1. Newton's Three laws of Motion:-
First law : A particle originally at rest, or moving in a straight line with constant velocity, tends to remain in this state provided the particle is not subjected to an unbalanced force.

2. Second law: A particle acted upon by an unbalanced force “F” experiences an acceleration “a” that has the same direction as the force and a magnitude that is directly proportional to the force.

3. Third Law: The mutual forces of action and reaction between two particles are equal, opposite, and collinear.

4. Newton’s law of gravitational attraction :

5. Characteristics of forces
External and Internal Forces
Forces acting on rigid bodies are divided into two groups:
External forces
Internal forces
External forces are shown in a free-body diagram.

6. External forces Internal forces
The external forces represent the action of other bodies on the rigid body under consideration. They are entirely responsible for the external behavior of the rigid body. They will either cause it to move or ensure that it remains at rest.
Internal forces
The internal forces are the forces which hold together the particles forming the rigid
body. If the rigid body is structurally composed of several parts, the forces holding the
components parts together are also defined as internal forces.

7. Example
From the image it is observed that, the line of action of the force F is a horizontal line passing through both the front and the rear bumpers of the truck.
Using the principle of transmissibility, we can therefore replace F by an
equivalent force F’ acting on the rear bumper.
In other words, the conditions of motion are unaffected, and all the other external forces acting on the truck ( W, R1, R2 ) remain unchanged.

8. Transmissibility This maintains the equivalent of the system as
the equilibrium remains unchanged.

9. Scalars and Vectors Scalar
– A quantity characterized by a positive or negative number
– Indicated by letters in italic such as A
Eg: Mass, volume and length

10. Vector – A quantity that has both magnitude and direction
Eg: Position, force and moment
– Represent by a letter with an arrow over it such as or A
– Magnitude is designated as or simply A

11. 2.1 Scalars and Vectors Vector – Represented graphically as an arrow
– Length of arrow = Magnitude of Vector
– Angle between the reference axis and arrow’s line of action = Direction of Vector
– Arrowhead = Sense of Vector

12. Magnitude of Vector = 4 units
Example
Magnitude of Vector = 4 units
Direction of Vector = 20° measured counterclockwise from the horizontal axis
Sense of Vector = Upward and to the right
The point O is called tail
of the vector and the point
P is called the tip or head

13. Vector Operations Multiplication and Division of a Vector by a Scalar
- Product of vector A and scalar a = aA
- Magnitude =
- If a is positive, sense of aA is the same as sense of A
- If a is negative sense of
aA, it is opposite to the
sense of A

14. Multiplication and Division of a Vector by a Scalar
- Negative of a vector is found by multiplying the vector by ( -1 )
- Law of multiplication applies
Eg: A/a = ( 1/a ) A, a≠0

15. Vector Addition
- Addition of two vectors A and B gives a resultant vector R by the parallelogram law
- Result R can be found by triangle construction
- Communicative
Eg: R = A + B = B + A

16. Vector Addition

17. Vector Addition
- Special case: Vectors A and B are collinear (both have the same line of action)

18. Vector Subtraction - Special case of addition
Eg: R’ = A – B = A + ( - B )
- Rules of Vector Addition Applies

19. Resolution of Vector
- Any vector can be resolved into two components by the parallelogram law
- The two components A and B are drawn such that they extend from the tail or R to points of intersection

20. Vector Addition of Forces
When two or more forces are added, successive applications of the parallelogram law is carried out to find the resultant
Eg: Forces F1, F2 and F3 acts at a point O
- First, find resultant of
F1 + F2
- Resultant,
FR = ( F1 + F2 ) + F3

21. Procedure for Analysis Parallelogram Law
- Make a sketch using the parallelogram law
- Two components forces add to form the resultant force
- Resultant force is shown by the diagonal of the parallelogram
- The components is shown by the sides of the parallelogram

22. Procedure for Analysis Trigonometry
- Redraw half portion of the parallelogram
- Magnitude of the resultant force can be determined by the law of cosines
- Direction if the resultant force can be determined by the law of sines

23. Procedure for Analysis Trigonometry
- Magnitude of the two components can be determined by the law of sines

24. Example
The screw eye is subjected to two forces F1
and F2. Determine the
magnitude and direction
of the resultant force.

