Forces and the Laws of Motion

1. Forces and the Laws of Motion Newton’s Laws

2. Newton’s First Law(Law of inertia) • An object at rest remains at rest, and an object in motion continues in motion with constant velocity unless the object experiences a net external force.

3. Inertia • Inertia is the tendency of an object not to accelerate. • In other words, Newton’s First Law says that when the net external force on an object is zero, the object’s acceleration (change in velocity) is zero.

8. Net force • The sum of forces action on an object is called the net force. F ground F resistance F forward F gravity A car moving at a constant velocity has a vector sum of zero.

10. Force Pairs • Forces always come in pairs 1. Applied Force…force ON an object 2. Frictional Force…opposes motion • Normal Force…upward forces, opposes weight • Gravity/Weight…acts downward • Fa Fn Ff Fg

14. Inertia • Inertia is proportional to an objects mass. • The greater the mass, the less the body accelerates under an applied force. • Ex. Basketball and bowling ball – Which has more inertia? • Bowling Ball! Why? Because it has more mass! • Therefore, mass can be defined as the amount of matter in an object and also as a measure of the inertia of an object.

15. Equilibrium • Definition: objects that are either at rest or moving with constant velocity are said to be in equilibrium. Equilibrium means the net force acting on a body in equilibrium must be equal to zero. • It the net force is zero, the body is in equilibrium. IF there is a net force, a second force equal and opposite to this net force will put the body in equilibrium.

16. Law of Acceleration – Newton’s Second Law • Relates acceleration of an object to the force applied on it and the mass of the object • A is directly proportional to F • A is indirectly proportional to M • Newton’s 2nd law of motion – which states: The acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to the object’s mass. Force = mass x acceleration

20. Formula • F = ma • Units: m….kg a….m/s2 F….kg-m/s2 or N (Newton)

21. Mass vs. Weight • Mass…the amount of matter • Weight…the gravitational attraction for matter • 1 N is the “weight” of one medium apple! • So, W = mg where g= 9.8 m/s2 • Is a vector … + is up - is down

23. VI. Net Force = Fnet = MAnet • Fnet is the sum of all of the forces acting on an object VERTICAL: Fnet = Fup + Fdown = Fa + W = Fa + mg HORIZONTAL: Fnet = Fforward + Ffriction = Fa + μW = Fa + μmg

25. Ex. 13. A 1090 N person, standing on a scale, is moving upward in an elevator at a constant velocity. Find the reading on the scale. + 1090 N

26. Fnet = Fup + Fdown MAnet = Fscale + W Fscale = MAnet – W Fscale = 0 – ( - 1090) = + 1090 N

27. Ex. 14. A 88.2 kg person standing on a scale in an elevator is moving upward with an acceleration of 1.55 m/s2. Find the reading on the scale. + 1.00 x 103N

28. Fnet = Fup + Fdown MAnet = Fscale + mg Fscale = MAnet – MG = M (a-g) = 88.2 [1.55 – (-9.80)] = 88.2 (11.35) = 1001.07 = + 1.00 x 103N

29. 15. A 36 kg object is accelerated upward at 2.0 m/s2 . Calculate the force applied on the object. 420 N

30. Fnet = MAnet = Fup + W Fup = MAnet – W = MAnet – mg = m (a-g) = 36 [2.0 – (-9.80)] = 36 (11.8) = 424.8 = 420 N

31. 16.. A rocketship ( m = 1.75 x106 kg) has a lift-off force of 3.50 x107 N. Calculate the net acceleration of the spacecraft. + 10.2 m/s2

32. Fnet = MAnet = Fup + W Anet = (Fup + W) / M = (Fup + mg) / m = [3.50 x107 N + (1.75 x106 kg)(-9.80)] / 1.75 x106 kg

33. 19.. A 4.55 kg cat is shot upwards with an overall acceleration of 15.5 m/s2. Calculate the thrust needed for the launch. 115 N

34. Fnet = MAnet = Fup + W Fup = Fnet – W = ma – mg = m (a-g) = 4.55 [15.5 –(-9.80)] = 4.55(25.3) = 115 N

35. Fnet = MAnet = Fup + W Fup = MAnet – W = MAnet – mg = m (a-g) = 36 [2.0 – (-9.80)] = 36 (11.8) = 424.8 = 420 N

36. 16.. A rocketship ( m = 1.75 x106 kg) has a lift-off force of 3.50 x107 N. Calculate the net acceleration of the spacecraft. + 10.2 m/s2

37. Fnet = MAnet = Fup + W Anet = (Fup + W) / M = (Fup + mg) / m = [3.50 x107 N + (1.75 x106 kg)(-9.80)] / 1.75 x106 kg

38. 17.. A 36 kg rock is pushed against a surface where μ = 0.800. What force must be applied to give an net acceleration of 0.25 m/s2? 290 N

39. Fnet = MAnet = Fa + Ff MAnet = Fa + μmg Fa = MAnet – μmg = m (a – μg) = 36 [0.25 – (0.800)(-9.80)] = 36 [0.25 + 7.84] = 36(8.09) = 291.24 = 290 N

40. 18.. A 788 kg rock is pushed along a road, using 6850 N of force to achieve 0.250 m/s2 of acceleration. Calculate the COF of the road and the rock. 0.862

41. Fnet = MAnet = Fa + μmg μmg = MAnet – Fa μ = (ma – Fa) / mg = [(788)(0.250) – 6850] / (788)(-9.80) = -6653/-7722.4 μ = 0.862

42. Friction Friction…any force that opposes the motion of an object • Due to bonding forces between the 2 surfaces…”microwelds”

44. Types of friction: 1. Static Friction…force necessary to start the motion of an object at rest …break the microwelds 2. Sliding Friction…F necessary to keep an object in motion at a constant velocity …less than the static friction

46. 3. Rolling Friction…friction between a rolling object and the surface it is rolling on 4. Fluid Friction…friction between a moving fluid and the surface of the object it is flowing within …”fluids” include liquids and gases

49. Coefficient of Friction = μ • The ratio of the applied force to the weight • High μ…high resistance to motion • Low μ…low resistance to motion • μ = -Fa(Ff) / Fgor -Fa(Ff)=μFg