activity5 1a x y components vectors anskey

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POE – Unit 5 – Activity 5.1a – Working with X and Y Components Vectors Answer Key –

Activity 5.1a - Working with X and Y Components of Vectors Answer Key

Purpose In order to be effective in design work it is necessary to be able to mathematically discover answers to questions. This exercise is designed to help you discover forces and what effect the angle that force is exerted from has on your design.

Equipment Ruler Calculator Pencil and paper

Procedure 1. Find the x and y components of vector V.

Solution hint: Think of the vector as pointing southeast. You will need to split it up into its y component (pointing south and its x component (pointing east).

1. ANSWER: VX = x-component of V VY = y-component of V To find VX, notice there is a right triangle. The value of the hypotenuse is 5 and the opposite angle is 30. Use sine:



sin = hypotenuse

opposite

sin 30 = 5

Vx

VX = 5 * sin 30 = 2.5 According to the coordinate system drawn above, “east” is positive and “west” is negative. Since the x-component of V is pointing “east”:

To find VY, notice there is a right triangle. The value of the hypotenuse is 5 and the adjacent angle is 30. Use cosine:

cos = hypotenuse

adjacent

cos 30 = 5

Vy

VY = 5 * cos 30 = 4.3

According to the coordinate system drawn above, “north” is positive and “south” is negative. Since the y-component of V is pointing “south”:

2. Find the x and y components of vector W. Can you predict the solution based on Problem 1?

VX = + 2.5 N

VY = - 4.3 N



Solution hint: Think of the vector as pointing southwest. You need to split it up into its y component (pointing south, and its x component (pointing west).

2. ANSWER:

WX = x-component of W WY = y-component of W

To find WX, notice there is a right triangle. The value of the hypotenuse is 5 and the opposite angle is 30. Use sine:

sin = hypotenuse

opposite

sin 30 = 5

Wx

WX = 5 * sin 30 = 2.5

According to the coordinate system drawn above, “east” is positive and “west” is negative. Since the x-component of W is pointing “west”:

WX = - 2.5 N



To find WY, notice there is a right triangle. The value of the hypotenuse is 5 and the adjacent angle is 30. Use cosine:

cos = hypotenuse

adjacent

cos 30 = 5

Wy

WY = 5 * cos 30 = 4.3

According to the coordinate system drawn above, “north” is positive and “south” is negative. Since the y-component of W is pointing “south”:

3. This picture is hung from a nail with wire. The nail supports two forces V = 5N and W = 5N. If the

resultant of these forces acts vertically downwards, find its X and Y components (Using your solutions from Problems 1 and 2).

WY = - 4.3 N



3. ANSWER:

Recall: VX = + 2.5 N VY = - 4.3 N WX = - 2.5 N WY = - 4.3 N

To find the x-component of the resultant force, add the x-components of the vectors V and W:

FX = VX + WX

FX = ( 2.5 N ) + ( - 2.5 N )

To find the y-component of the resultant force, add the y-components of the vectors V and W:

FY = VY + WY

FY = ( - 4.3 N ) + ( - 4.3 N )

*The forces in the x-direction cancel each other out. The resultant force is only the 8.6 N pulling downward.

4. Find the x and y components of vector A.

FX = 0

FY = - 8.6 N



Solution hint: Think of the vector as pointing northeast. You need to split it up into the y component (pointing north, and the x component pointing east).

4. ANSWER:

AX = x-component of A AY = y-component of A

To find AX, notice there is a right triangle. The value of the hypotenuse is 50 and adjacent angle is 20. Use cosine:

cos = hypotenuse

adjacent

cos 20 = 50

A x

AX = 50 * cos 20 = 47

AX = +47N



To find AY, notice there is a right triangle. The value of the hypotenuse is 50 and opposite angle is 20. Use sine:

To find AY, notice there is a right triangle. The value of the hypotenuse is 50 and opposite angle is 20. Use sine:

sin = hypotenuse

opposite

sin 20 = 50

A y

AY = 50 * sin 20 = 17

5. Find the x and y components of vector B.

Solution: Think of the vector as pointing northeast. You need to split it up into the y component (pointing north, and the x component pointing east).

