2. CONTINUATION OF COMPONENTS OF FORCES

3. Realize in these problems that a right triangle will represent a FORCE and the COMPONENTS of the force, when the force is not parallel with either of the X and Y axes. The hypotenuse of the triangle (the longest side) will represent the SLOPE of the force, in the same proportion that the legs of the triangle represent the components of the force. The angle the force makes with the X and Y axes will be defined by the length of the legs of the triangle. The FORCE will be directly proportional to the hypotenuse of the triangle. The COMPONENTS of the force will be directly proportional to the respective legs of the triangle.

4. Find the Fx and Fy component values by proportions of a right triangle. Fx = 80 ; 5 Fx = 80 x 4 Fx = 80 ; 5 Fx = 80 x 4 4 5 4 5 Fx = 80 x 4 ; Fx = 64 lb. 5 Fy = 80 ; 5 Fy = 80 x 3 3 5 3 5 Fy = 80 x 3 Fy = 48 lb. Fy = 80 x 3 Fy = 48 lb. 5 REVIEW OF COMPONENTS TO FORCE The slope of a vector that represents a force of 80 lbs is shown as 4 units horizontal to 3 units vertical.

5. The hypotenuse of slope is the square root of the sum of 7 and 8 squared. Hypotenuse = 64 + 49 = 10.63 Fx = 7 ; 10.63Fx = 90 x 7 90 10.63 90 10.63 Fx = 90 x 7 ; Fx = 59.27 10.63 10.63 Fy = 8 ; 10.63 Fy = 90 x 8 90 10.63 90 10.63 Fy = 90 x 8 ; Fy = 67.73 10.63 10.63 Now consider a force with a slope of 8 units horizontal for each 9 vertical – First find the corresponding hypotenuse

6. Remember that to solve any problem concerning the equilibrium of a body acted upon by forces, it is essential that you consider the three equations of equilibrium. They are: The sum of all forces that are horizontal must equal zero. The sum of all forces that are vertical must equal zero. The sum of all moments taken about any point must equal zero.

7. PROBLEM: Find a single force that is the resultant of the three forces. SOLUTION: Find the X and Y components of each force, add them together algebraically. The result will be one X component and one Y component. Remember the sign convention; up is positive, down is negative, to the right is positive, and to the left is negative. Force 1 will have positive X & Y Force 2 will have negative X, positive Y Force 3 will have negative X and negative Y NOW CONSIDER 3 FORCES THAT BEGIN AT THE SAME POINT Force 1 = 50 lb Force 2 = 70 lb Force 3 = 40 lb

8. Force X component Y component 50 4/5 x 50 = + 40 3/5 x 50 = +30 70 5/13 x 70 = -26.92 12/13 x 70 = +64.62 40 1/1.414 x 40 = -28.28 1/1.414 x 40 = -28.28 MAKE A CHART AND CALCULATE THE COMPONENT VALUES OF EACH FORCE

9. In the X column the algebraic sum equals, +40 – 26.92 – 28.28 = -26.91 = resulting X component In the Y column the algebraic sum equals, In the Y column the algebraic sum equals, +30 +64.62 – 28.28 = +66.34 = resulting Y component So here we have two remaining forces that can both be components of a single force. The value of the single force is found by the Pythagorean law: Force = 26.91 2 + 66.34 2 = 71.59 So the 71.59 value is actually the resultant of the original 3 forces So the 71.59 value is actually the resultant of the original 3 forces - 26.91 + 66.34 71.59

10. CONSIDER THE DIAGRAM And visualize the assembly as a free body diagram. The Force at D must be resisted by a force at A, which is the horizontal component of the force in AB. The only place for an upward vertical force is at A, which is the vertical component of the tension in AB --- and it must equal 500 lb, because there is no other vertical force. A CORD IS FASTENED TO A WALL AT POINT A. IT SUPPORTS A WEIGHT OF 500 POUNDS. FIND THE TENSION IN CORD AB AND THE FORCE REQUIRED AT D TO MAINTAIN THE SHAPE.

11. The force in the cord AB can be found by proportion since the value of the vertical component is known. The slope of the force is known by the component lengths, 4’ and 5’, so the slope length of the slope triangle is 5 2 + 4 2 = 6.40’ By proportion: AB / 6.40 = Ay / 5 ; and AB = 500 x 6.40 / 5, and AB = 640 Ax / 640 = 4 / 6.40 ; and Ax = 640 x 4 / 6.4, and Ax = 400 To satisfy the equilibrium of horizontal forces, the force at D must be equal and opposite to Ax, so D = 400 lb

12. Now use what you have learned and work the following problem

13. A cantilevered truss is loaded as shown - - - Find The horizontal force at point A The horizontal force at point F The force in member FE

14. Moments at F : - 400(4) – 300(8) – 200(12) + Ax(6) = 0 : Ax = 1066.67 lb Sum of horizontal forces = 0, so Fx – 1066.67 lb Then Force in FE = 1066.67 ( 4.47 / 4 ) = 1192 lb

16.  now

17. As a free body diagram the directions of the components at B are correct because the tension in the cable is an axial load – through its center. Also since the horizontal component at B is to the right, the horizontal component at A must be to the left, and is equal in magnitude to that at B since there are no other horizontal components.

18. But, the direction of Ay is not obvious because the force in the member ACD is not necessarily axial, because there are component forces in the body of the boom at C. The boom has a tendency to bend at C, so the compressive force in the member is not axial. The solution is to first find the force in cable BC, then the vertical component, By – then Ay can be found.

19. Since A and B have two unknowns, each, the sum of forces in the X and Y directions do not yield a solution. Take moments about point A to find Bx, since the line of action of By goes through A. +2000(70) –Bx(70) = 0 So Bx = 2000 lb. BC = 58.31 / 50 x 2000 BC = 2332.4 lb ; and By = 2332.4 x 30 / 58.31 By = 1200 Distance AC = 50 2 + 40 2 = 64.03’ Distance BC = 50 2 + 30 2 = 58.31 1200 2000

20. 1200 2000 800 By summing vertical and horizontal forces, Ax = 2000 left, and Ay = 2000 – 1200 = 800 upward By illustrating Ax & Ay to scale, notice the line of action of the resultant does not align with member ACD, since it is not a two-force member.

22. NOW ON A SEPARATE SHEET OF PAPER TO HAND IN, DO THIS EXERCISE... In consideration for beginning a solution to finding the forces in cords AD and BD, Write two equations of the things you know simply by observing the assembly as a free body diagram. D