1. 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

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

12. V = P x Q V = PQ sin q Q x P = - (P x Q) V = P x Q
The vector product of two
vectors is defined as Q
V = P x Q q P
The vector product of P and
Q forms a vector which is
perpendicular to both P and Q, of magnitude
V = PQ sin q
This vector is directed in such a way that a person located at the
tip of V observes as counterclockwise the rotation through q
which brings vector P in line with vector Q. The three vectors P,
Q, and V - taken in that order - form a right-hand triad. It
follows that
Q x P = - (P x Q)

13. P = Px i + Py j + Pz k Q = Qx i + Qy j + Qz k
It follows from the definition of the vector
product of two vectors that the vector
products of unit vectors i, j, and k are k i
i x i = j x j = k x k = 0
i x j = k , j x k = i , k x i = j , i x k = - j , j x i = - k , k x j = - i
The rectangular components of the vector product V of two
vectors P and Q are determined as follows: Given
P = Px i + Py j + Pz k Q = Qx i + Qy j + Qz k
The determinant containing each component of P and Q is
expanded to define the vector V, as well as its scalar
components

14. P = Px i + Py j + Pz k Q = Qx i + Qy j + Qz k
V = P x Q =
= Vx i + Vy j + Vz k
where
Vx = Py Qz - Pz Qy
Vy = Pz Qx - Px Qz
Vz = Px Qy - Py Qx

15. MO = r x F MO = rF sin q = Fd Mo The moment of force F
about point O is defined
as the vector product
MO = r x F F
where r is the position
vector drawn from point O
to the point of application
of the force F. The angle
between the lines of action
of r and F is q. O r q d A
The magnitude of the moment of F about O can be expressed as
MO = rF sin q = Fd
where d is the perpendicular distance from O to the line of action
of F.

16. Mx = y Fz - z Fy My = zFx - x Fz Mz = x Fy - y Fx
Fy j
A (x , y, z )
y j
The rectangular
components of the
moment Mo of a force F are determined by
expanding the
determinant of r x F.
Fx i r
Fz k O
x i x
z k z i x Fx j y Fy k z Fz
Mo = r x F =
= Mx i + My j + Mzk
where
Mx = y Fz - z Fy My = zFx - x Fz
Mz = x Fy - y Fx

17. rA/B = xA/B i + yA/B j + zA/B k
Fy j
A (x A, yA, z A)
B (x B, yB, z B)
In the more general case of the moment about an arbitrary point B of a force F applied at A, we have
Fx i r
Fz k O x z i
xA/B Fx j
yA/B Fy k
zA/B Fz
MB = rA/B x F =
rA/B = xA/B i + yA/B j + zA/B k
where
xA/B = xA- xB yA/B = yA- yB zA/B = zA- zB
and

18. MB = (xA- xB )Fy + (yA- yB ) Fx
In the case of problems involving only two dimensions, the force F can be assumed to lie in the xy plane. Its moment about point B is perpendicular to that plane. It can be completely defined by the scalar
Fy j A
rA/B
Fx i
(yA - yB ) j B
(xA - xB ) i O x z
MB = MB k
MB = (xA- xB )Fy + (yA- yB ) Fx
The right-hand rule is useful for defining the direction of the
moment as either into or out of the plane (positive or negative
k direction).

19. P Q = PQ cos q P Q = Px Qx + Py Qy + Pz Qz The scalar product of two
vectors P and Q is denoted as P Q ,and is defined as Q q
P Q = PQ cos q
where q is the angle between
the two vectors P
The scalar product of P and Q is expressed in terms of the
rectangular components of the two vectors as
P Q = Px Qx + Py Qy + Pz Qz

20. POL = Px cos qx + Py cos qy + Pz cos qz
The projection of a vector
P on an axis OL can be obtained by forming the scalar product of P and the unit vector l along OL. qy A l P qx x O qz z
POL = P l
Using rectangular components,
POL = Px cos qx + Py cos qy + Pz cos qz

21. Sx Px Qx Sy Py Qy Sz Pz Qz S (P x Q ) =
The mixed triple product of three vectors S, P, and Q is Sx Px Qx Sy Py Qy Sz Pz Qz
S (P x Q ) =
The elements of the determinant are the rectangular components of the three vectors.

22. l x x Fx l y y Fy l z z Fz MOL = l MO = l (r x F) = y
The moment of a force F about an axis OL is the projection OC on OL of the moment MO of the force F. This can be written as a mixed triple product. MO F C l
A (x, y, z) r O x z
l x x Fx
l y y Fy
l z z Fz
MOL = l MO = l (r x F) =
lx, ly , lz = direction cosines of axis OL
x, y , z = components of r
Fx, Fy , Fz = components of F

23. M
- F d F
Two forces F and -F having the same magnitude, parallel lines of action, and opposite sense are said to form a couple.
The moment of a couple is independent of the point about which it is computed; it is a vector M perpendicular to the plane of the couple and equal in magnitude to the product Fd.

24. M y y
- F y
(M = Fd) My M d F O Mx O O x x x z z z Mz
Two couples having the same moment M are equivalent (they have the same effect on a given rigid body).

25. F F MO r A A O O
Any force F acting at a point A of a rigid body can be replaced
by a force-couple system at an arbitrary point O, consisting
of the force F applied at O and a couple of moment MO equal to
the moment about point O of the force F in its original position.
The force vector F and the couple vector MO are always
perpendicular to each other.

26. Any system of forces can be reduced to a force-couple
system at a given point O. First, each of the forces of the
system is replaced by an equivalent force-couple system at O.
Then all of the forces are added to obtain a resultant force R,
and all of couples are added to obtain a resultant couple vector
MO. In general, the resultant force R and the couple vector MO
will not be perpendicular to each other. R R

27. S F = S F’ S Mo = S Mo’ F3 F1 R A3 A1 r1 r3 r2 O A2 O F2 M
As far as rigid bodies are concerned, two systems of forces, F1, F2, F , and F’1, F’2, F’ , are equivalent if, and
only if,
S F = S F’
and
S Mo = S Mo’

28. If the resultant force R and the resultant couple vector MO are
perpendicular to each other, the force-couple system at O can be further reduced to a single resultant force. R A3 R r1 A1 r2 O A2 O R O F2 M
This is the case for systems consisting of either
(a) concurrent forces,
(b) coplanar forces, or
(c) parallel forces.
If the resultant force and couple are directed along the same
line, the force-couple system is termed a wrench.