2. CLASS DAY 24 STRUCTURES 2 THE CONCEPT OF COMPONENTS Components are the parts of a Whole Distance, or a Whole Force, that help to define the magnitude and direction of that Distance or Force Distances and Forces can be reduced to their components to aid in the analysis of equilibrium, and in like manner, components can be used to identify magnitude and direction of distance and force. A component will be either VERTICAL or HORIZONTAL

3.  Cartesian Coordinates consist of a pair of axes 90 degrees apart, used as reference for identifying components of forces that are not parallel or perpendicular to either the vertical or horizontal A vector whose direction is at an angle with the reference axes will have COMPONENTS – one for each of the reference axes, which will define their relationship with the vector X X Y Y 0

4.  Imagine the shape as a box, and the green line is a diagonal of the box, and represents a vector, so that two sides of the box are represented by the yellow and cyan lines. The two lines are COMPONENTS of the green line, just as two sides of a box are COMPONENTS of the diagonal of a box. Just as two sides of a box have a mathematical relationship with the diagonal of the box, so do the COMPONENTS of a vector have a mathematical relationship to the value of the vector. X X Y Y 0

5.  Now consider a box whose side dimensions are 4’ and 3’. This automatically makes the direction of the diagonal at a slope of 4 to 3  In other words the diagonal is at an angle defined by 3 units vertical for each 4 units horizontal. These are COMPONENTS of the diagonal. The length of the diagonal can be found by the Pythagorean solution of a right triangle, which defines the diagonal as the square root of the sums of the squares of the two sides. In this case, the diagonal is 5’. 3 4 This says that for any vector whose slope is known (the relationship of the components) the magnitude of the vector can be found if the value of the components are known – the proportional relationships are the same.

6. LIKEWISE, FORCES CAN HAVE COMPONENTS

7. FOR ANY FORCE WHOSE SLOPE ANGLE IS KNOWN

8.  COMPONENTS OF DISTANCE If one were to fly directly from Lubbock to Oklahoma City, a distance of 300 miles, the path would be at a direction of 30 degrees above an East/West line through Lubbock. In other words, Oklahoma City is Northeast of Lubbock. To traverse the 300 miles, part of the path would be north, at the same time part of it is east. Because of the 30 degree angle of the direct route, Oklahoma City is 150 miles NORTH of Lubbock, and it is also 269.8 miles EAST of Lubbock. The NORTH and EAST distances are COMPONENTS of the 300 mile direct route.

9. In this example, the component distances are calculated by trigonometry, since the north and east distances form a 30 degree right triangle, with a hypotenuse of 300 miles. The short leg (north) is simply 300 x sin 30 = 300 x.5 = 150 miles. The long leg (east) is 300 x cos 30 = 300 x.866 = 259.8 miles. In a similar manner, COMPONENTS of forces can be calculated. But it is necessary to adopt a standard for sign convention and direction. The simple Cartesian Coordinates system is valuable in determining components in terms of X and Y values.

10.  Cartesian Coordinates consist of a pair of axes 90 degrees apart, used as reference for identifying components of forces that are not parallel or perpendicular to either the vertical or horizontal Values on the X plane to the right of 0 are positive, and values on the X plane to the left of 0 are negative. Values on the X plane to the right of 0 are positive, and values on the X plane to the left of 0 are negative. Values on the Y plane above 0 are positive, and values on the Y plane below 0 are negative. Values on the Y plane above 0 are positive, and values on the Y plane below 0 are negative. X X Y Y 0

11.  FINDING COMPONENTS OF A FORCE USING A SLOPE TRIANGLE - - - If the direction of a Force has a known relationship with both the X and Y axes, such that the force acts in a direction of Y distance for a given X amount of distance, the X and Y components of the force become the value that defines the SLOPE of the Force - - - the measure of the direction the FORCE varies from the X and Y axes. With the Y distance and X distances as legs of a right triangle, the hypotenuse can be found by the Pythagorean Theorem, and the resulting triangle becomes the Slope Triangle that defines the direction of the Force.

