1. Chapter 2 Statics of Particles
CE 102 Statics
Chapter 2
Statics of Particles

2. Contents Introduction Resultant of Two Forces Vectors
Addition of Vectors
Resultant of Several Concurrent Forces
Sample Problem 2.1
Sample Problem 2.2
Rectangular Components of a Force: Unit Vectors
Addition of Forces by Summing Components
Sample Problem 2.3
Equilibrium of a Particle
Free-Body Diagrams
Sample Problem 2.4
Sample Problem 2.5
Rectangular Components in Space
Sample Problem 2.6

3. Introduction
The objective for the current chapter is to investigate the effects of forces on particles:
- replacing multiple forces acting on a particle with a single equivalent or resultant force,
- relations between forces acting on a particle that is in a state of equilibrium.
The focus on particles does not imply a restriction to miniscule bodies. Rather, the study is restricted to analyses in which the size and shape of the bodies is not significant so that all forces may be assumed to be applied at a single point.

4. Resultant of Two Forces
force: action of one body on another; characterized by its point of application, magnitude, line of action, and sense.
Experimental evidence shows that the combined effect of two forces may be represented by a single resultant force.
The resultant is equivalent to the diagonal of a parallelogram which contains the two forces in adjacent legs.
Force is a vector quantity.

5. Vectors
Vector: parameters possessing magnitude and direction which add according to the parallelogram law. Examples: displacements, velocities, accelerations.
Scalar: parameters possessing magnitude but not direction. Examples: mass, volume, temperature
Vector classifications:
Fixed or bound vectors have well defined points of application that cannot be changed without affecting an analysis.
Free vectors may be freely moved in space without changing their effect on an analysis.
Sliding vectors may be applied anywhere along their line of action without affecting an analysis.
Equal vectors have the same magnitude and direction.
Negative vector of a given vector has the same magnitude and the opposite direction.

6. Addition of Vectors Trapezoid rule for vector addition
Triangle rule for vector addition B C
Law of cosines,
Law of sines,
Vector addition is commutative,
Vector subtraction

7. Addition of Vectors
Addition of three or more vectors through repeated application of the triangle rule
The polygon rule for the addition of three or more vectors.
Vector addition is associative,
Multiplication of a vector by a scalar

8. Resultant of Several Concurrent Forces
Concurrent forces: set of forces which all pass through the same point. A set of concurrent forces applied to a particle may be replaced by a single resultant force which is the vector sum of the applied forces.
Vector force components: two or more force vectors which, together, have the same effect as a single force vector.

9. Sample Problem 2.1 SOLUTION:
Graphical solution - construct a parallelogram with sides in the same direction as P and Q and lengths in proportion. Graphically evaluate the resultant which is equivalent in direction and proportional in magnitude to the the diagonal.
The two forces act on a bolt at A. Determine their resultant.
Trigonometric solution - use the triangle rule for vector addition in conjunction with the law of cosines and law of sines to find the resultant.

10. Sample Problem 2.1
Graphical solution - A parallelogram with sides equal to P and Q is drawn to scale. The magnitude and direction of the resultant or of the diagonal to the parallelogram are measured,
Graphical solution - A triangle is drawn with P and Q head-to-tail and to scale. The magnitude and direction of the resultant or of the third side of the triangle are measured,

11. Sample Problem 2.1
Trigonometric solution - Apply the triangle rule. From the Law of Cosines,
From the Law of Sines,

12. Sample Problem 2.2 SOLUTION:
Find a graphical solution by applying the Parallelogram Rule for vector addition. The parallelogram has sides in the directions of the two ropes and a diagonal in the direction of the barge axis and length proportional to 5000 N.
A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is 5000 N directed along the axis of the barge, determine
Find a trigonometric solution by applying the Triangle Rule for vector addition. With the magnitude and direction of the resultant known and the directions of the other two sides parallel to the ropes given, apply the Law of Sines to find the rope tensions.
the tension in each of the ropes for a = 45o,
the value of a for which the tension in rope 2 is a minimum.
The angle for minimum tension in rope 2 is determined by applying the Triangle Rule and observing the effect of variations in a.

