1. Free-Body Diagrams ForceSymbol/FormulaDescription WeightF w = mgAlways directed downward Tension ForceTPulling forces directed away from the body Normal ForceFnFn Force exerted perpendicularly outward by a flat surface on an object pressed against it Frictional Force F f =  F n Force which acts opposite the motion or an impending motion and parallel to the surface of contact

2. Angle of Repose the (minimum) angle a plane makes with the horizontal at which a body on the plane will begin to slide down depends on the roughness of the surfaces the rougher the surfaces, the greater the angle of repose  tan -1 

3. A crate weighing 562 N rests on an inclined plane 30 o above the ground. Find the components of the force parallel and perpendicular to the plane. Chris is going down a water slide sloped at 37 o. If he weighs 62kg, how fast is Chris going 5.0s after starting from rest?  k 

4. Things to remember Finding the Resultant Force – draw a free body diagram – use the length of the arrow to indicate the vector's magnitude and the direction of the arrow to show its direction – simply find the sum of the vectors – a negative answer means that the force acts in the opposite direction to the one that you chose to be positive – 0 net force = balanced forces = 0 acceleration

5. Equilibrium – An object in equilibrium has both the sum of the forces acting on it and the sum of the moments of the forces equal to zero. – If a resultant force acts on an object then that object can be brought into equilibrium by applying an additional force that exactly balances this resultant. Equilibrant – The equilibrant of any number of forces is the single force required to produce equilibrium, and is equal in magnitude but opposite in direction to the resultant force.

7. Friction forces – When the object is not moving static friction: F f ≤ μ s F N – When the object is moving kinetic friction: F f = μ k F N

8. Newton's Law of Universal Gravitation Every point mass attracts every other point mass by a force directed along the line connecting the two. This force is proportional to the product of the masses and inversely proportional to the square of the distance between them. The magnitude of the attractive gravitational force between the two point masses, F is given by: F = Gm 1 m 2 r 2 where: G is the gravitational constant, m 1 is the mass of the first point mass, m 2 is the mass of the second point mass and r is the distance between the two point masses.

9. For example, consider a man of mass 80 kg standing 10 m from a woman with a mass of 65 kg. The attractive gravitational force between them would be: F = Gm 1 m 2 r 2 = (6.67×10 −11 N ⋅ m 2 /kg 2 )(80kg)(65kg) (10m) 2 = 3.47 × 10 −9 N G = 6.67 x 10 -11 N m 2 /kg 2

10. If the man and woman are 1 m apart, then the force is: F = Gm 1 m 2 r2r2 = (6,67×10 −11 N ⋅ m 2 /kg 2 )(80kg)(65kg) (1m) 2 = 3.47×10 −7 N

11. Gravitational force between the Earth and the Moon The mass of the Earth is 5.98×10 24 kg, the mass of the Moon is 7.35×10 22 kg and the Earth and Moon are 3.8×10 8 m apart. The gravitational force between the Earth and Moon is: F = Gm 1 m 2 r 2 = (6.67×10 −11 N ⋅ m 2 /kg 2 )(5.98×10 24 kg)(7.35×10 22 kg) (0.38×10 9 m) 2 = 2.03×10 20 N

12. Comparative problems involve calculation of something in terms of something else that we know. For example, if you weigh 490 N on Earth and the gravitational acceleration on Venus is 0.903 that of the gravitational acceleration on the Earth, then you would weigh 0.903 x 490 N = 442.5 N on Venus.

13. Principles for answering comparative problems Write out equations and calculate all quantities for the given situation Write out all relationships between variable from first and second case Write out second case Substitute all first case variables into second case Write second case in terms of first case

14. A man has a mass of 70 kg. The planet Zirgon is the same size as the Earth but has twice the mass of the Earth. What would the man weigh on Zirgon, if the gravitational acceleration on Earth is 9.8 m/s 2 ?

