 15 kg x 3 m/s 2 Horizontal Force = 45 N But…. Question: If the horizontal component of force is 45 N, what was the magnitude of the force the boy exerted at an angle of 40 degrees? Which trig function can be used to find the hypotenuse when the angle and adjacent side are known? Cos  = adj / hyp hyp = adj / cos  The boy’s force = 58.74 N 45 N 40 o Force = ?  F = ma