Statics Review PresentationTough Problems from Chapter 3

and 5!!

3-56

FBD

+x

+y

+z FW

6m

4m

6m12m

4m

6m

4m

FB

FCFD

Force vectors – Books Approach:

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

3-42

FBD

2.83m

2.00m 2.00m

2.00m 2.00m

TC TD

100N/m

2FW

Use 2FW because thereAre two cylinders

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

3D Rigid Body Equilibrium (Chapter 5)

FBDObjectAxisForce/MomentsDistance

Z

yxFM

FP

A

B

BZ

BX

AX

AZ

0.1m

0.6m

0.1m

0.1m

0.2m

0.5m

5-68

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

5-71

FBD

Z

XY

1ftAZ

AxAY

FCD

EZ

EX

1ft1ft

1.5ft

FW=250lb

1ft

A

E

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

∑My=0;1.5(250)-FDC

FDC=375lbs

∑Mx=0;0=-3(250)+2EZ-FDC

0=-750+2EZ-3751125=2EZ

562.5lb=EZ

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

5-82

FBDZ

X

Y

4ft

8ft

6ft

12ft

FBC

A

B

MAZ

AX

Ay

MAY

MAX

Sum Forces

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 Continued

FBD

Z

X Y

A

B

CAXAZ

BZ

BX

45O

450N

CZ

CY

0.6m0.4m

0.8m

0.4m

5-72

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

FBD-1

C

60O

E EX

EY

1ft

0.25ft

1200lb

1.50ft

FAC

This is an FBD of just the bucket

6-106

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

FBD-2FBD for hydraulic cylinderY

X

45O

60O

FAC = 2512.2lbs

FAD

FAB

Solution

∑Fx=0;0=FAD-FAB(sin45O)-2512.2(cos60O)

∑Fy=0;0=2512.2(sin60O)-FAB(cos45O)0.707FAB = 2175.6FAB = 3077.2lbs

∑Fx=0;0=FAD-FAB(sin45O)-2512.2(cos60O)FAD=3077.2(0.707)+1256.1FAD = 3431.7lbs

There are no distances now so we know there are no moments