Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

1. Engineering 36 Chp2: ForceResultants (1) Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. Force: Action Of One Body OnAnother; Characterized By Its Point Of Application Magnitude (Intensity) Direction Experimental Evidence Shows That The Combined Effect Of Two Forces May Be Represented By A Single Resultant Force Resultant of Two Forces

3. CONCURRENT FORCES Set Of Forces WhichAll Pass Through The Same Point A Set Of Concurrent Forces Applied To A body May Be Replaced By A Single Resultant Force Which Is The Vector Sum Of The Applied Forces Concurrent Force Resultant • VECTOR FORCE COMPONENTS Two Or More Force VectorsWhich, Together, HaveThe Same Effect As An Original, Single Force Vector

4. The Resultant Is Equivalent To The Diagonal Of A Parallelogram Which Contains The Two Forces In Adjacent Legs As Forces are VECTOR Quantities and they Add as Such Resultant cont. 3D Vector 2D Vector

5. Find the Force Resultant • Find the Resultant of multiple Forces by Vector Addition • The two basic forms of Vector addition • Decomposition • Decompose all vectors into Axes Components • Combine Like Components to Obtain Resultant • Graphical → Mag & Dir to SCALE • Tip-to-Tail (a.k.a. Head-to-Tail) • Parallelogram

6. Consider a Vector Set in the XY Plane: Graphical Vector Addition • All vector lengths are SCALED relative to their magnitudes Place V1 at ANY convenient location Place V2 at the TIP of V1 Place V3 at the TIP of V2 • Add in Tip-to-Tail Fashion

7. Connect the TAIL of V1 to the TIP of V3 to reveal the RESULTANT, VR Graphical Vector Addition 3 4 2 1

8. Consider a Vector Set in the XY Plane: Graphical Vector Addition • Addition Of Three Or More Vectors proceeds Through RepeatedApplication Of The Parallelogram Rule • Note that Parallelogram vector addition proceeds in TAIL-to-TAIL fashion • Add by Parallelogram Rule

9. Layout scaled vectors V1 & V2 in Tail-to-Tail Fashion Draw a “Construction Line” (a.k.a. “XL”) from Tip of V1 that is Parallel (a.k.a. ||) to V2 Graphical Vector Addition • Draw an XL from the tip of V2 that is || to V1. • The Two XL’s will intersect if V1 & V2 are NOT Parallel • Connect the tail-pt of V1 & V2 to the XL intersection to reveal the intermediate, Vector, Vinter

10. Construction of Vinter Graphical Vector Addition Draw an XL from the tip of Vinter that is || to V3 Draw an XL from the tip of V3 that is || to Vinter Connect the tail-pt of Vinter & V3 to the XL intersection to reveal the Resultant Vector, VR 4 3 1 2 1 Start a NEW dwg and LayOutVinter& V3 Tail-to-Tail

11. Graphical Vector Addition 7 6 8 5 5

12. Parallelogram Vector Addition • Any Number of Vectors may be added by the parallelogram ruleby the repeatedConstruction of Intermediate Vectors • Generally parallelogram addition is more cumbersome than Tip-to-Tail

13. Example: Graphical Force Add • Consider Spring & Cable Supported Wt • The Weight is not moving; i.e., it’s in Static Equilibrium Spring Supports

14. Spring Example Notes • Springs • Free-Length for BOTH = 450 mm • kAB = 1.5 N/mm • kAD = 0.5 N/mm • Find • Tension in Cord, TAC • Weight of Block, W

15. Example: Solution Plan • Use Spring Constants and Extension to Find TAB & TAD • Draw Vector Force PolyGon Noting that the PolyGon Must Close for a system in Equilibrium • Draw Force/Vector PolyGon in Tip-to-Tail form to reveal TAC & W • Note that the Directions are known for W & TAC; i.e., the Force LoA is CoIncident with Geometry • Solve by Hand & AutoCAD Scaling

16. Spring Digression: Hooke’s Law • Robert Hooke (1635-1703) formulated the relationship between the force applied to, and extension of, a Linear Elastic structural member. For a Spring: • Where • Fs≡ Spring Force (N or lb) • k ≡ Spring Constant (N/m or lb/in) • ΔL ≡ Spring Extension from Free-Length (m or in)

17. Example: FBD on Ring/Eye • Angles by ATAN • LAB & LAD by Pythagoras

18. Recall Stretched Lengths LAB = 550 mm LAD = 680 mm Free Length for Both Springs is 450mm And the Spring Constants kAB = 1.5 N/mm kAD = 0.5 N/mm Calc Spring-Cable Tensions • Use Hook’s law to Calc the Tension • Thus

19. Graphical Solution Hand Select Scaling Factor of 4in/150N Draw Known Force TAB with Direction & Scaled-Mag From Tip of TAB Draw Known Force TAD Make XL’s for Known LoA’s For W and TAC • XL for TAC LoA from Tip of TAD • XL for W LoA from Tail of TAB

20. Graphical Solution → Hand Scaled

21. Graphical Solution → AutoCAD • Let’s use AutoCAD (c.f. EGNR22) to GREATLY improve the accuracy of our graphical Solution

22. AutoCAD Graphical Solution • Scaling Up using 4” = 150N TAC TAD W TAB

23. Check the Pencil & Paper Solution to mathematically precise ACAD soln Compare: Hand vs. ACAD • TAC: 61/65.6 → Hand Soln 9.3% Low • W: 197/207 → 4.83% low • Not Bad for Engr Comp-Pad, Ruler, and Protractor

24. WhiteBoard Work • Do: • Graphically • Analytically Let’s WorkThis NiceProblem • Determine the design angle φ (0° ≤ φ ≤ 90°) between struts AB and AC so that the 400-lb horizontal force has a component of 600 lb which acts up to the left, in the same direction BA. Also find the magnitude of the force directed along line AC. Take θ = 30°.

25. Strut-City • Note the structure is composed to SLENDER RODS in in tension, with connecting pts at the ends of the rods • In this case member forces are CoIncident with Geometry

28. WhiteBoard Work • Do: • Graphically • Analytically Let’s WorkThis NiceProblem • Resolve F1 into components along the u & v axes and determine the magnitudes of these components.

31. Engineering 36 Appendix Bruce Mayer, PE Registered Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

32. WhiteBoard Work Let’s WorkThe SpringProblem byDeComp