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

Statics Review Presentation Tough Problems from Chapter 3 and 5!!. 3-56. FBD. F W. +z. 6m. 4m. F D. 12m. F C. 6m. 4m. +y. 4m. F B. 6m. +x. Force vectors – Books Approach:. Or Use Professor Robert Michael’s Table Method. Sum The Forces. ∑ Fx =0; 0=0.286F B -0.429F C -0.286F D

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

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  1. Statics Review PresentationTough Problems from Chapter 3 and 5!!

  2. 3-56

  3. FBD FW +z 6m 4m FD 12m FC 6m 4m +y 4m FB 6m +x

  4. Force vectors – Books Approach:

  5. Or Use Professor Robert Michael’s Table Method

  6. Sum The Forces ∑Fx=0; 0=0.286FB-0.429FC-0.286FD ∑Fy=0; 0=-0.429FB-0.286FC+0.429FD ∑Fz=0; 0=-0.857FB-0.857FC-0.857FD+1472 Solution ∑Fx=0; 0=0.286FB-0.429FC-0.286FD 0.286FD=0.286FB-0.429FC FD=FB-1.5FC

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

  8. 3-42

  9. FBD TC TD 2.83m 2.00m 2.00m 100N/m 2.00m 2.00m 2FW Use 2FW because there Are two cylinders

  10. Solution FirstFind θ θ=tan-1 (2/2) θ=45o Sum Forces ∑Fx= 0; -TC(cos45o)+TD(cos45o)=0 TC=TD ∑Fy=0; TD(sin45o)+TC(sin45o)-2FW=0

  11. Determine TC and TD TD=TC 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 Fs=(spring constant)x(distance) Figure out Tension TD TC TC=TD=(100N/m)(2.83m-2.50m)=32.8N Sum of Forces in Y ∑Fy=0; FW=mg 0=TD(sin45o)+TC(sin45o)-2mg 0=46.4-2(9.81)m m=2.36kg

  13. 3D Rigid Body Equilibrium (Chapter 5)

  14. 5-68 FBD Object Axis Force/Moments Distance Z AZ A BZ AX 0.1m BX x B y FM 0.6m 0.5m 0.1m 0.2m FP 0.1m

  15. Sum of Forces FM=75(9.81)=735 ∑Fx=0; 0=AX+FP-BX ∑Fy=0; 0=AY ∑Fz=0; 0=AZ+BZ-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 ∑MBY=0; 0= (735)(0.1)-0.2FP 0.2FP=73.5 FP=367.5 ∑MBZ=0; 0=1.1AX-0.2FP 0= 1.1AX-0.2(367.5) 1.1AX=73.5 AX=66.8 ∑Fx=0; 0=AX+FP-BX 66.8+367.5=BX BX=434.3 ∑MBX=0; 0=-1.1AZ+(735)(0.5) 1.1AZ=367.5 AZ=334.1 ∑Fz=0; 0=AZ+BZ-735 0=334.1+BZ-735 BZ=400.9 AX=66.8N AY=0N AZ=334.1N BX=434.4 BZ=400.9 FP=367.5N

  17. 5-71

  18. FBD Z 1ft AZ 1ft 1ft AY Ax A FCD EZ E 1ft EX Y 1.5ft X FW=250lb

  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)-FDC FDC=375lbs ∑Mx=0; 0=-3(250)+2EZ-FDC 0=-750+2EZ-375 1125=2EZ 562.5lb=EZ

  21. Solution Continued ∑Fx=0; 0=AX+EX 0=AX+0 AX=0 ∑Fz=0; 0=AZ+EZ-FDC-250 0=AZ+562.5-375-250 62.5=AZ AX=0 AY=0 AZ=62.5lb EX=0 EZ=562.5lb FDC=375lbs

  22. 5-82

  23. FBD Z 8ft 6ft MAZ Y MAY Ay FBC AX A MAX B 12ft 4ft X

  24. Sum Forces

  25. Solution ∑Fz=0; 0=-75+0.43FBC 0.43FBC=75 FBC=174.4 lb ∑Fy=0; 0=0.29FBC-AY-40 AY=0.29(174.4)-40 AY=10.58 lb ∑Fx=0; 0=AX+20-0.86FBC AX+20-150=0 AX=130

  26. Solution Continued

  27. FBD 5-72 Z CZ AZ AX C 450N CY BZ 45O A BX 0.6m B 0.4m 0.8m X Y 0.4m

  28. Solution ∑Fx=0; 0=AX-BX AX=BX ∑Fy=0; 0=450cos45O-CY CY=450(0.707) CY=318.2N ∑Fz=0; 0=AZ-BZ+CZ-450sin45O ∑Mx=0; 0=-0.8BZ-450cos45O(0.4)-450sin45O(1.2)+318.2(0.4)+1.2CZ ∑My=0; 0=0.6CZ-300 CZ=500N ∑Mz=0; 0=-0.8BX-318.2(0.6) BX=238.7N

  29. Solution Continued ∑Mx=0; 0=-0.8BZ-450cos45O(0.4) 450sin45O(1.2)+318.2(0.4)+1.2CZ 0=0.8BZ-450cos45O(0.4)-450sin45O(1.2)+318.2(0.4)+1.2(500) 0=-0.8BZ-127.3-381.8+127.3+600 BZ=272.8N ∑Fx=0; 0=AX-BX AX=BX=238.7N ∑Fz=0; 0=AZ-BZ+CZ-450sin45O 0=AZ-272.8+500-318.2 AZ=91.0N AX= 238.7N AZ= 91.0N BX= 238.7N BZ= 272.8N CY= 318.2N CZ=500N

  30. FBD-1 6-106 This is an FBD of just the bucket EY EX E FAC 1200lb 1ft 60O C 0.25ft 1.50ft

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

  32. FBD-2 FBD for hydraulic cylinder Y FAB 45O FAD 60O X FAC = 2512.2lbs

  33. Solution There are no distances now so we know there are no moments ∑Fx=0; 0=FAD-FAB(sin45O)-2512.2(cos60O) ∑Fy=0; 0=2512.2(sin60O)-FAB(cos45O) 0.707FAB = 2175.6 FAB = 3077.2lbs ∑Fx=0; 0=FAD-FAB(sin45O)-2512.2(cos60O) FAD=3077.2(0.707)+1256.1 FAD = 3431.7lbs

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