Ay = +17N



5. ANSWER:

BX = x-component of B BY = y-component of B

To find BX, notice there is a right triangle. The value of the hypotenuse is 100 and opposite angle is 15. Use sine:

sin = hypotenuse

opposite

sin 15 = 100

Bx

BX = 100 * sin 15 = 26

To find BY, notice there is a right triangle. The value of the hypotenuse is 100 and adjacent angle is 15. Use cosine:

cos = hypotenuse

adjacent

cos 15 = 100

BY

BY = 100 * cos 15 = 97

BY = +97N

Bx = +26N



6. Two ropes are attached to the screw eye hook in this picture. Robe B is being pulled with a force of 100N at an angle of 15 degrees to the y-axis. Rope A is being pulled with a force of 50 N at an angle of 20 degrees to the x-axis. Use your solutions from Problems 4 and 5 to find x and y components of the resultant force, F:

6. ANSWER:

Recall: AX = 47 N AY = 17 N BX = 26 N BY = 97 N

To find the x-component of the resultant force, add the x-components of the vectors A and B: FX = AX + BX FX = 47N + 26N

To find the y-component of the resultant force, add the y-components of the vectors A and B:

FY = AY + BY

FY = 17N + 97N

FX = 73 N

F = resultant



Conclusion 1. When you are opening a classroom door what are the forces acting on it and how are they balanced

out so the door does not slam shut when released?

2. You have been asked to hang a mirror ball in the center of the gym, ten feet from the ceiling. There are beams that run across the gym but they are not centered. One beam is 10 feet off center in one direction and the other is 7 feet from the center. The mirror ball weighs 67 pounds. How long should each cable be and how much force should each one be able to hold?

Conclusion 2 - ANSWERS:

Diagram of Apparatus:

2.1 How long should each cable be?

Length AB:

22 10 10 AB 200

Length CB:

4114 .

FY = 114 N



22 10 7 CB 149

2.2 How much force should each cable be able to hold?

Solution Option #1: Find FAB and FAC by solving simultaneous equations.

Free Body Diagram of Ball:

Find the value of A and C

tan = adjacent

opposite tan =

adjacent

opposite

tan A = 10

10 = 1 tan C =

7

10 = 1.43

A = 45 C = 55

Find the x & y components of force vectors FAB and FBC

FABy = FAB sin A FBCy = FBC sin C

FABx = FAB cos A FBCx = FBC cos C

1212 .



The sum of the forces in the x-direction is equal to zero

Fx = 0

- FABx + FBCx = 0

FBCx = FABx

FBC cos C = FAB cos A

Equation – 1

The sum of the forces in the y-direction is equal to zero

Fy = 0

FABy + FBCy – W = 0

FABy + FBCy = W

Equation – 2

Substitute Equation–1 into Equation–2

FAB sin A + FAB C

A

cos

cos· sin C = W

FAB (sin A + cos A tan C) = W

Substitute results into Equation-1

FBC = CAAC

A

tancos sincos

cos W

FAB = CAA tancos sin

W

FBC = FAB C

A

cos

cos

FAB sin A + FBC sin C = W



For: W = 67 A = 45 C = 55

FAB = 55tan45cos54 sin

67

FBC = 39.0

55cos

54cos

Solution Option #2: Find FAB and FAC by calculating the reactive forces at A and C using sum of moments formulas.

A moment of a force is a measure of its tendency to cause a body to rotate about a point or axis. Using an equation for the sum of the moments about points A and C, the reactive forces at points A and C can be calculated.

Draw a Free Body Diagram of the whole system (cables & ball).

FAB = 39.0 lbs. FBC = 48.1 lbs.



The forces FAB and FBC can be broken into their X & Y components:

Redraw the Free Body Diagram:

Sum the moments about point A to calculate the reactive force at point C (FBCy). Note: FABx, FABy and FBCx pass through point A so they do not contribute to a moment.

0AM

-W(10) + FBCy(17) = 0 -67(10) + FBCy (17) = 0 -670 + FBCy (17) = 0 FBCy (17) = 670

Sum the moments about point C to calculate the reactive force at point A (FABy). Note: FABx, FABy and FBCx pass through point C so they do not contribute to a moment.

0CM

-W(7) + FABy(17) = 0 -67(7) + FABy (17) = 0 -469 + FABy (17) = 0 FABy (17) = 469

Solving FAB: Solving FBC:



AB

AByA F

Fsin

BC

BCyC F

Fsin

A

AByAB sin

F F

C

BCyBC sin

F F

sin45

27.6 F

YAB sin55

39.4 FBC

activity5 1a x y components vectors anskey

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