1. Statics Review Presentation Tough Problems from Chapter 3 and 5!!

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3. FBD +x +y +z FWFW 6m 4m 6m 12m 4m 6m 4m FBFB FCFC FDFD

4. Force vectors – Books Approach:

5. Or Use Professor Robert Michael’s Table Method ForcedxdydzdF X =(dx/d)FF Y =(dy/d)FF Z =(dz/d)F FBFB 4-6-12140.286F B -0.429F B -0.857F B FCFC -6-4-1214-0.429F C -0.286F C -0.857F C FDFD -46-1214-0.286F D 0.429F D -0.857F D FWFW 000000(150kg)(9.8m /s 2 )=1472N

6. Sum The Forces ∑Fx=0; 0=0.286F B -0.429F C -0.286F D ∑Fy=0; 0=-0.429F B -0.286F C +0.429F D ∑Fz=0; 0=-0.857F B -0.857F C -0.857F D +1472 Solution ∑Fx=0; 0=0.286F B -0.429F C -0.286F D 0.286F D =0.286F B -0.429F C F D =F B -1.5F C

7. Solution Continued ∑Fy=0; 0=-0.429F B -0.286F C +0.429F D 0=-0.429F B -0.286F C +0.429(F B -1.5F C ) 0=-0.429F B -0.286F C +0.429F B -0.644F C 0=-0.93F C FC=0 ∑Fz=0; 0=-0.857F B -0.857F C -0.857F D +1472 0=-0.857F B -0.857(F B -1.5F C )-0.857F C +1,472 0=-0.857F B -0.857F B +1,472 1.71F B =1,472 F B =860.8N F D =F B -1.5F C F D =F B =860.8N

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9. FBD 2.83m 2.00m TCTC TDTD 100N/m 2F W Use 2F W because there Are two cylinders

10. Solution First Find θ θ=tan -1 (2/2) θ=45 o Sum Forces ∑Fx= 0; -T C (cos45 o )+T D (cos45 o )=0 T C =T D ∑Fy=0; T D (sin45 o )+T C (sin45 o )-2F W =0

11. Determine T C and T D T D =T C When the load is applied the distance is 2.83m With no load the distance is 2.83m 2.50m 1.50m 2.00m

12. Final Solution Spring Force Equation F s =(spring constant)x(distance) Figure out Tension T D T C T C =T D =(100N/m)(2.83m-2.50m)=32.8N Sum of Forces in Y ∑Fy=0; F W =mg 0=T D (sin45 o )+T C (sin45 o )-2mg 0=46.4-2(9.81)m m=2.36kg

13. 3D Rigid Body Equilibrium (Chapter 5)

14. FBD Object Axis Force/Moments Distance Z yx FMFM FPFP A B BZBZ BXBX AXAX AZAZ 0.1m 0.6m 0.1m 0.2m 0.5m 5-68

15. Sum of Forces F M =75(9.81)=735 ∑Fx=0; 0=A X +F P -B X ∑Fy=0; 0=A Y ∑Fz=0; 0=A Z +B Z -735 Moments About B ∑MBX=0; 0=-1.1AZ+(735)(0.5) ∑MBY=0; 0= (735)(0.1)-0.2FP ∑MBZ=0; 0=1.1AX-0.2FP

16. Solution ∑M BY =0; 0= (735)(0.1)-0.2F P 0.2F P =73.5 F P =367.5 ∑M BZ =0; 0=1.1A X -0.2F P 0= 1.1A X -0.2(367.5) 1.1A X =73.5 A X =66.8 ∑Fx=0; 0=A X +F P -B X 66.8+367.5=B X B X =434.3 ∑M BX =0; 0=-1.1A Z +(735)(0.5) 1.1A Z =367.5 A Z =334.1 ∑Fz=0; 0=A Z +B Z -735 0=334.1+B Z -735 B Z =400.9 A X =66.8N A Y =0N A Z =334.1N B X =434.4 B Z =400.9 F P =367.5N

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18. FBD Z X Y 1ft AZAZ AxAx AYAY F CD EZEZ EXEX 1ft 1.5ft F W =250lb 1ft A E