25. Solution
Parallelogram Law
Unknown: magnitude of
FR and angle θ

26. Solution
Trigonometry
Law of Cosines

27. Solution
Trigonometry
Law of Sines

28. Solution
Trigonometry
Direction Φ of FR measured from the horizontal

29. Q = 60N P = 40N Sample Problem The two forces act on bolt at A
Determine their resultant

30. Solution
Q = 60N
P = 40N

31. Solution
Q = 60N
P = 40N

32. Law of Cosines c A B a b C Solution a2 = b2 + c2 – 2bc(cosA)
P = 40N
Q = 60N c
Law of Cosines A B
a2 = b2 + c2 – 2bc(cosA)
b2 = a2 + b2 – 2ac(cosB)
c2 = a2 + b2 – 2ab(cosC) a b C

33. Law of Sinus Q = 60N P = 40N Q = 60N P = 40N
Solution
Law of Sinus
Q = 60N
P = 40N
a/sin A = b/sin B = c/sin C
Q = 60N
P = 40N

34. Sample Problem
SP 2.2
A barge is pulled by two tugboats.
If the resultant of the forces exerted by the tugboats is a 25 kN force
directed along the axis of the barge,
determine
the tension in each rope given α = 45,
(b) the value of α for which the tension in rope 2 is minimum.

35. Solution
25kN
Tension for Graphical Solution
The parallelogram law is used; the diagonal (resultant) is known to be equal 25kN
and to be directed to the right. This sides are drawn paralled to the ropes.If the
drawing is done to scale, we measure
T1 = 18.5 kN
T2= 13.0 kN

36. 25kN
18.30kN
25kN
12.94kN

37. 25kN
12.5kN
25kN
25kN
21.65kN
25kN

38. Rectangular Components of a Force
x- y components
Perpendicular to each other
Fx = F cos θ Fy = F sin θ y Fy F j θ O x i Fx

39. Example 1
A force of 800N is exerted on a bold A as shown
in the diagram.
Determine the horizontal and vertical components
of the force.
Fx = -F cosα= -(800 N ) cos 35º = -655N
Fy = +F sinα= +(800 N ) sin 35º = +459N

40. Example 2 (a) (2.9) (b) Fx = + (300N) cos α Fy = - (300N) sin α
A man pulls with a force of 300N on a rope attached
to a building as shown in the picture.
What are the horizontal and vertical components of the
force exerted by the rope at point A?
Fx = + (300N) cos α
Fy = - (300N) sin α
Observing that AB = 10m
cos α = 8m = 8m = 4
AB m
sin α = 6m = 6m = 3
We thus obtain
Fx = + (300N) = +240 N 4 5
Fy = - (300N) = -180 N 3 5
We write
F = + (240N) i - (180 N) j
tan θ= Fx Fy
(2.9)
F =

41. A force F = (3.1 kN)i + (6.7 kN)j is applied to bolt A.
Example　3
6.7
A force F = (3.1 kN)i + (6.7 kN)j is applied to bolt A.
Determine the magnitude of the force and the angle θit forms with the horizontal.

42. First we draw the diagram showing the two
Solution
First we draw the diagram showing the two
rectangular components of the force and the angle θ.
From Eq.(2.9), we write
6.7
θ= 65.17˚
Using the formula given before,
Fx = F cos θ Fy = F sin θ

43. Pyj P
Syj
Ryi R S
Pxi
Sxi O O
Qxi
Rxi Q
Qyj

44. ADDITION OF FORCE BY SUMMING X AND Y COMPONENTS
How we calculate the force ?

45. 2.8 ADDITION OF FORCE BY SUMMING X AND Y COMPONENTS

46. SAMPLE PROBLEM 2.3 Four forces act on A as shown.
F2 = 80 N
F1 = 150 N
F4 = 100 N
F3 = 110 N
Four forces act on A as shown.
Determine the resultant of the forces on the bolt.

47. Solution ( F2 cos 20˚ ) j ( F1 sin 30˚ ) j F2 = 80 N F1 = 150 N
- F3 j
- ( F2 sin 20˚ ) i
( F2 cos 20˚ ) j
- ( F4 sin 15˚ ) j
( F4 cos 15˚ ) i
( F1 sin 30˚ ) j
( F1 cos 30˚ ) i
F2 = 80 N
F1 = 150 N
F4 = 100 N
F3 = 110 N

48. Solution ( F2 cos 20˚ ) j ( F1 sin 30˚ ) j F2 = 80 N F1 = 150 N
- F3 j
- ( F2 sin 20˚ ) i
( F2 cos 20˚ ) j
- ( F4 sin 15˚ ) j
( F4 cos 15˚ ) i
( F1 sin 30˚ ) j
( F1 cos 30˚ ) i
F2 = 80 N
F1 = 150 N
F4 = 100 N
F3 = 110 N

49. - F3 j
- ( F2 sin 20˚ ) i
( F2 cos 20˚ ) j
- ( F4 sin 15˚ ) j
( F4 cos 15˚ ) i
( F1 sin 30˚ ) j
( F1 cos 30˚ ) i
Ry = (14.3 N) j
Rx = (199.1 N) i

50. PROBLEMS

51. 2.21
Determine the x and y component of each
of the forces shown.

