1. Chapter 2 STATICS OF PARTICLES
Forces are vector quantities; they add according to the
parallelogram law. The magnitude and direction of the
resultant R of two forces P and Q can be determined either
graphically or by trigonometry. R P A Q
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2. Any given force acting on a particle can be resolved into two or more components, i.e.., it can be replaced by two or more forces which have the same effect on the particle.
A force F can be resolved
into two components P
and Q by drawing a
parallelogram which has
F for its diagonal; the
components P and Q
are then represented by
the two adjacent sides
of the parallelogram
and can be determined
either graphically or by
trigonometry. Q F A P

3. F = Fx i + Fy j Fx = F cos q Fy = F sin q Fy tan q = Fx F = Fx + Fy 2
A force F is said to have been resolved into two rectangular
components if its components are directed along the coordinate
axes. Introducing the unit vectors i and j along the x and y axes,
F = Fx i + Fy j y
Fx = F cos q Fy = F sin q
tan q = Fy Fx
Fy = Fy j F j
F = Fx + Fy 2 q x i
Fx = Fx i

4. Rx = S Rx Ry = S Ry Ry tan q = R = Rx + Ry 2 Rx
When three or more coplanar forces act on a particle, the
rectangular components of their resultant R can be obtained
by adding algebraically the corresponding components of the
given forces.
Rx = S Rx Ry = S Ry
The magnitude and direction of R can be determined from
tan q = Ry Rx
R = Rx + Ry 2

5. Fx = F cos qx Fy = F cos qy Fz = F cos qz Fy Fy qy F F qx Fx Fx Fz FzB B Fy Fy qy A F A F D D O O qx Fx Fx x x Fz Fz E E C C z z y
A force F in three-dimensional space can be resolved into components B Fy F
Fx = F cos qx Fy = F cos qy A D O
Fz = F cos qz Fx x qz E Fz C z

6. F = F (cosqx i + cosqy j + cosqz k )
l (Magnitude = 1)
The cosines of
qx , qy , and qz
are known as the
direction cosines of the force F. Using the unit vectors i , j, and k, we write
Fy j
F = F l
cos qy j
Fx i
cos qz k x
Fz k
F = Fx i + Fy j + Fz k
cos qx i z or
F = F (cosqx i + cosqy j + cosqz k )

7. l = cosqx i + cosqy j cos2qx + cos2qy F = Fx + Fy + Fz 2 cosqx = Fx F
l (Magnitude = 1)
cos qy j
l = cosqx i + cosqy j
+ cosqz k
Fy j
cos qz k
F = F l
Since the magnitude
of l is unity, we have
Fx i x
cos2qx + cos2qy
+ cos2qz = 1
Fz k z
cos qx i
In addition,
F = Fx + Fy + Fz 2
cosqx = Fx F
cosqy = Fy F
cosqz = Fz F

8. MN = dx i + dy j + dz k MN 1 l = = ( dx i + dy j + dz k ) d y
A force vector F
in three-dimensions
is defined by its
magnitude F and
two points M and
N along its line of
action. The vector
MN joining points
M and N is
N (x2, y2, z2) F
dy = y2 - y1 l
dz = z2 - z1
< 0
dx = x2 - x1
M (x1, y1, z1) x z
MN = dx i + dy j + dz k
The unit vector l along the line of action of the force is
l = = ( dx i + dy j + dz k ) MN 1 d

9. F = F l = ( dx i + dy j + dz k ) F d Fdx d Fdy d Fdz d Fx = Fy = Fz =
d = dx + dy + dz 2 2 2
N (x2, y2, z2)
dy = y2 - y1
A force F is
defined as the
product of F and
l. Therefore,
dz = z2 - z1
< 0
dx = x2 - x1
M (x1, y1, z1) x z
F = F l = ( dx i + dy j + dz k ) F d
From this it follows that
Fdx d
Fdy d
Fdz d
Fx =
Fy =
Fz =

10. When two or more forces act on a particle in three-dimensions, the rectangular components of their resultant R is obtained by adding the corresponding components of the given forces.
Rx = S Fx
Ry = S Fy
Rz = S Fz
The particle is in equilibrium when the resultant of all forces acting on it is zero.

11. S Fx = 0 S Fy = 0 S Fz = 0 S Fx = 0 S Fy = 0
To solve a problem involving a particle in equilibrium, draw a free-body diagram showing all the forces acting on the particle. The conditions which must be satisfied for particle equilibrium are
S Fx = S Fy = S Fz = 0
In two-dimensions , only two of these equations are needed
S Fx = S Fy = 0