13. Sample Problem 2.2
Graphical solution - Parallelogram Rule with known resultant direction and magnitude, known directions for sides.
Trigonometric solution - Triangle Rule with Law of Sines

14. Sample Problem 2.2
The angle for minimum tension in rope 2 is determined by applying the Triangle Rule and observing the effect of variations in a.
The minimum tension in rope 2 occurs when T1 and T2 are perpendicular.

15. Rectangular Components of a Force: Unit Vectors
May resolve a force vector into perpendicular components so that the resulting parallelogram is a rectangle are referred to as rectangular vector components and
Define perpendicular unit vectors which are parallel to the x and y axes.
Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components. Fx and Fy are referred to as the scalar components of

16. Addition of Forces by Summing Components
Wish to find the resultant of 3 or more concurrent forces,
Resolve each force into rectangular components
The scalar components of the resultant are equal to the sum of the corresponding scalar components of the given forces.
To find the resultant magnitude and direction,

17. Sample Problem 2.3 SOLUTION:
Resolve each force into rectangular components.
Determine the components of the resultant by adding the corresponding force components.
Calculate the magnitude and direction of the resultant.
Four forces act on bolt A as shown. Determine the resultant of the force on the bolt.

18. Sample Problem 2.3 SOLUTION:
Resolve each force into rectangular components.
Determine the components of the resultant by adding the corresponding force components.
Calculate the magnitude and direction.

19. Equilibrium of a Particle
When the resultant of all forces acting on a particle is zero, the particle is in equilibrium.
Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or will continue at constant speed in a straight line.
Particle acted upon by two forces:
equal magnitude
same line of action
opposite sense
Particle acted upon by three or more forces:
graphical solution yields a closed polygon
algebraic solution

20. Free-Body Diagrams
Free-Body Diagram: A sketch showing only the forces on the selected particle.
Space Diagram: A sketch showing the physical conditions of the problem.

21. Sample Problem 2.4 SOLUTION:
Construct a free-body diagram for the particle at the junction of the rope and cable.
Apply the conditions for equilibrium by creating a closed polygon from the forces applied to the particle.
Apply trigonometric relations to determine the unknown force magnitudes.
In a ship-unloading operation, a 3500-N automobile is supported by a cable. A rope is tied to the cable and pulled to center the automobile over its intended position. What is the tension in the rope?

22. Sample Problem 2.4 SOLUTION:
Construct a free-body diagram for the particle at A.
Apply the conditions for equilibrium.
Solve for the unknown force magnitudes.

23. Sample Problem 2.5 SOLUTION:
Choosing the hull as the free body, draw a free-body diagram.
Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero.
It is desired to determine the drag force at a given speed on a prototype sailboat hull. A model is placed in a test channel and three cables are used to align its bow on the channel centerline. For a given speed, the tension is 40 N in cable AB and 60 N in cable AE.
Determine the drag force exerted on the hull and the tension in cable AC.
Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions.

24. Sample Problem 2.5 SOLUTION:
Choosing the hull as the free body, draw a free-body diagram.
Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero.

25. Sample Problem 2.5 ( ) ( ) ( ) ( ) ( ) ( ) ( )
Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions. r r r ( ) ( ) T = - 40 N
sin 60 . 26 i + 40 N
cos 60 . 26 j AB r r ( ) ( ) = - 34 . 73 N i + 19 . 84 N j r r r T = T
sin 20 . 56 i + T
cos 20 . 56 j AC AC AC r r = .
3512 T i + .
9363 T j AC AC r r ( ) T = - 6 N j r r F = F i D D r R = r ( ) = - 34 . 73 + .
3512 T + F i AC D r ( ) + 19 . 84 + .
9363 T - 60 j AC

26. Sample Problem 2.5
This equation is satisfied only if each component of the resultant is equal to zero.

27. Rectangular Components in Space
The vector is contained in the plane OBAC.
Resolve into horizontal and vertical components.
Resolve into rectangular components. f q
sin
cos y h z x F =

28. Rectangular Components in Space
With the angles between and the axes,
is a unit vector along the line of action of and are the direction cosines for

29. Rectangular Components in Space
Direction of the force is defined by the location of two points,

30. Sample Problem 2.6 SOLUTION:
Based on the relative locations of the points A and B, determine the unit vector pointing from A towards B.
Apply the unit vector to determine the components of the force acting on A.
Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.
The tension in the guy wire is 2500 N. Determine:
a) components Fx, Fy, Fz of the force acting on the bolt at A,
b) the angles qx, qy, qz defining the direction of the force

31. Sample Problem 2.6 SOLUTION:
Determine the unit vector pointing from A towards B.
Determine the components of the force.

