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

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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

13. Vector Addition

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

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

16. 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

17. 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

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

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

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

21. Solution
Trigonometry
Law of Cosines

22. Solution
Trigonometry
Law of Sines

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

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

25. Solution
Q = 60N
P = 40N

26. Solution
Q = 60N
P = 40N

27. 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

28. 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

29. 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.

30. 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

31. 25kN
18.30kN
25kN
12.94kN

32. 25kN
12.5kN
25kN
25kN
21.65kN
25kN

33. 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

34. 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

35. 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 =

36. 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.

37. 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

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

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

40. 2.8 ADDITION OF FORCE BY SUMMING X AND Y COMPONENTS

41. 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.

42. 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

43. 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

44. - 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

45. PROBLEMS

46. 2.21
Determine the x and y component of each
of the forces shown.

47. Solution

48. Solution

49. 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

50. Solution
2.59m
2.69m
2.5 m

51. Solution
2.59m
2.69m
2.5 m

52. THE END