1. Newton’s Second Law
Chapter 6

2. Force = mass X acceleration
The Second Law
Force = mass X acceleration
SF = ma
SF = 0 or SF = ma
-Still object -Accelerating object
-Obj. at constant velocity
Sum of all the forces acting on a body
Vector quantity

3. The Second Law Situation One: Non-moving Object Still has forces
Force of the material of the rock
Force of gravity

4. The Second Law Situation Two: Moving Object: Constant Velocity
SF = 0
Fpedalling = Fair + Ffriction
Fpedalling
Fair
Ffriction

5. The Second Law Situation Two: Moving Object: Accelerating Fpedalling
SF = ma
ma = Fpedalling – Fair - Ffriction
Fpedalling
Fair
Ffriction

6. The Second Law Unit of Force = the Newton
SF=ma
SF = (kg)(m/s2)
1 N = 1 kg-m/s2  (MKS)
1 Newton can accelerate a 1 kg object from rest to 1 m/s in 1 s.

7. BOTH INDICATE NO ACCELERATION
Equilibrium
No motion
Constant velocity
BOTH INDICATE NO ACCELERATION
SF=0

8. Three ropes are tied together for a wacky tug-of-war
Three ropes are tied together for a wacky tug-of-war. One person pulls west with 100 N of force, another south with 200 N of force. Calculate the magnitude and direction of the third force. ?
100 N
200 N

9. A car with weight 15,000 N is being towed up a 20o slope (smooth) at a constant velocity. The tow rope is rated at 6000 N. Will it break?

10. Accelerating Systems
SF=ma
Must add up all forces on the object

11. What Force is needed to accelerate a 5 kg bowling ball from 0 to 20 m/s over a time period of 2 seconds?

12. Calculate the net force required to stop a 1500 kg car from a speed of 100 km/h within a distance of 55 m.
100 km/h = 28 m/s
v2 = vo2 + 2a(x-xo)
a = (v2 - vo2)/2(x-xo)
a = 02 – (28 m/s)2/2(55m) = m/s2

13. A 1500 kg car is pulled by a tow truck
A 1500 kg car is pulled by a tow truck. The tension in the rope is 2500 N and the 200 N frictional force opposes the motion. The car starts from rest.
a. Calculate the net force on the car
b. Calculate the car’s speed after 5.0 s

14. A 500. 0 gram model rocket (weight = 4
A gram model rocket (weight = 4.90 N) is launched straight up from rest by an engine that burns for 5 seconds at 20.0 N.
Calculate the net force on the rocket
Calculate the acceleration of the rocket
Calculate the height and velocity of the rocket after 5 s
Calculate the maximum height of the rocket even after the engine has burned out.

15. Calculate the sum of the two forces acting on the boat shown below. (53.3 N, +11.0o)

16. Mass vs. Weight Mass Weight
The amount of matter in an object/INTRINSIC PROPERTY
Independent of gravity
Measured in kilograms
Weight
Force that results from gravity pulling on an object
Weight = mg (g = 9.8 m/s2)

17. Mass vs. Weight Weight = mg is really a re-write of F=ma.
Weight is a force
g is the acceleration (a) of gravity
Metric unit of weight is a Newton
English unit is a pound

18. A 60. 0 kg person weighs 1554 N on Jupiter
A 60.0 kg person weighs 1554 N on Jupiter. What is the acceleration of gravity on Jupiter?

19. Elevator at Constant Velocity
SF = FN – mg
ma = FN – mg
0 = FN – mg
FN = mg
Suppose Chewbacca has a mass of 102 kg:
FN = mg = (102kg)(9.8m/s2)
FN = 1000 N FN mg
a is zero

20. Elevator Accelerating Upward
a = 4.9 m/s2
SF = FN – mg
ma = FN – mg
FN = ma + mg
FN = m(a + g)
FN=(102kg)(4.9m/s2+9.8 m/s2)
FN = 1500 N FN mg
a is upward

21. Elevator Accelerating Downward
a = -4.9 m/s2
SF = FN – mg
ma = FN – mg
FN = ma + mg
FN = m(a + g)
FN=(102kg)(-4.9m/s2+9.8 m/s2)
FN = 500 N FN mg
a is down

22. At what acceleration will he feel weightless?
FN = 0
SF = FN – mg
ma = FN – mg
ma = 0 – mg
ma = -mg
a = -9.8 m/s2
Apparent weightlessness occurs if a > g FN mg

23. A 10. o kg present is sitting on a table
A 10.o kg present is sitting on a table. Calculate the weight and the normal force. FN
Fg = W

24. Suppose someone leans on the box, adding an additional 40.0 N of force. Calculate the normal force.

25. Now your friend lifts up with a string (but does not lift the box off the table). Calculate the normal force.

26. What happen when the person pulls upward with a force of 100 N?
SF = FN+ Fp – mg
SF = N – 98N = 2.0N
ma = 2N
a = 2N/10.0 kg = 0.2 m/s2
Fp = N
Fg = mg = 98.0 N