32. Sample Problem 2.6
Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.

33. Problem 2.7 The direction of the 75-lb forces may vary, but the 240 lb
angle between the forces
is always 50o. Determine
the value of a for which
the resultant of the forces
acting at A is directed
horizontally to the left.

34. Solving Problems on Your Own
240 lb
30o a
50o
75 lb A
The direction of the 75-lb
forces may vary, but the
angle between the forces
is always 50o. Determine
the value of a for which
the resultant of the forces
acting at A is directed
horizontally to the left.
1. Determine the resultant R of two or more forces.
2. Draw a parallelogram with the applied forces as two adjacent
sides and the resultant as the included diagonal.
3. Set the resultant, or sum of the forces, directed horizontally.

35. R1 = 2(75 lb) cos25o = 135.95 lb R1 = 135.95 lb a +25o 240 lb
Problem 2.7 Solution
Determine the resultant R of two
or more forces.
We first Replace the two 75-lb
forces by their resultant R1, using
the triangle rule. a
R1 = 2(75 lb) cos25o = lb
25o
50o
25o R1
R1 = lb
a +25o

36. sin(a+25o) 240 lb sin(30o) 135.95 lb a + 25o = 61.97o a = 37.0o
Problem 2.7 Solution
Draw a parallelogram with the applied forces as two adjacent
sides and the resultant as the included diagonal.
Set the resultant, or sum of the forces, directed horizontally. R2
Consider the resultant R2
of R1 and the 240-lb force
and recall that R2 must be
horizontal and directed to
the left.
30o
a+25o
R1 = lb
240 lb
Law of sines:
sin(a+25o)
240 lb =
sin(30o) lb
sin(a+25o) =
(240 lb) sin(30o) =
a + 25o = 61.97o
a = 37.0o

37. Problem 2.8 y C D A container of weight W = 1165 N is supported O
600 mm
320 mm
450 mm
360 mm
500 mm D O C B A x y z
A container of weight
W = 1165 N is supported
by three cables as shown.
Determine the tension in
each cable.

38. F d y Solving Problems on Your Own C D A container of weight O
600 mm
320 mm
450 mm
360 mm
500 mm D O C B A x y z
Problem 2.8
Solving Problems on Your Own
A container of weight
W = 1165 N is supported
by three cables as shown.
Determine the tension in
each cable.
1. Draw a free-body diagram of the particle. This diagram shows
the particle and all the forces acting on it.
2. Resolve each of the forces into rectangular components.
Follow the method outlined in the text.
F = F l = (dx i + dy j + dz k) F d

39. Solving Problems on Your Own
600 mm
320 mm
450 mm
360 mm
500 mm D O C B A x y z
Problem 2.8
Solving Problems on Your Own
A container of weight
W = 1165 N is supported
by three cables as shown.
Determine the tension in
each cable.
3. Set the resultant, or sum, of the forces exerted on the particle
equal to zero. You will obtain a vectorial equation consisting
of terms containing the unit vectors i, j, and k. Three scalar
equations result, which can be solved for the unknowns.

40. S F = 0 TAB + TAC + TAD + W = 0 AB = (450 mm)i + (600 mm)j AB = 750 mm
Problem 2.8 Solution
600 mm
320 mm
450 mm
360 mm
500 mm D C B A x y z
TAD
TAB
TAC
W = _ (1165 N) j O
Draw a free-body diagram of the particle.
S F = 0
TAB + TAC + TAD + W = 0
AB = (450 mm)i + (600 mm)j AB = 750 mm
AC = (600 mm)j _ (320mm)k AC = 680 mm
AD = (_500 mm)i + (600 mm)j + (360 mm)k AD = 860 mm

41. ( ) ( ) ( ) ( ) ( ) 450 750 i 600 j TAB = (0.6 i + 0.8 j) TAB
600 mm
320 mm
450 mm
360 mm
500 mm D C B A x y z
TAD
TAB
TAC
W = _ (1165 N) j O
Problem 2.8 Solution
Resolve each of the forces
into rectangular components.
( )
450
750 i +
600 j
TAB
= (0.6 i j) TAB
TAB = TAB lAB = AB =
TAC AC
( )
600
320
680 k
( ) 15 8
TAC = TAC lAC = = j
TAC = j k
TAC
680 _ _ 17 17
TAD AD
( )
500
860 i +
600
j +
360 k
TAD = TAD lAD = =
TAD =
( ) 25 43 i + 30
j + 18 k =
TAD