19. Sum forces and Moments About A ∑Fx=0; 0=AX+EX ∑Fy=0; 0=AY ∑Fz=0; 0=AZ+EZ-FDC-250 ∑Mx=0; 0=-3(250)+2EZ-FDC ∑My=0; 1.5(250)-FDC ∑Mz=0; 0=-2EX 0=EX

20. Solution ∑My=0; 1.5(250)-F DC F DC =375lbs ∑Mx=0; 0=-3(250)+2EZ-F DC 0=-750+2E Z -375 1125=2E Z 562.5lb=E Z

21. Solution Continued ∑Fx=0; 0=A X +E X 0=A X +0 A X =0 ∑Fz=0; 0=A Z +E Z -F DC -250 0=A Z +562.5-375-250 62.5=A Z A X =0 A Y =0 A Z =62.5lb E X =0 E Z =562.5lb F DC =375lbs

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23. FBD Z X Y 4ft 8ft 6ft 12ft F BC A B M AZ AXAX AyAy M AY M AX

24. Sum Forces

25. Solution ∑Fz=0; 0=-75+0.43F BC 0.43F BC =75 F BC =174.4 lb ∑Fy=0; 0=0.29F BC -A Y -40 A Y =0.29(174.4)-40 A Y =10.58 lb ∑Fx=0; 0=A X +20-0.86F BC A X +20-150=0 A X =130

26. Solution Continued

27. FBD Z X Y A B C AXAX AZAZ BZBZ BXBX 45 O 450N CZCZ CYCY 0.6m 0.4m 0.8m 0.4m 5-72

28. Solution ∑Fx=0; 0=A X -B X A X =B X ∑Fy=0; 0=450cos45 O -C Y C Y =450(0.707) C Y =318.2N ∑Fz=0; 0=A Z -B Z +C Z -450sin45 O ∑Mx=0; 0=-0.8B Z -450cos45 O (0.4)-450sin45 O (1.2)+318.2(0.4)+1.2C Z ∑My=0; 0=0.6C Z -300 C Z =500N ∑Mz=0; 0=-0.8B X -318.2(0.6) B X =238.7N

29. Solution Continued ∑Mx=0; 0=-0.8B Z -450cos45 O (0.4) 450sin45 O (1.2)+318.2(0.4)+1.2C Z 0=0.8B Z -450cos45 O (0.4)-450sin45 O (1.2)+318.2(0.4)+1.2(500) 0=-0.8B Z -127.3-381.8+127.3+600 B Z =272.8N ∑Fx=0; 0=A X -B X A X =B X =238.7N ∑Fz=0; 0=A Z -B Z +C Z -450sin45 O 0=A Z -272.8+500-318.2 A Z =91.0N A X = 238.7N A Z = 91.0N B X = 238.7N B Z = 272.8N C Y = 318.2N C Z =500N

30. FBD-1 C 60 O E EXEX EYEY 1ft 0.25ft 1200lb 1.50ft F AC This is an FBD of just the bucket 6-106

31. Sum Forces and Moments about E ∑Fx=0; 0=F AC (cos60O)-E X ∑Fy=0; 0=-F AC (sin60O)+E Y -1200 ∑M E =0; 0=F AC (cos60O)(1)-(1200)(1.5)+F AC (sin60O)(0.25) 0=0.5F AC -1800+0.217F AC 1800=0.717F AC F AC =2512.2lbs For this problem finding E X and E Y is not needed

32. FBD-2 FBD for hydraulic cylinder Y X 45 O 60 O F AC = 2512.2lbs F AD F AB

33. Solution ∑Fx=0; 0=F AD -F AB (sin45 O )-2512.2(cos60 O ) ∑Fy=0; 0=2512.2(sin60 O )-F AB (cos45 O ) 0.707F AB = 2175.6 F AB = 3077.2lbs ∑Fx=0; 0=F AD -F AB (sin45 O )-2512.2(cos60 O ) F AD =3077.2(0.707)+1256.1 F AD = 3431.7lbs There are no distances now so we know there are no moments