# Essay about Trigonometry and Lb

Submitted By Iamareallygoodstuden
Words: 1059
Pages: 5

Force Vectors

Principles Of Engineering

Vectors
Vector Quantities
Have both a magnitude and direction
Examples: Position, force, moment

Vector Notation
Vectors are given a variable, such as A or B
Handwritten notation usually includes an r ur arrow, such as u

A or B

Illustrating Vectors
Vectors are represented by arrows
Include magnitude, direction, and sense
Magnitude: The length of the line segment
Magnitude = 3

30°

+X

Illustrating Vectors
Vectors are represented by arrows
Include magnitude, direction, and sense
Direction: The angle between a reference axis and the arrow’s line of action
Direction = 30° counterclockwise from the positive x-axis

30°

+x

Illustrating Vectors
Vectors are represented by arrows
Include magnitude, direction, and sense
Sense: Indicated by the direction of the tip of the arrow
Sense = Upward and to the right

30°

+x

Sense

+y (up)

+y (up)

-x (left)

+x (right)
(0,0)

-y (down)

-x (left)

-y (down)

+x (right)

Trigonometry Review
Right Triangle
A triangle with a 90° angle
Sum of all interior angles = 180°
Pythagorean Theorem: c2 = a2 + b2

H

us n te o p y p) y h e( 

Opposite Side
(opp)
90°

Trigonometry Review
Trigonometric Functions soh cah toa sin θ° = opp / hyp cos θ° = adj / hyp tan θ° = opp / adj

H

us n te o p y p) y h e( 

Opposite Side
(opp)
90°

Trigonometry Application
The hypotenuse is the Magnitude of the
Force, F
In the figure here,
The adjacent side is the x-component, Fx
The opposite side is the y-component, Fy

H

se u ten o yp

F



Opposite Side

Fy
90°

Trigonometry Application sin θ° = Fy / F

Fy= F sin θ°

cos θ° = Fx / F

Fx= F cos θ°

tan θ° = Fy / Fx
H

ten o yp

use

F



Opposite Side

Fy
90°

Fx and Fy are negative if left or down, respectively.

VectorurX and Y Components
Vector A

Magnitude = 75.0 lb
Direction = 35.0°CCW from positive x-axis
+y

Sense = right, up ur A  75.0 lb opp = FAy
35.0°

-x
-y

+x

Vector X and Y Components
Solve for FAx
FAx
cos35.0 
75.0 lb

adj cos   hyp FAx  75.0 lb cos 35.0 up

+y

ur
A  75.0 lb

FAx  61.4 lb opp = FAy

35.0°

-x

-y

+x

Vector X and Y Components
Solve for FAy
FAy

opp sin   hyp F sin   ur
A

+y

FAY  75.0 lb sin 35.0 up

Ay

sin 35.0 

FAy  43.0 lb

ur
A  75.0 lb opp = FAy
35.0°

-x

-y

75.0 lb

+x

Vector X and Y Components – Your Turn ur Vector B
Magnitude =
Direction =
+y

-x

75.0 lb
35.0°CW from positive x-axis

right, down
+x

35.0° opp = FBy

-y

ur
B  75.0 lb

Vector X and Y Components – Your Turn
Solve for FBx adj cos   hyp FBx cos35.0 
75.0 lb

+y

FBx  75.0 lb cos 35.0 right adj = FBx

-x

+x

35.0° opp = FBy

-y

ur
B  75.0 lb

FBx  61.4 lb

Vector X and Y Components – Your Turn
Solve for FBY opp sin   hyp sin 35.0 

+x

35.0° opp = FBy

-y

75.0 lb

FBy  75.0 lb sin 35.0 down

+y

-x

 FBy

ur
B  75.0 lb

FBy  43.0 lb

Resultant Force
Two people are pulling a boat to shore.
They are pulling with the same magnitude. ur A  75.0 lb

  35.0
  35.0

ur
B  75.0 lb

Resultant Force ur A  75 lb

FAy = 43.0 lb

  35

FAx = 61.4 lb

  35

FBx = 61.4 lb

List the forces according to sense. Label right and up forces as positive, and label left and down forces as negative. Fx
FAx = +61.4 lb
FBx = +61.4 lb
Fy

FBy= -43.0 lb

ur
B  75 lb

FAy = +43.0 lb
FBy = -43.0 lb

Resultant Force
Sum () the forces

Fx
FAx = +61.4 lb

Fx = FAx + FBx

FBx = +61.4 lb

Fx = 61.436 lb + 61.436 lb
Fx = 122.9 lb (right)

Fy
FAy = +43.0 lb
FBy = -43.0 lb

Fy = FAy + FBy

Fy = 43.018 lb + (-43.018 lb) = 0
Magnitude is 122.9 lb
Direction is…