2.4 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
ˆ
2 vectors, i 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 = 4 iˆ
-a vector 5 units long in the negative y-direction V = 5(- ĵ ) = -5 ĵ
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Is P=.06 iˆ + 0.8 ĵ A unit vector? Yes |P| = (0.62  0.82 ) 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 ĵ
What is the resultant, F?
 

F  Fx  Fy  Fx iˆ  Fy ˆj
 

Remember: Fx , Fy and Fx are vectors
Fx , Fy and F are the magnitude of vectors, which are scalars

What does F look like?
y
What is the magnitude of F?

Fy
F  / FX / 2  / F / 2

F
What is ?
 / F
 / FX
  tan 1 

ĵ
iˆ
Lecture03.doc

Fx
/

/ 
x
1



Given F what is Fx and Fy

Fx  / F / cos

Fy  / F / sin 
iˆ
ˆj
/ FX /  / F / cos 
What is / FX / and / F / ?
/ F /  / F / sin 
Coplanar force resultants

 
   
Given 3 forces, P,Q, and S their resultant is R  P  Q  S
y

P

S
x

Q

Q  / Qx /( iˆ)  / Q y /(  ˆj )

Q   / Qx / iˆ  / Q y / ˆj

P  / Px / iˆ  / Py / ˆj

S  / S x / iˆ  / S y / ˆj
   
R  P  Q  S  / Px / 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  [/ Py /  ( / Q y /)  / S y /] ˆj
/ RX /
/ R /

R  / Rx / iˆ  / R y / ˆj
So:
/ R x /  Fx
/ R y /  Fy
Sign convention! You have either + or - components.
From now on, drop magnitude (/ /) sign for all scalars. All vectors will have arrows.

P Vector
P  Magnitude of P.
Once you have the resultant components, the resultant vector can be sketched and found
using:
Homework: Problems: 2-31, 2-38, 2-45, 2-46, 2-55, 2-56, 2-57
Lecture03.doc
2
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

x

F4 x

F4
40o
30o

F6 x

F4 y
y

F6

F6 y
x
 

R  F6  F4

F6

F6

F6

F6

F6


 F6 x  F6 y

F4

F4

F4

F4

F4


 F4 x  F4 y
 F6 x iˆ  F6 y ˆj
 F6 cos 30  iˆ  F6 sin 30  ˆj
 6000 cos 30  iˆ  6000 sin 30  ˆj
 5200 iˆ  3000 ˆj
 F4 x iˆ  F4 y ˆj
 F4 sin 40  ( iˆ)  F4 cos 40  ˆj
 4000 sin 40  ( iˆ)  4000 cos 40  ˆj
 2570 iˆ  3060 ˆj

R  5200 iˆ  3000

R  2630 iˆ  6060
Lecture03.doc
ˆj  2570 iˆ  3060 ˆj
ˆj N
3

2). Given: Previous problem. Find R using scalar notation.
4kN
y
40

6kN
30

x
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
Lecture03.doc
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