Upload
virginia-tyler
View
37
Download
8
Embed Size (px)
DESCRIPTION
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 - PowerPoint PPT Presentation
Citation preview
Or Use Professor Robert Michael’s Table Method
Force dx dy dz d FX=(dx/d)F FY=(dy/d)F FZ=(dz/d)F
FB 4 -6 -12 14 0.286FB -0.429FB -0.857FB
FC -6 -4 -12 14 -0.429FC -0.286FC -0.857FC
FD -4 6 -12 14 -0.286FD 0.429FD -0.857FD
FW 0 0 0 0 0 0 (150kg)(9.8m/
s2)=1472N
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
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+14720=-0.857FB-0.857(FB-1.5FC)-0.857FC+1,4720=-0.857FB-0.857FB+1,4721.71FB=1,472FB=860.8N
FD=FB-1.5FC
FD=FB=860.8N
Solution First Find θ
θ=tan-1 (2/2)θ=45o
Sum Forces∑Fx= 0;-TC(cos45o)+TD(cos45o)=0TC=TD
∑Fy=0;TD(sin45o)+TC(sin45o)-2FW=0
Determine TC and TD
TD=TC
When the load is applied the distance is 2.83m
With no load the distance is
2.83m
2.50m1.50m
2.00m
Final SolutionSpring 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=mg0=TD(sin45o)+TC(sin45o)-2mg0=46.4-2(9.81)mm=2.36kg
Sum of ForcesFM=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
Solution∑MBY=0; 0= (735)(0.1)-0.2FP
0.2FP=73.5FP=367.5
∑MBZ=0;0=1.1AX-0.2FP
0= 1.1AX-0.2(367.5)1.1AX=73.5AX=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=367.5AZ=334.1
∑Fz=0;0=AZ+BZ-7350=334.1+BZ-735BZ=400.9
AX=66.8NAY=0NAZ=334.1NBX=434.4BZ=400.9FP=367.5N
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=-2EX0=EX
Solution Continued∑Fx=0;0=AX+EX
0=AX+0AX=0
∑Fz=0;0=AZ+EZ-FDC-2500=AZ+562.5-375-25062.5=AZ
AX=0AY=0AZ=62.5lbEX=0EZ=562.5lbFDC=375lbs
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
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-300CZ=500N
∑Mz=0;0=-0.8BX-318.2(0.6)BX=238.7N
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+600BZ=272.8N
∑Fx=0;0=AX-BX
AX=BX=238.7N
∑Fz=0;0=AZ-BZ+CZ-450sin45O
0=AZ-272.8+500-318.2AZ=91.0N
AX= 238.7NAZ= 91.0NBX= 238.7NBZ= 272.8NCY= 318.2NCZ=500N
Sum Forces and Moments about E∑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
For this problem finding EX and EY is not needed