12. A force with a magnitude of 50 pounds lies in the XY plane, and the slope of the line that defines the force is Y = 3 units for each X = 4 units. The 3, 4 slope will form a right triangle whose hypotenuse = 5 units. Imagine the line of force of 50 is the hypotenuse of a right triangle whose X and Y components form a similar right triangle. The COMPONENT values are directly proportional to the legs of the Slope Triangle, and may be calculated by simple proportion:

13. Say the Y component of Force is proportional to the 3 of the slope triangle; and the X component of the Force is proportional to the 4 of the slope triangle; so it follows that the FORCE must be proportional to the 5 of the slope triangle. So, Y = Force ; cross multiply, 3 5 3 5 And 5Y = 50 x 3, and Y = 3 x 50 5 And Y = 30 lb, So the Y COMPONENT of the 50 pounds is 30 pounds.

14. Likewise, the X component of the force of 50 lb is proportional to the 4 of the slope triangle; So, X = Force ; cross multiply, 4 5 4 5 And 5X = 50 x 4 ; and X = 4 x 50 5 And X = 40 lb, So, the X COMPONENT of the 50 pounds is 40 pounds

15.  If the components of a force can be determined from the magnitude and direction of the force, then it follows that if the magnitude of a force is not known, but either the X or Y component can be found, then by the known direction of the force, its magnitude can be found.  Example: From the previous exercise, since the Y component of the 50 lbs equals 3 / 5 of the Force, then the force must equal 5 / 3 times the component. 5 / 3 x 30 = 150 / 3 = 50 lbs. Since the X component of the 50 lbs equals 4 / 5 of the Force, then the force must equal 5 / 4 times the component. 5 / 4 x 40 = 200 / 4 = 50 lbs.

16. CONSIDER THIS EXERCISE CONSIDER THIS EXERCISE 1A force of 60 lbs is at a direction such that the y direction of the slope triangle is 5, and the x direction of the slope triangle is 12, then the hypotenuse is 13. Find the X and Y Components of the force. 2If the Y component of a force of the same slope as above equals 50 pounds, what is the amount of the Force?

17. Remember the Y component of force is proportional to the 5 in the force triangle, the same way the X component of the force is proportional to the 12 in the force triangle – and the same way the 60 lb force is proportional to the hypotenuse, or 13 in the force triangle. A really handy realization... So it follows that 60/13 = Y component / 5, and Y component = 5/13 x 60 Y component = 23.07 lb and 60/13 = X component / 12, and X component = 12/13 x 60 = 55.38 lb

18. Consider the assembly shown: A cable, AC supports a rigid bar, AB, which holds a 1200 lb load at end A, attached to a wall at B and C. Find the tensile FORCE in the cable, and the horizontal FORCE exerted at B. Consider that the force in AC must be pure tension, as the cable is flexible. Also consider that the force in the bottom member, AB must be in pure compression, since the member is horizontal, and no other forces occur on the member. First, Show the assembly as a free body diagram - - - Realize that since the cable is slanted with respect to vertical and horizontal – the force in the cable must have vertical and horizontal components. Also realize the rigid member AB is horizontal and does not have components – only an axial force.

19. A free body diagram shows the assembly free of its support, replaced by the forces that hold it in equilibrium. The two forces that support the assembly are the Resisting Force at B, and the Resisting Force at C, which is equal and opposite to the tension in the cable AC. The force at B has no components in the XY plane because it is parallel with the X axis. The force at C has a direction as shown by the slope triangle, and the force has an X component and a Y component. Realize that all forces in the X direction must = 0, as must all forces in the Y direction also = 0 – in order for the assembly to remain in equilibrium.

20. Since the X and Y forces at C are components of the Resisting force at C, the resisting force can be replaced by the components. Note that there are only two forces in the Y direction, one of which is 1200 lb. downward, So the Component AC y must be 1200 lb. upward. Note also there are only two forces in the X direction, neither of which are known. Realize that all forces in the X direction must = 0, as must all forces in the Y direction also = 0 – in order for the assembly to remain in equilibrium.

