Dynamics

1. Dynamics

2. Things to Remember from Last Year Newton's Three Laws of Motion Inertial Reference Frames Mass vs. Weight Forces we studied:  weight / gravity  normal force  tension  friction (kinetic and static) Drawing Free Body Diagrams Problem Solving

3. Newton's Laws of Motion 1. An object maintains its velocity (both speed and direction) unless acted upon by a nonzero net force. 2. Newton’s second law is the relation between acceleration and force. 3. Whenever one object exerts a force on a second object, the second object exerts an equal force in the opposite direction on the first object. ΣF = ma

4. 1 Which of Newton's laws best explains why 
motorists should buckle-up? A First Law B Second Law C Third Law D Law of Gravitation

5. 2 You are standing in a moving bus, facing forward, 
and you suddenly fall forward. You can infer from 
this that the bus's A velocity decreased. B velocity increased. C speed remained the same, but it is turning 
to the right. D speed remained the same, but it is turning 
to the left.

6. 3 You are standing in a moving bus, facing forward, 
and you suddenly fall forward as the bus comes to 
an immediate stop. What force caused you to fall 
forward? A gravity B normal force due to your contact with the floor 
of the bus C force due to friction between you and the floor 
of the bus D There is not a force leading to your fall.

7. 4 When the rocket engines on the spacecraft are 
suddenly turned off, while traveling in empty 
space, the starship will A stop immediately. B slowly slow down, then stop. C go faster. D move with constant speed

8. 5 A net force F accelerates a mass m with an 
acceleration a. If the same net force is applied to 
mass 2m, then the acceleration will be A 4a B 2a C a/2 D a/4

9. 6 A net force F acts on a mass m and produces an 
acceleration a. What acceleration results if a net 
force 2F acts on mass 4m? A a/2 B 8a C 4a D 2a

10. 7 An object of mass m sits on a flat table. The Earth 
pulls on this object with force mg, which we will 
call the action force. What is the reaction force? A The table pushing up on the object with force 
mg B The object pushing down on the table with 
force mg C The table pushing down on the floor with force 
mg D The object pulling upward on the Earth with 
force mg

11. 8 A 20-ton truck collides with a 1500-lb car and 
causes a lot of damage to the car. Since a lot of 
damage is done on the car A the force on the truck is greater then the 
force on the car B the force on the truck is equal to the force 
on the car C the force on the truck is smaller than the 
force on the car D the truck did not slow down during the 
collision

12. Inertial Reference Frames Newton's laws are only valid in inertial reference 
frames: An inertial reference frame is one in which Newton’s 
first law is valid. This excludes rotating and 
accelerating frames.

13. Mass and Weight MASS is the measure of the inertia of an object, the 
resistance of an object to accelerate. WEIGHT is the force exerted on that object by 
gravity. Close to the surface of the Earth, where the 
gravitational force is nearly constant, the weight is: Mass is measured in kilograms, weight Newtons. FG = mg

14. Normal Force and Weight FN The Normal Force, 
FN, is ALWAYS perpendicular to 
the surface. Weight, mg, is 
ALWAYS directed 
downward. mg

15. 9 The acceleration due to gravity is lower on the 
Moon than on Earth. Which of the following is 
true about the mass and weight of an astronaut on 
the Moon's surface, compared to Earth? A Mass is less, weight is same B Mass is same, weight is less C Both mass and weight are less D Both mass and weight are the same

16. 10 A 14 N brick is sitting on a table. What is the 
normal force supplied by the table? A 14 N upwards B 28 N upwards C 14 N downwards D 28 N downwards

17. Kinetic Friction Friction forces are 
ALWAYS parallel to 
the surface exerting 
them. Kinetic friction is 
always directed 
opposite to 
direction the object 
is sliding and has 
magnitude: fK = μkFN v fK

18. 11 A 4.0kg brick is sliding on a surface. The 
coefficient of kinetic friction between the surfaces 
is 0.25. What it the size of the force of friction? A 8.0 B 8.8 C 9.0 D 9.8

19. 12 A brick is sliding to the right on a horizontal 
surface. What are the directions of the two surface forces: 
the friction force and the normal force? A right, down B right, up C left, down D left, up

20. Static Friction Static friction is 
always equal and 
opposite the Net 
Applied Force 
acting on the object 
(not including 
friction). Its magnitude is: fS ≤ μSFN fS FAPP

21. 13 A 4.0 kg brick is sitting on a table. The coefficient 
of static friction between the surfaces is 0.45. 
What is the largest force that can be applied 
horizontally to the brick before it begins to slide? A 16.33 B 17.64 C 17.98 D 18.12

22. 14 A 4.0kg brick is sitting on a table. The coefficient 
of static friction between the surfaces is 0.45. If a 
10 N horizontal force is applied to the brick, what 
will be the force of friction and will the brick 
move? A 16.12, no B 17.64, no C 16.12, yes D 17.64, yes

23. Tension Force When a cord or rope pulls on an 
object, it is said to be under 
tension, and the force it exerts is 
called a tension force, FT. FT a mg

24. 15 A crane is lifting a 60 kg load at a constant 
velocity. Determine the tension force in the cable. A 568 N B 578 N C 504 N D 600 N

25. 16 A system of two blocks is accelerated by an 
applied force of magnitude F on the frictionless 
horizontal surface. The tension in the string 
between the blocks is: A 6F F 6 kg 4 kg B 4F C 3/5 F D 1/6 F E 1/4 F

26. Two Dimensions

27. Adding Orthogonal Forces Since forces are vectors, they add as vectors. The simplest case is if the forces are perpendicular 
(orthogonal) with another. Let's add a 50 N force at 0o with a 40 N force at 90o.

