1. Chp 2: Force Resultants (1)
Engineering 36
Chp 2: Force Resultants (1)
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer

2. Resultant of Two Forces
Force: Action Of One Body On Another; 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

3. Concurrent Force Resultant
CONCURRENT FORCES  Set Of Forces Which All 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
VECTOR FORCE COMPONENTS  Two Or More Force Vectors Which, Together, Have The Same Effect As An Original, Single Force Vector

4. Resultant cont.
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
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. Graphical Vector Addition
Consider a Vector Set in the XY Plane:
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. Graphical Vector Addition
Connect the TAIL of V1 to the TIP of V3 to reveal the RESULTANT, VR 3 4 2 1

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

9. Graphical Vector Addition
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
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. Graphical Vector Addition
Construction of Vinter
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 LayOut Vinter & V3 Tail-to-Tail

11. Graphical Vector Addition7 6 8 5 5

12. Parallelogram Vector Addition
Any Number of Vectors may be added by the parallelogram rule by the repeated Construction 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 Find 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 ( ) 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. Calc Spring-Cable Tensions
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
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
Do in ACAD to FULL SCALE; i.e., 1” = 1N => need to change dimscale in DimStyle

22. AutoCAD Graphical Solution
Scaling Up using 4” = 150N
TAC
TAD W
TAB
Do in ACAD to FULL SCALE; i.e., 1” = 1N => need to change dimscale in DimStyle to 50X for force-arrows. Layers => GREEN for XL, BLUE for Force-Arrows

23. Compare: Hand vs. ACAD
Check the Pencil & Paper Solution to mathematically precise ACAD soln
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. Let’s Work This Nice Problem
WhiteBoard Work
Let’s Work This Nice Problem
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°.
H13e 2-18
Do:
Graphically
Analytically

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. Let’s Work This Nice Problem
WhiteBoard Work
Let’s Work This Nice Problem
Resolve F1 into components along the u & v axes and determine the magnitudes of these components.
H13e 2-18
Do:
Graphically
Analytically

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

32. Let’s Work The Spring Problem by DeComp
WhiteBoard Work
Let’s Work The Spring Problem by DeComp