21. Since AC y = 1200 lb, then the resisting force at C can be found by the slope triangle, since it is the same configuration as the assembly. By proportion: 1200 = Resisting force C ; 1200 = Resisting force C ; 3 5 3 5 Cross multiply: 3 x Force C = 1200 x 5 ; 3 x Force C = 1200 x 5 ; Force C = 1200 x 5 ; and 3 Force C = 2000 lb, which is the amount of tension in the cable AC. Realize that all forces in the X direction must = 0, as must all forces in the Y direction also = 0 – in order for the assembly to remain in equilibrium.

22. Since the force at C = 2000 lb, then the horizontal component can be found by the slope triangle. ACx = 4 x 2000 = 8,000 ; 5 5 5 5 And ACx = 1,600 lb to the right Since ACx = 1,600 lb, then the resisting force at B must be 1,600 lb, to the left in order to maintain equilibrium. Realize that all forces in the X direction must = 0, as must all forces in the Y direction also = 0 – in order for the assembly to remain in equilibrium.

23. Another way to calculate the tensile force in AC is by summation of moments created by the external loads. Since there are two unknowns, the force at C and the force at B, use point B as the moment center. The resisting force at B will not cause moment about B, since the distance equals 0. But, perpendicular distance, BD, from B to the line of action of the tensile force in AC (shown by the dashed line) will have to be found. D Triangle BCD is the same proportion as the slope triangle. So BD / 4 = 6 / 5 ; BD = 4/5 x 6 = 4.8 ft.

24. With BD = 4.8 ft. ; Take moments about point B : The line of action of the resisting force AC rotates clockwise about B. The line of action of the 1200 lb load rotates counter clockwise about B, so - (AC x 4.8) + (1200 x 8) = 0 4.8 AC = 9600, and 4.8 AC = 9600, and AC = 2000 lb D Triangle BCD is the same proportion as the slope triangle. So BD / 4 = 6 / 5 ; BD = 4/5 x 6 = 4.8 ft.

25. EXAMPLE PROBLEM TWO Two cords are fastened to a ceiling that support a 500 lb weight. Find the amount of tension in the two cords, AC, and BC. First show the assembly as a free body diagram, replacing the restraints at A and B with forces that will hold it in equilibrium. Realize that the two vertical Y components at A and B must be upward and total 500 lbs. Also realize that the horizontal components at A and B must be the same magnitude, but opposite in direction.

26. By observation one can say that A y + B y = 500, and that A x = B x. and that A x = B x. If point A or point B is used to sum moments, the horizontal components A x and B x will not cause rotation about either point, since the distance = 0. So if moments are summed about point A, then only one unknown remains, which is B y, since A y will not cause rotation about A since the distance = zero. Observation reveals that the shortest solution is to find at least one of the components, then solve for the tension in both cords by what is known above.

27. As a starting place, select point A, and sum the moments of the forces that can cause rotation, REPLACING the forces with their components Write the equation: -(500 x 12) + (B y x18.66) = 0 And - 6,000 + 18.66 B y = 0 B y = 6,000 = 321.54 lb. B y = 6,000 = 321.54 lb. 18.66 18.66 Since B y + A y = 500, then A y = 500 – 321.54 = 178.46 With both component forces known, the tension in both cords can be calculated, BUT, slope of the forces must be known. The slopes are given by the dimensions, but the length of the cords, (the hypotenuse of the slope triangles) must be calculated - - -

28. By the Theorem of Pythagoras, The length of the cords are thus: AC = (5x5) + (12x12) = 13’ BC = (5x5) + (6.66x6.66) = 8.33’ And, since the forces and the components are directly proportional to the components of the force triangle: AC is proportional to 13’ as Ay is proportional to 5’, so AC = 13 x 178.46 = 464 lb, and since 5 BC is proportional to 8.33’ as By is proportional to 5’, then BC is proportional to 8.33’ as By is proportional to 5’, then BC = 8.33 x 321.54 = 535.69 lb 5

29.  To check the calculations, the statement was made at the beginning that Ax must equal Bx. So Ax is proportional to 12’ as AC is proportional to 13’, so Ax = 12 x 464 = 428.30 lb, and 13 13 Bx is proportional to 6.66’ as BC is proportional to 8.33’ so, Bx = 6.66 x 535.69 = 428.30 lb 8.33 8.33  Some minute discrepancy in calculations will occur because of rounding off numbers to only two place accuracy