28. Adding Orthogonal Forces 1. Draw the first force 
vector, 50 N at 0o, 
beginning at the origin. 90o 180o 0o 270o

29. Adding Orthogonal Forces 1. Draw the first force 
vector, 50 N at 0o, beginning at the origin. 2. Draw the second force 
vector, 40 N at 90o, with its tail at the tip of the first 
vector. 90o 180o 0o 270o

30. Adding Orthogonal Forces 1. Draw the first force 
vector, 50 N at 0o, beginning at the origin. 2. Draw the second force 
vector, 40 N at 90o, with its tail at the tip of the first 
vector. 3. Draw Fnet from the tail of the first force to the tip of 
the last force. 90o 180o 0o 270o

31. Adding Orthogonal Forces We get the magnitude of the resultant from the 
Pythagorean Theorem. c2 = a2 + b2 Fnet2 = (50N)2 + (40N)2 = 2500N2 + 1600N2 = 4100N2 Fnet = (4100N2)1/2 = 64 N 90o 180o 0o 270o

32. Adding Orthogonal Forces We get the direction of the net Force from the inverse tangent. tan(θ) = opp / adj tan(θ) = (40N) / (50N) tan(θ) = 4/5 tan(θ) = 0.8 θ = tan-1(0.80) = 39o Fnet = 64 N at 39o 90o 180o 0o 270o

33. Decomposing Forces into Orthogonal 
Components 90o Let's decompose a 120 N 
force at 34o into its x and y 
components. F θ 180o 0o 270o

34. Decomposing Forces into Orthogonal 
Components 90o We have Fnet, which is the hypotenuse, so let's first 
find the adjacent side by 
using cosθ = adj / hyp cosθ = Fx / Fnet Fx = Fnet cosθ = 120N cos34o = 100 N F θ 180o 0o Fx 270o

35. Decomposing Forces into Orthogonal 
Components 90o Now let's find the opposite 
side by using sinθ = adj / hyp sinθ = Fy / Fnet Fy = Fnet sinθ = 120N sin34o = 67 N Fy F θ 180o 0o Fx 270o

36. Adding non-Orthogonal Forces Now that we know how to add orthogonal forces. And we know now to break forces into orthogonal 
components, so we can make any force into two 
orthogonal forces. We combine those two steps to add any number of 
forces together at any different angles.

37. Adding non-Orthogonal Forces Let's add together these three forces: F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o First, we will do it graphically, to show the 
principle for how we will do it analytically. Then, we will do it analytically, which is easy once 
you see why it works that way.

38. Adding non-Orthogonal Forces 90o First, here are the 
vectors all drawn 
from the origin. F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o 180o 0o 270o

39. Adding non-Orthogonal Forces 90o Now we arrange 
them tail to tip. F3 F2 F1 180o 0o F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o 270o

40. Adding non-Orthogonal Forces 90o Then we draw in the 
Resultant, the sum of the vectors. F3 F2 F F F1 F2 F3 F1 180o 0o F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o 270o

41. Adding non-Orthogonal Forces 90o F3 Now, we graphically 
break each vector 
into its orthogonal 
components. F3y F3x F F2 F2y F2x F1 F1y F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o 180o 0o F1x 270o

42. Adding non-Orthogonal Forces 90o Then, we remove 
the original vectors 
and note that the 
sum of the 
components is the 
same as the sum of 
the vectors. F3y F3x F F2y F2x F1y 180o 0o F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o F1x 270o

43. Adding non-Orthogonal Forces 90o This is the key result: The sum of the x-
components of the 
vectors equals the x-
component of the 
Resultant. As it is for the y-
components. F3y F F2y Fy F1y F3x F1x F2x F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o 180o 0o Fx 270o

44. Adding non-Orthogonal Forces This explains how we will add vectors analytically. 1. Write the magnitude and direction (from 0 to 360 degrees) 
of each vector. 2. Compute the values of the x and y components. 3. Add the x components to find the x component of the 
Resultant. 4. Repeat for the y-components. 5. Use Pythagorean Theorem to find the magnitude of the 
Resultant. 6. Use Inverse Tangent to find the Resultant's direction.

45. Adding non-Orthogonal Forces First make a table and enter what you know. F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o

46. Adding non-Orthogonal Forces Then calculate the x and y components. F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o

47. Adding non-Orthogonal Forces Then add up the x and y components. F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o

48. Adding non-Orthogonal Forces Then use Pythagorean Theorem and tan-1 to determine Fnet. F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o

49. Adding non-Orthogonal Forces Let's add together these three forces: F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o

50. Adding non-Orthogonal Forces Now use the same procedure to add these forces. DO NOT IGNORE NEGATIVE SIGNS F1 = 21 N at 60o F2 = 65 N at 150o F3 = 8.4 N at 330o