52. Solution

53. Solution

54. Problems 2.23
Determine the x and y components of each of the forces shown.
1.2 m
1.4 m
2.3 m
1.5 m
2.0 m
1800 N
950N
900N

55. Solution
2.59m
2.69m
2.5 m

56. Solution
2.59m
2.69m
2.5 m

57. EQUILIBRIUM OF A PARTICLE
The resultant force acting on a particle is zero
R = ΣF = 0
ΣFx = 0 ΣFy = 0 A
20 N
The forces of 20 N acting on the line but in opposite direction, passing through point A having the same magnitude. This produces a resultant of R = 0.

58. Equilibrium of Forces
F4 = 400 N
F1 = 300 O
30
F2 = 173.2
F1 = 300 O
F4 = 400 N
F3 = 200 N
F3 = 200
F2 = N
30
An equilibrium system of forces produces a closed force polygon

59. Example Figure 2.27 shows four forces acting on A.
In figure 2.28, the resultant of the given forces
is determined by the polygon rule. Starting from point
O with F1 and arranging the forces in tip-to-tail fashion,
we find that the tip of F4 coincides with the
starting point O.
Thus the resultant R of the given system of forces is zero,
and the particle is in equilibrium.
The closed polygon drawn in Fig 2.28 provides a
graphical expression of the equilibrium of A.
To express algebraically the condition for the
equilibrium of a particle, we write
Resolving each force F into rectangular components,
we have
We conclude that the necessary and sufficient
conditions for the equilibrium of a particle are

60. Returning to the particle shown in figure 2.27,
we check that the equilibrium conditions are
satisfied. We write

61. Sample Problem 2.4
In a ship-unloading operation, a 15.6 kN automobile is supported by a cable. A rope is tied to the cable at A and pulled in order to centre the automobile over its intended position. The angle between the cable and the vertical is 2, while the angle between the rope and the horizontal is 30. What is the tension in the rope?

62. 15.6kN
TAB
TAC

63. Sample problem 2.6
2.13m
0.46m
1.22m
180N
270N
As part of the design of a new sailboat, it is desired to determine the drag force which
may be expected at a given speed.
To do so, a model of the proposed hull is placed in a test channel and three cables
are used to keeps its bow on the centerline of the channel.
Dynamometer reading indicate that for a given speed, the tension is 180N in cable AB
and 270N in cable AE.
Determine the drag force exerted on the hull and the tension in cable AC.

64. Solution tanα 1.22m 2.13m = 1.75 α 60.26˚ tanβ 1.22m 0.46m = 0.375 β
Determine of the Angles
First the angles α and βdefining the
direction of cables AB and AC are determined.
2.13m
0.46m
1.22m
180N
270N
tanα
1.22m
2.13m =
1.75 α
60.26˚
tanβ
1.22m
0.46m =
0.375 β
20.56˚
TAC
TAE = 270 N FD
TAB = 180 N
β= 20.56˚
α= 60.26˚
Free body diagram
Choosing the hull as a free body, we draw the
free-body diagram shown. It includes the forces
exerted by the three cables an the hull, as well as
the drag force FD exerted by the flow.

65. Equilibrium Condition
Solution
2.13m
0.46m
1.22m
180N
270N
Equilibrium Condition
We express that the hull is in
the equilibriumby writing that the
resultant of all forces is zero
R = TAB + TAC + TAE + FD
TAC
TAE = 270 N FD
TAB = 180 N
β= 20.56˚
α= 60.26˚

66. R = TAB + TAC + TAE + FD TAC FD Y X
TAE = 270 N FD
β= 20.56˚
α= 60.26˚
TAB = 180 N
Since more than three forces are involved,
we resolved the forces into X and Y components
TAB = -(180N) sin 60.26˚ i + (180N) cos 60.26˚ j
TAC = TAC sin 20.56˚ i TAC cos 20.56˚ j
TAE = - (270N) j
= -(156.3N) i (89.3N) j
= TAC i TAC j
FD = FD i
(180N) cos 60.26˚ j
- (180N) sin 60.26˚ i
- (270N) j X Y
R = TAB + TAC + TAE + FD
from,
(-156.3N + TAC FD ) i + (89.3N + TAC – 270 N) j = 0

67. R = TAB + TAC + TAE + FD TAC = 193 N FD = 88.5 N
TAB = -(180N) sin 60.26˚ i + (180N) cos 60.26˚ j
TAC = TAC sin 20.56˚ i TAC cos 20.56˚ j
TAE = - (270N) j
= -(156.3N) i (89.3N) j
= TAC i TAC j
FD = FD i
R = TAB + TAC + TAE + FD
from,
(-156.3N + TAC FD ) i + (89.3N + TAC – 270 N) j = 0
: N + TAC FD = 0
89.3N + TAC – 270 N = 0
TAC = 193 N
FD = 88.5 N

68. THE END