42. 0.6 TAB _ 25 43 TAD = 0 TAB = 0.9690 TAD 15 17 30 43 TAD 0.8 TAB + TAC
600 mm
320 mm
450 mm
360 mm
500 mm D C B A x y z
TAD
TAB
TAC
W = _ (1165 N) j O
Problem 2.8 Solution
Set the resultant, or sum, of
the forces exerted on the
particle equal to zero.
Substitution into S F = 0,
factor i, j, k and set their
coefficients to zero:
0.6 TAB _ 25 43
TAD = 0
TAB = TAD
(1) 15 17 30 43
TAD
0.8 TAB +
TAC +
_ 1165 N = 0 +
(2) 8 17 18 43
TAD = 0
_ TAC +
TAC = TAD
(3)

43. 600 mm
320 mm
450 mm
360 mm
500 mm D C B A x y z
TAD
TAB
TAC
W = _ (1165 N) j O
Problem 2.8 Solution
Substitution for TAB and
TAC from (1) and (3)
into (2): 15 17 30 43
( 0.8 x x )TAD _ 1165 N = 0
TAD _ 1165 N = 0
TAD = 516 N
TAB = 500 N
TAC = 459 N
From (1): TAB = (516 N)
From (3): TAC = (516 N)

44. Problem 2.9 y A Cable AB is 65 ft long, and 56 ft
the tension in that cable is
3900 lb. Determine (a) the
x, y, and z components of
the force exerted by the
cable on the anchor B, (b)
the angles qx, qy, and qz
defining the direction of
that force. D a O B
20o
50o z C x

45. Solving Problems on Your Ownx y z
56 ft D B C A
50o
20o a
Problem 2.9
Solving Problems on Your Own
Cable AB is 65 ft long, and
the tension in that cable is
3900 lb. Determine (a) the
x, y, and z components of
the force exerted by the
cable on the anchor B, (b)
the angles qx, qy, and qz
defining the direction of
that force.
1. Determine the rectangular components of a force defined by
its magnitude and direction. If the direction of the force F is
defined by the angles qy and f, projections of F through these
angles or their components will yield the components of F.

46. Solving Problems on Your Ownx y z
56 ft D B C A
50o
20o a
Problem 2.9
Solving Problems on Your Own
Cable AB is 65 ft long, and
the tension in that cable is
3900 lb. Determine (a) the
x, y, and z components of
the force exerted by the
cable on the anchor B, (b)
the angles qx, qy, and qz
defining the direction of
that force.
2. Determine the direction cosines of the line of action of a force.
The direction cosines of the line of action of a force F are
determined by dividing the components of the force by F.
cos qx= Fx F
cos qy= Fy F
cos qz= Fz F

47. Fy = + F cos qy = (3900 lb)(0.86154) Fy = + 3360 lb
Problem 2.9 Solution A
Determine the direction cosines of
the line of action of a force.
65 ft qy
From triangle AOB:
cos qy =
56 ft
65 ft =
qy = 30.51o F Fy
56 ft Fx O B
(a) Fx = _ F sin qy cos 20o
= _ (3900 lb) sin 30.51o cos 20o
Fx = _1861 lb
20o Fz z x
Fy = + F cos qy = (3900 lb)( ) Fy = lb
Fz = + (3900 lb) sin 30.51o sin 20o Fz = lb

48. Fx F = _1861 lb 3900 lb cos qx = _ 0.4771 qx = 118.5o cos qz = Fz Fy
Problem 2.9 Solution O x B A Fy Fz F Fx
20o qy
65 ft
Determine the direction cosines
of the line of action of a force.
(b) cos qx = Fx F =
_1861 lb
3900 lb
56 ft
cos qx = _
qx = 118.5o
From above (a): qy = 30.5o
cos qz = Fz F
= +
677 lb
3900 lb
= qz = 80.0o