15. the mass of the man, m the mass of the planet Zirgon (m Z ) in terms of the mass of the Earth (m E ), m Z =2m E the radius of the planet Zirgon (r Z ) in terms of the radius of the Earth (r E ), r Z =r E

16. Situation on Earth w E = mg E = G m E ⋅ m r E 2 = (70kg)(9.8m/s 2 ) = 686 N

17. Situation on Zirgon in terms of situation on Earth w Z = mg Z = Gm Z ⋅ m r Z 2 = G 2m E ⋅ m r E 2 = 2 Gm E ⋅ m r E 2 = 2 w E = 2 (686N) = 1372N

18. Exercise Two objects of mass 2m and 3m respectively exert a force F on each other when they are a certain distance apart. What will be the force between two objects situated the same distance apart but having a mass of 5m and 6m respectively? A: 0.2 F B: 1.2 F C: 2.2 F D: 5 F

19. As the distance of an object above the surface of the Earth is greatly increased, the weight of the object would A: increase B: decrease C: increase and then suddenly decrease D: remain the same

20. A satellite circles around the Earth at a height where the gravitational force is 4 times less than at the surface of the Earth. If the Earth's radius is R, then the height of the satellite above the surface is: A: R B: 2 R C: 4 R D: 16 R

21. A satellite experiences a force F when at the surface of the Earth. What will be the force on the satellite if it orbits at a height equal to the diameter of the Earth: A: 1F B: 12F C: 13F D: 19F

22. The weight of a rock lying on surface of the Moon is W. The radius of the Moon is R. On planet Alpha, the same rock has weight 8W. If the radius of planet Alpha is half that of the Moon, and the mass of the Moon is M, then the mass, in kg, of planet Alpha is: A: M2 B: M4 C: 2 M D: 4 M

23. Impulse and Momentum Momentum – the product of a moving body’s mass and its velocity – the tendency of an object to continue to move in its direction of travel p= mv F= ma F= m(v f -v i ) t Ft=mv f -mv i impulse-momentum eq. Ft = impulse

24. The change in momentum is equal to the impulse acting. Ft = m(v f –v i ) The change in momentum divided by time is equal to the force acting. F= m(v f -v i ) t

25. A 1.0kg ball is kicked from rest, giving it a velocity of 12.0 m/s. 1.Find the impulse 2.If the ball and foot were in contact for 0.22s, what was the average force exerted by the foot?

26. Calculating the Total Momentum of a System Two billiard balls roll towards each other. They each have a mass of 0.3 kg. Ball 1 is moving at v 1 =1m/s to the right, while ball 2 is moving at v 2 =0.8m/s to the left. Calculate the total momentum of the system. p total = p 1 + p 2 = m 1 v 1 + m 2 v 2 = 0.06 kg m/s to the right

27. Change in Momentum Consider a tennis ball (mass = 0.1 kg) that is dropped at an initial velocity of 5 m/s and bounces back at a final velocity of 3 m/s. p i = m v i p f = m v f  p = p f –p i  p = 0.8 kg m/s upwards

28. Impulse and Change in momentum A 150 N resultant force acts on a 300 kg trailer. Calculate how long it takes this force to change the trailer's velocity from 2 m/s to 6 m/s in the same direction. Assume that the forces act to the right. F net  t = m (v f – v i )  t = 8 sec

29. Conservation of Momentum The total linear momentum of an isolated system is constant. An isolated system has no forces acting on it from the outside. p i =p f m 1 v i1 +m 2 v i2 =m 1 v f1 +m 2 v f2

30. A toy car of mass 2 kg moves westwards with a speed of 2 m/s. It collides head-on with a toy train. The train has a mass of 1.5 kg and is moving at a speed of 1.5 m/s eastwards. If the car rebounds at 2.05 m/s, calculate the velocity of the train. v f1 = 3.9 m/s westwards 2 kg

31. Initial momentum Before collision Final momentum After collision Change in momentum (final minus initial) 2.0 kg toy car(2.0 kg) (-2 m/s) = -4.0 kg m/s (2.0 kg) (2.05 m/s) = 4.1 kg m/s 4.1 – -4.0 = 8.1 kg m/s 1.5 kg toy train(1.5 kg) (1.5 m/s) = 2.25 kg m/s (1.5 kg) (-3.9 m/s) = -5.85 kg m/s -5.85 – 2.25 = -8.1 kg m/s Combined-1.75 kg m/s

32. Related concepts Transportation safety – Seatbelts – Airbags – Crumple zones Padding as protection Follow through in sports Recoil

33. Ohm’s Law V = IR V – voltage, I – current, R – resistance R = V/I 1 ohm = 1 volt/1 ampere

34. Series circuit R = R 1 + R 2 + R 3 I s = I R1 = I R2 = IR3 V s = V R1 + V R2 + V R3

35. Parallel Circuit V s = V R1 = V R2 = V R3 I s = I R1 + I R2 +I R3

36. Power & Energy P = IV = (V/R) V P = V 2 /R P – power 1 watt = 1 ampere x 1 volt E = P T E – electric energy, P – power rating, T – time 1 kilowatt hour = 1 kilowatt x 1 hour