Force Vectors

Principles of Engineering © 2012 Project Lead The Way, Inc. 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 arrow, such as 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) -y (down)

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

Opposite Side (opp)

 90° Adjacent Side (adj) Trigonometry Review Trigonometric Functions soh cah toa sin θ° = opp / hyp cos θ° = adj / hyp tan θ° = opp / adj

Opposite Side (opp)

 90° Adjacent Side (adj) 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

Opposite Side

Fy  90°

Adjacent Side Fx Trigonometry Application

sin θ° = Fy / F Fy= F sin θ°

cos θ° = Fx / F Fx= F cos θ°

tan θ° = Fy / Fx

Opposite Side

Fy  90°

Adjacent Side Fx

Fx and Fy are negative if left or down, respectively. Vector X and Y Components Vector A Magnitude = 75.0 lb Direction = 35.0°CCW from positive x-axis

+y Sense = right, up

A = 75.0 lb

opp = FAy

35.0° -x +x

adj = FAx -y Vector X and Y Components

Solve for FAx

adj FA FA cos θ = cos θ = x cos 35.0° = x hyp A 75.0 lb

FA = 75.0 lb cos 35.0° up +y x

FAx = 61.4 lb A = 75.0 lb

opp = FAy

35.0° -x +x

adj = FAx -y Vector X and Y Components

Solve for FAy opp F F sin θ = Ay Ay hyp sin θ = sin 35.0° = A 75.0 lb

FA = 75.0 lb sin 35.0° up +y y

FAy = 43.0 lb A = 75.0 lb

opp = FAy

35.0° -x +x

adj = FAx -y Vector X and Y Components – Your Turn Vector B Magnitude = 75.0 lb Direction = 35.0°CW from positive x-axis +y Sense = right, down

adj = FBx -x +x 35.0°

opp = FBy B = 75.0 lb -y Vector X and Y Components – Your Turn

Solve for FBx

adj FB FB cos θ = cos θ = x cos 35.0° = x hyp B 75.0 lb +y FBx = 75.0 lb cos 35.0° right

adj = F F = 61.4 lb Bx Bx -x +x 35.0°

opp = FBy B = 75.0 lb -y Vector X and Y Components – Your Turn

Solve for FBy opp −FB −FBy sin θ = sin θ = y sin 35.0° = hyp B 75.0 lb +y FBy = 75.0 lb sin 35.0° down

adj = F F = −43.0 lb Bx By -x +x 35.0°

opp = FBy B = 75.0 lb -y Resultant Force Two people are pulling a boat to shore. They are pulling with the same magnitude.

A = 75.0 lb

 =35.0   =35.0 

B = 75.0 lb Resultant Force List the forces according to sense. F = 43.0 lb A = 75.0 lb Ay Label right and up forces as positive, and label left and down forces as negative.

F = 61.4 lb  =35  Ax Fx FA = +61.4 lb  =35  x FB = 61.4 lb x FBx = +61.4 lb Fy

FAy = +43.0 lb

FBy = −43.0 lb F = −43.0 lb By B = 75.0 lb Resultant Force Sum (S) the forces Fx ෍ Fx = FA + FB FAx = +61.4 lb x x

FBx = +61.4 lb ෍ Fx = 61.436 lb + 61.436 lb Fy F = +43.0 lb Ay ෍ Fx = 122.9 lb (right)

FBy = −43.0 lb

෍ Fy = FAy + FBy

෍ Fy = 43.018 lb + (− 43.018 lb) = 0

Magnitude is 122.9 lb Direction is 0° from the positive x-axis Sense is right Resultant Force

Draw the resultant force (FR) Magnitude is 123 lb Direction is 0° from the positive x-axis

FAy = 43.0 lb Sense is right

FAx = 61.4 lb

FR = 122.9 lb

FBx = 61.4 lb

FBy = −43.0 lb Resultant Force

C = 300 lb

Determine the sense, magnitude, and direction for the  =60.  resultant force.

 =30. 

D = 400 lb Resultant Force Find the x and y components of vector C.

FCx = 300 lb cos 60° right

C = 300 lb FCx = 150 lb

FCy

FCy = 300 lb sin 60° up 60. FCy = 260 lb FCx Resultant Force Find the x and y components of vector D.

FD = 400 lb cos 30° right FDx x F = 350 lb 30. Dx

FDy D = 400 lb FDy = 400 lb sin 30° down

FDy = −200 lb Resultant Force List the forces according to sense. Label right and up forces F = 259.8 lb Cy C = 300 lb as positive, and label left and down forces as negative.

 =60  Fx FCx = 150.0 lb FCx = +150.0 lb F = 346.4 lb  =30  Dx FDx = +346.4 lb Fy FD = −200.0 lb D = 400 lb y FCy = +259.8 lb

FDy = −200.0 lb Resultant Force Sum (S) the forces

Fx

FCx = +150.0 lb ෍ F = F + F FDx = +346.4 lb x Cx Dx

෍ Fx = 150.0 lb + 346.4 lb = 496.4 lb (right) Fy ෍ F = F + F FCy = +259.8 lb y Cy Dy

FDy = −200.0 lb ෍ Fy = 259.8 lb + −200.0 lb = 59.8 lb (up)

Sense is right and up. Resultant Force Draw the x and y components of the resultant force.

σ Fx = 496.4 lb (right) σ Fy = 59.8 lb up

Two equivalent ways to draw the X and Y components

FR 496.4 lb 59.8 lb 59.8 lb FR 496.4 lb Resultant Force Solve for magnitude.

a2 + b2 = c2 FR 59.8 lb 2 2 2 (59.8 lb) +(496.4 lb) = FR

496.4 lb 2 2 59.8 lb + 496.4 lb = FR

FR = 500.0 lb

Magnitude is 500. lb Resultant Force Solve for direction. opp tan= adj

59.8 lb 500. lb tan = 59.8 lb 496.4 lb  −1 59.8 =tan  496.4 lb 496.4

 =7 

Direction is 7° counterclockwise from the positive x-axis. Resultant Force

Draw the resultant force (FR) Magnitude = 500. lb Direction = 7° CCW from positive x-axis Sense = right and up

500. lb

7°