Chapter 2 – Statics of Particles
Scalars and Vectors
Vectors – A mathematical quantity possessing magnitude and direction.
Scalar – A mathematical quantity possessing magnitude only.
Name some vectors: forces, velocity, displacement
Name some scalars: Area, volume, mass energy
Representation of vector
Bold R – Word Processors Book uses this.
Arrow R – Long Hand, Word Processors
Underline R – Long Hand, Typewriter, Word Processors
Magnitude of a Vector
R Book uses italics for all scalars
Types of Vectors
1). Fixed (or bound) vectors – a vector for which a unique point of
application is specified and thus cannot be moved without
modifying
the conditions of the problem.
2). Free vector – a vector whose action is not confined to or associated
with a unique line in space. (couple)
3). Sliding vector – a vector for which a unique line in space (line of
action) must be maintained.
For 2 vectors to be equal they must have the same:
P P
1). Magnitude
2). Direction
They do not need to have the same point of application.
A negative vector of a given vector has same magnitude but opposite direction.
P -P
P and –P are equal and opposite P + (-P) = 0
Vector Operations
Product of a scalar and a vector
P + P + P = 4P (the number 4 is a scalar)
This is a vector in the same direction as P but 4 times as long.
(+n)P = vector same direction as P, n times as long
(-n)P = vector opposite direction as P, n times as long
Vector Addition
The sum of 2 vectors can be obtained by attaching the 2 vectors to the
same point and constructing a parallelogram – Parallelogram law.
R
Q
R=P+Q
P
R = resultant vector
Note: The magnitude of P + Q is not usually equal to P Q .
Addition of vectors is communative: P + Q = Q + P
Triangle Rule
P
Q R R Q
Let’s add 3 vectors!
Parallelogram Law R1 R1
Q Q
R
Q
P P P
S S S
R1 = Q + P R = R1 + S = Q + P + S
Triangle Rule
Q Q P P S
R Q R
1 1
P R
S S
R = R1 + S = Q + P + S
R1 = Q + P
Polygon Rule – Successive applications of triangle rule.
Q
Q P
S
P R
S
Note: P +Q + S = (P + Q) + S = P + (Q + S) vector addition is associative
Vector Subtraction – the addition of the corresponding negative vector
P
P – Q = P + (-Q)
-Q
R
Resolution of vector into components
A single vector can be represented by 2 or more vectors. These vectors
are components of the original vector. Finding these is called
resolving the vector into its components.
There is an infinite number of ways to resolve one vector.
P 1 P 2
P P
2 1 etc.
P P
2 cases of particular interest are:
1). One of the 2 components is known. Easy (see above)
2). The line of action of both components is know.
When would #2 happen? When you are given a coordinate system!
y
P
x
What are the x and y components of P if P = 1000 lbs, = 30o
Px = P cos 30o = 866 lbs Py = P sin 30o = 500 lbs
Note: Given Px and Py, what is P?
P2 = Px2 + Py2=8662 + 5002 = 1000 lbs
1). Given: The fixed structure shown below.
B
P
P = 500 N
T = 200 N T
5m
75 o
A
C D
3m
Find: Combine P and T into a single force R
P = 500
T = 200 R
BD 5 sin 75
tan
AD 3 5 cos 75
48.4
Law of cosines:
c 2 a 2 b 2 2ab cos(c )
R 2 200 2 500 2 2( 200)(500) cos(48.4 )
R 396.5 N
Law of sines:
200 396.5
22.2
sin sin 48.4
R 396.5 N 22.2
2.) Given: A barge is pulled by 2 tugboats. The resultant of the forces exerted by the
tugboats is a 5000 pound force directed along the center axis of the barge.
A
1
30
B
C
2
Find: a). tension in each rope if =45 degrees
b). value of such that the tension in rope 2 is minimum.
5000
a).
45 30
T2 T1
T1 T2 5000
sin 45 sin 30 sin 105
T1 3660 lbs
T2 2590 lbs
b). 5000
30o dir of T1
Dir of T2?
