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Force Vectors. Vectors Have both a magnitude and direction Examples: Position, force, moment Vector Quantities Vector Notation Handwritten notation usually.

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Presentation on theme: "Force Vectors. Vectors Have both a magnitude and direction Examples: Position, force, moment Vector Quantities Vector Notation Handwritten notation usually."— Presentation transcript:

1 Force Vectors

2 Vectors Have both a magnitude and direction Examples: Position, force, moment Vector Quantities Vector Notation Handwritten notation usually includes an arrow, such as Vectors are given a variable, such as A or B

3 Illustrating Vectors Vectors are represented by arrows Include magnitude, direction, and sense 30° +X Magnitude: The length of the line segment Magnitude = 3

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

5 Illustrating Vectors Vectors are represented by arrows Include magnitude, direction, and sense 30° +X Sense: Indicated by the direction of the tip of the arrow Sense = Upward and to the right

6 Sense +x (right) -x (left) +y (up) -y (down) (0,0)

7 Trigonometry Review Hypotenuse (hyp) 90° Opposite Side (opp) Adjacent Side (adj) Right Triangle A triangle with a 90° angle Sum of all interior angles = 180° Pythagorean Theorem: A 2 + B 2 = C 2

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

9 Trigonometry Application The hypotenuse is the Force, F The adjacent side is the x-component, F x The opposite side is the y-component, F y Hypotenuse F 90° Opposite Side F y Adjacent Side F x

10 Trigonometry Application sin θ° = F y / F cos θ° = F x / F tan θ° = F y / F x Hypotenuse F 90° Opposite Side F y Adjacent Side F x F y = F sin θ° F x = F cos θ°

11 Vector X and Y Components 35.0° Vector Magnitude = 75.0 lb Direction = 35.0° from the horizontal Sense = right, up +X +Y -Y -X opp = F AY adj = F AX

12 Vector X and Y Components Solve for F AX 35.0° +X +Y -Y -X opp = F AY adj = F AX

13 Vector X and Y Components Solve for F AY 35.0° +X +Y -Y -X opp = F AY adj = F AX

14 Vector X and Y Components - Your Turn 35.0° Vector Magnitude = Direction = Sense = +X +Y -Y -X 75.0 lb 35.0° from the horizontal right, down opp = F BY adj = F BX

15 Solve for F BX Vector X and Y Components – Your Turn 35.0° +X +Y -Y -X opp = F BY adj = F BX

16 Vector X and Y Components – Your Turn Solve for F BY 35.0° +X +Y -Y -X opp = F BY adj = F BX

17 Resultant Force Two people are pulling a boat to shore. They are pulling with the same magnitude.

18 Resultant Force F AX = 61.4 lb F BX = 61.4 lb F AY = 43.0 lb F BY = 43.0 lb F X F AX = +61.4 lb F BX = +61.4 lb F Y F AY = +43.0 lb F BY = -43.0 lb List the forces according to sense. Label right and up forces as positive, and label left and down forces as negative.

19 Resultant Force Sum (S) the forces F X F AX = +61.4 lb F BX = +61.4 lb F Y F AY = +43.0 lb F BY = -43.0 lb SF X = F AX + F BX SF X = 61.436 lb + 61.436 lb SF X = 122.9 lb (right) SF Y = F AY + F BY SF Y = 43.018 lb + (-43.018 lb) = 0 Magnitude is 122.9 lb Direction is 0° from the x axis Sense is right

20 Resultant Force Draw the resultant force (F R ) Magnitude is 123 lb Direction is 0° from the x axis Sense is right F R = 122.9 lb F AX = 61.4 lb F BX = 61.4 lb F AY = 43.0 lb F BY = 43.0 lb

21 Resultant Force Determine the sense, magnitude, and direction for the resultant force.

22 Resultant Force F CY F CX Find the X and Y components of vector C. F CX = 300. lb cos60.° F CX = 150 lb F CY = 300. lb sin60.° F CY = 260 lb

23 Resultant Force F DY F DX Find the X and Y components of vector D. F DX = 400. lb cos30.° F DX = 350 lb F DY = 400. lb sin30.° F DY = 2.0x10 2 lb

24 Resultant Force List the forces according to sense. Label right and up forces as positive, and label left and down forces as negative. FXFX F CX = +150.0 lb F DX = +346.4 lb FYFY F CY = +259.8 lb F DY = -200.0 lb F CX = 150.0 lb F DX = 346.4 lb F CY = 259.8 lb F DY = 200.0 lb

25 Resultant Force Sum (S) the forces F X F CX = +150.0 lb F DX = +346.4 lb F Y F CY = +259.8 lb F DY = -200.0 lb SF X = F CX + F DX SF X = 150.0 lb + 346.4 lb = 496.4 lb (right) SF Y = F CY + F DY SF Y = 259.8 lb + (-200.0 lb) = 59.8 lb (up) Sense is right and up.

26 Resultant Force Draw the X and Y components of the resultant force. SF X = 496.4 lb (right) SF Y = 59.8 (up) 496.4 lb 59.8 lb 496.4 lb 59.8 lb FRFR FRFR Two ways to draw the X and Y components

27 Resultant Force Solve for magnitude. 496.4 lb 59.8 lb FRFR a 2 + b 2 = c 2 (59.8 lb) 2 +(496.4 lb) 2 =F R 2 F R = 500 lb or 5.0x10 2 lb Magnitude is 5.0x10 2 lb

28 Resultant Force Solve for direction. 496.4 lb 59.8 lb 500 lb Direction is 7° counterclockwise from the positive X axis.

29 Resultant Force Draw the resultant force (F R ) Magnitude is 5.0x10 2 lb Direction is 7° counterclockwise from the positive x axis Sense is right and up 7° 5.0x10 2 lb


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