49. Problem 2.10 B A Two cables are tied together at C and
9 ft
5 ft
8.5 ft
12 ft
7.5 ft
Two cables are tied
together at C and
loaded as shown.
determine the tension
(a) in cable AC,
(b) in cable BC. B C
396 lb

50. Solving Problems on Your OwnA
9 ft
5 ft
8.5 ft
12 ft
7.5 ft B C
396 lb
Solving Problems on Your Own
Two cables are tied
together at C and
loaded as shown.
determine the tension
(a) in cable AC,
(b) in cable BC.
1. Draw a free-body diagram of the particle. This diagram shows
the particle and all the forces acting on it.
2. Set the resultant, or sum, of the forces exerted on the particle
equal to zero. You will obtain a vectorial equation consisting
of terms containing the unit vectors i, j, and k. Three scalar
equations result, which can be solved for the unknowns.

51. TBC = 0 S Fx = 0 : + TBC = 1.088 TAC S Fy = 0 : TBC 396 lb = 04
7.5 12
3.5
8.5
12.5
396 lb
FREE BODY C:
TBC = 0
S Fx = 0 : +
TBC = TAC
S Fy = 0 :
TBC lb = 0
Problem Solution
Draw a free-body diagram
of the particle.
Set the resultant, or sum, of
the forces exerted on the
particle equal to zero.

52. y
Problem Solution
TBC
TAC
12.5
8.5
3.5 4 x 12
7.5
396 lb
(a) Substitute for TBC:
3.5 4
TAC +
(1.088 TAC ) _ 396 lb = 0
12.5
8.5
( ) TAC _ 396 lb = 0
TAC = 500 lb
(b)
TBC = (500 lb)
TBC = 544 lb

53. Problem 2.11 y x Collars A and B are connected P
20 in A B O z P Q
Collars A and B are connected
by a 25-in.-long wire and can
slide freely on frictionless rods.
If a 60-lb force Q is applied to
collar B as shown, Determine
(a) the tension in the wire when
x = 9 in., (b) the corresponding
magnitude of the force P required
to maintain the equilibrium of
the system.

54. Solving Problems on Your Ownx
20 in A B O z P Q
Solving Problems on Your Own
Collars A and B are connected
by a 25-in.-long wire and can
slide freely on frictionless rods.
If a 60-lb force Q is applied to
collar B as shown, Determine
(a) the tension in the wire when
x = 9 in., (b) the corresponding
magnitude of the force P required
to maintain the equilibrium of
the system.
Problem 2.11
1. Draw a free-body diagram of the particle. This diagram shows
the particle and all the forces acting on it.

55. Solving Problems on Your Ownx
20 in A B O z P Q
Solving Problems on Your Own
Collars A and B are connected
by a 25-in.-long wire and can
slide freely on frictionless rods.
If a 60-lb force Q is applied to
collar B as shown, Determine
(a) the tension in the wire when
x = 9 in., (b) the corresponding
magnitude of the force P required
to maintain the equilibrium of
the system.
Problem 2.11
2. Set the resultant, or sum, of the forces exerted on the particle
equal to zero. You will obtain a vectorial equation consisting
of terms containing the unit vectors i, j, and k. Three scalar
equations result, which can be solved for the unknowns.

56. S F = 0: P i + Ny j + Nz k + TAB lAB = 0
Problem Solution x P
_ x i _ (20 in) j + z k AB
lAB = = AB
25 in A
Draw a free-body diagram of the particle. O
20 in Q z B x z
Free Body: Collar A
Ny j
Nz k
TAB lAB
P i
S F = 0: P i + Ny j + Nz k + TAB lAB = 0 A
Substitute for lAB and set coefficients
of i equal to zero:
P _
TAB x 25
= (1)

57. S F = 0: (60 lb) k + Nx i + Ny j _ TAB lAB = 0
Problem Solution
Ny j
Nx i
_ TAB lAB
Q = (60 lb) k
Free Body: Collar B
S F = 0: (60 lb) k + Nx i + Ny j _ TAB lAB = 0 B
Substitute for lAB and set coefficients
of k equal to zero:
60 _
TAB z 25
= (2)
(a) Since x = 9 in.: (9 in)2 + (20 in) 2 + z 2 = (25 in) z = 12 in
From eq. (2):
60 _
TAB (12) 25
= TAB = lb
(b) From eq. (1): P =
(125.0 lb)(9 in)
25 in
P = 45.0 lb