Dir of T2?
Dir of T2? 60
T1 5000 cos 30 4330 lbs
T2 5000 sin 30 2500 lbs
3). Given: The vertical force F of 350 lbs acts downward at A on the two-membered
frame.
B
o
45
A
o
30
F C
Find: The magnitudes of the two components of F directed along AB and AC.
45
75
60
350 lbs
FAB FAC 350
sin 60 sin 45 sin 75
F AB 314 lbs
F AC 256 lbs
Addition of a System of Coplanar Forces
Vector Notation
In many problems it will be necessary to resolve a force into 2
components that are perpendicular to each other.
y
ĵ
iˆ x
O
2 vectors, iˆ and ĵ that have the direction shown and magnitude 1 - unit
vectors.
The iˆ and ĵ provide direction!
-suppose I want a vector 4 units long in the x-direction V 4iˆ
-a vector 5 units long in the negative y-direction
V 5 ˆj 5 ˆj
Is P 0.6iˆ 0.8 ˆj a unit vector? Yes P 0.6 2 0.8 2 1
Suppose we have a force, with magnitude Fx , that lies on the x-axis.
Fx Fx iˆ and one on the y-axis Fy Fy iˆ .
What is the
resultant,
F ?
F Fx Fy Fx iˆ Fy ˆj
Remember: Fx , F y and Fx are vectors
Fx , Fy and F are magnitude of vectors, which are
scalars
What does F look like?
y
What is the magnitude of F
?
2 2
F Fx Fy
Fy F What is ?
Fy
tan 1
Fx
ĵ
iˆ Fx x
Given F what is Fx and Fy
Fx F cos iˆ
Fy F sin ˆj
FX F cos
What is F X and F ?
F F sin
Coplanar force resultants
Given 3 forces, P, Q, and S their resultant is R PQS
y
P
S
x
Q
Q Q x ( iˆ) Q y ( ˆj )
P Px iˆ Py ˆj S S x iˆ S y ˆj
Q Q x iˆ Q y ˆj
R P Q S P x iˆ Py ˆj Q x iˆ Q y ˆj S x iˆ S y ˆj
R Px Q x S x iˆ Py Q y S y ˆj
R Px Q x S x iˆ P Q
y y S ˆj
y
RX R
R R x iˆ R y ˆj
So:
R x Fx
Sign convention! You have either + or - components.
R y F y
From now on, drop magnitude (| |) sign for all scalars. All vectors will
have
arrows.
P Vector P Magnitude of P.
Once you have the components, the resultant vector can be sketched and
found using:
2
2 2 Ry
R R x R y tan
2
Rx
1). Given: Replace the 6 kN and 4 kN forces by a single force, expressed in vector
notation.
4 kN y
40
6 kN
30
F4 x y
R F6 F4 F4
o
40 F4 y
F6
30 o
F6 y
F6 x
x
F6 F6 x F6 y
F6 F6 x iˆ F6 y ˆj
F6 F6 cos 30 iˆ F6 sin 30 ˆj
F6 6000 cos 30 iˆ 6000 sin 30 ˆj
F6 5200 iˆ 3000 ˆj
F4 F4 x F4 y
F4 F4 x iˆ F4 y ˆj
F4 F4 sin 40 ( iˆ) F4 cos 40 ˆj
F4 4000 sin 40 ( iˆ) 4000 cos 40 ˆj
F4 2570 iˆ 3060 ˆj
R 5200 iˆ 3000 ˆj 2570 iˆ 3060 ˆj
R 2630 iˆ 6060 ˆj N
2). Given: Previous problem. Find R using scalar notation.
4kN
y
40
6kN
30
R x Fx R y Fy
R x F6 x F4 x R y F6 y F4 y
R x 6 cos 30 4 sin 40 R y 6 sin 30 4 cos 40
R x 2.63 kN R y 6.06 kN
R R x iˆ R y ˆj
R 2.63 iˆ 6.06 ˆj kN