27. Free Body Diagrams: Ex. 3
A person pulls on the box (10.0 kg) at an angle as shown below. Calculate the acceleration of the box and the normal force. (78.0 N)
Fp = 40.0 N
30o FN mg

28. Friction Always opposes the direction of motion.
Proportional to the Normal Force (more massive objects have more friction) FN
Ffr
Ffr Fa FN Fa mg mg

29. Friction Static -opposes motion before it moves (ms)
Generally greater than kinetic friction
Fmax = Force needed to get an objct moving
Fmax = msFN
Kinetic - opposes motion while it moves (mk)
Generally less than static friction
Ffr = mkFN

30. Friction and Rolling Wheels
Rolling uses static friction
A new part of the wheel/tire is coming in contact with the road every instant B A

31. Braking uses kinetic friction
Point A gets drug across the surface A

32. A 50. 0 kg wooden box is pushed across a wooden floor (mk=0
A 50.0 kg wooden box is pushed across a wooden floor (mk=0.20) at a steady speed of 2.0 m/s.
How much force does she exert? (98 N)
If she stops pushing, calculate the acceleration. (-1.96 m/s2)
Calculate how far the box slides until it stops. (1.00 m)

33. A 100 kg box is on the back of a truck (ms = 0. 40)
A 100 kg box is on the back of a truck (ms = 0.40). The box is 50 cm X 50 cm X 50 cm. a. Calculate the maximum acceleration of the truck before the box starts to slip.

34. Your little sister wants a ride on her sled
Your little sister wants a ride on her sled. Should you push or pull her?

36. Inclines
What trigonometric function does this resemble?

37. Inclines FN mg q

38. Inclines FN q
mgcosq mg q
mgsinq

39. Inclines FN
Ffr
mgsinq
mgcosq q

40. A 50.0 kg file cabinet is in the back of a dump truck (ms = 0.800).
Calculate the magnitude of the static friction on the cabinet when the bed of the truck is tilted at 20.0o (170 N)
Calculate the angle at which the cabinet will start to slide. (39o)

41. Given the following drawing:
Calculate the acceleration of the skier. (snow has a mk of 0.10) (4.0 m/s2)
Calculate her speed after 4.0 s? (16 m/s)

42. First we need to resolve the force of gravity into x and y components:

43. FGy = mgcos30o
FGx = mgsin30o
The pull down the hill is:
The pull up the hill is:
Ffr= mkFN
Ffr= (0.10)(mgcos30o)

44. SF = pull down – pull up
SF = mgsin30o– (0.10)(mgcos30o)
ma = mgsin30o– (0.10)(mgcos30o)
a = gsin30o– (0.10)(gcos30o)
a = 4.0 m/s2
(note that this is independent of the skier’s mass)

45. To find the speed after 4 seconds:
v = vo + at
v = 0 + (4.0 m/s2)(4.0 s) = 16 m/s

46. Suppose the snow is slushy and the skier moves at a constant speed
Suppose the snow is slushy and the skier moves at a constant speed. Calculate mk
SF = pull down – pull up
ma = mgsin30o– (mk)(mgcos30o)
a = gsin30o– (mk)(gcos30o)

47. Since the speed is constant, acceleration =0
0 = gsin30o– (mk)(gcos30o)
(mk)(gcos30o) = gsin30o
mk= gsin30o = sin30o = tan30o =0.577
gcos30o cos30o

48. Drag D ≈ ¼Av2 D = drag force A = Area V = velocity Fails for
Very small particles (dust)
Very fast (airplanes)
Water and dense fluids

49. Finding Acceleration with Drag
Derive the formula for the acceleration of a freefalling object including the drag force.

50. Terminal Speed
Find the formula for terminal speed (a=0) for a freefalling body
Calculate the terminal velocity of a person who is 1.8 m tall, 0.40 m wide, and 75 kg. (64 m/s)

51. A 1500 kg car is travelling at 30 m/s when the driver slams on the brakes (mk = 0.800). Calculate the stopping distance:
On a level road. (57.0 m)
Up a 10.0o incline (48.0 m)
Down a 10.0o incline (75.0 m)

52. A dogsled has a mass of 200 kg. The sled reaches cruising speed, 5
A dogsled has a mass of 200 kg. The sled reaches cruising speed, 5.0 m/s in 15 m. Two ropes are attached to the sled at 10.0o, one on each side connected to the dogs. (mk = 0.060)
Calculate the acceleration of the sled. (0.833 m/s)
Calculate T1 and T2 during the acceleration period. (140 N)

53. Formula Wrap-Up SF=ma Weight = mg (g = 9.8 m/s2) Fmax = msFN
Ffr = mkFN