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

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 Michaels 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
6
Sum The Forces

• ?Fx0
• 00.286FB-0.429FC-0.286FD
• ?Fy0
• 0-0.429FB-0.286FC0.429FD
• ?Fz0
• 0-0.857FB-0.857FC-0.857FD1472
• Solution
• ?Fx0
• 00.286FB-0.429FC-0.286FD
• 0.286FD0.286FB-0.429FC
• FDFB-1.5FC

7
Solution Continued

• ?Fy0
• 0-0.429FB-0.286FC0.429FD
• 0-0.429FB-0.286FC0.429(FB-1.5FC)
• 0-0.429FB-0.286FC0.429FB-0.644FC
• 0-0.93FC
• FC0
• ?Fz0
• 0-0.857FB-0.857FC-0.857FD1472
• 0-0.857FB-0.857(FB-1.5FC)-0.857FC1,472
• 0-0.857FB-0.857FB1,472
• 1.71FB1,472
• FB860.8N
• FDFB-1.5FC
• FDFB860.8N

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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

• First Find ?
• ?tan-1 (2/2)
• ?45o
• Sum Forces
• ?Fx 0
• -TC(cos45o)TD(cos45o)0
• TCTD
• ?Fy0
• TD(sin45o)TC(sin45o)-2FW0

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Determine TC and TD

• TDTC
• When the load is applied the distance is 2.83m
• With no load the distance is

2.83m
2.50m
1.50m
2.00m
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Final Solution

• Spring Force Equation
• Fs(spring constant)x(distance)
• Figure out Tension TD TC
• TCTD(100N/m)(2.83m-2.50m)32.8N
• Sum of Forces in Y
• ?Fy0 FWmg
• 0TD(sin45o)TC(sin45o)-2mg
• 046.4-2(9.81)m
• m2.36kg

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3D Rigid Body Equilibrium (Chapter 5)
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FBD
5-68
Object Axis Force/Moments Distance
Z
AZ
A
BZ
AX
0.1m
BX
y
x
B
FM
0.6m
0.5m
0.1m
0.2m
FP
0.1m
15
Sum of Forces

• FM75(9.81)735
• ?Fx0
• 0AXFP-BX
• ?Fy0
• 0AY
• ?Fz0
• 0AZBZ-735
• Moments About B
• ?MBX0
• 0-1.1AZ(735)(0.5)
• ?MBY0
• 0 (735)(0.1)-0.2FP
• ?MBZ0

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Solution

• ?MBY0
• 0 (735)(0.1)-0.2FP
• 0.2FP73.5
• FP367.5
• ?MBZ0
• 01.1AX-0.2FP
• 0 1.1AX-0.2(367.5)
• 1.1AX73.5
• AX66.8
• ?Fx0
• 0AXFP-BX
• 66.8367.5BX
• BX434.3
• ?MBX0
• 0-1.1AZ(735)(0.5)
• 1.1AZ367.5

AX66.8N AY0N AZ334.1N BX434.4 BZ400.9 FP367.
5N
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5-71
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FBD
Z
1ft
AZ
1ft
1ft
AY
Ax
A
FCD
EZ
E
1ft
EX
Y
1.5ft
X
FW250lb
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Sum forces and Moments About A

• ?Fx0
• 0AXEX
• ?Fy0
• 0AY
• ?Fz0
• 0AZEZ-FDC-250
• ?Mx0
• 0-3(250)2EZ-FDC
• ?My0
• 1.5(250)-FDC
• ?Mz0
• 0-2EX
• 0EX

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Solution

• ?My0
• 1.5(250)-FDC
• FDC375lbs
• ?Mx0
• 0-3(250)2EZ-FDC
• 0-7502EZ-375
• 11252EZ
• 562.5lbEZ

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Solution Continued

• ?Fx0
• 0AXEX
• 0AX0
• AX0
• ?Fz0
• 0AZEZ-FDC-250
• 0AZ562.5-375-250
• 62.5AZ
• AX0
• AY0
• AZ62.5lb
• EX0
• EZ562.5lb
• FDC375lbs

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5-82
23
FBD









Z
8ft
6ft
MAZ
Y
MAY
Ay
FBC
AX
A
MAX
B
12ft
4ft
X
 
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Sum Forces

•  

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Solution

• ?Fz0
• 0-750.43FBC
• 0.43FBC75
• FBC174.4 lb
• ?Fy0
• 00.29FBC-AY-40
• AY0.29(174.4)-40
• AY10.58 lb
• ?Fx0
• 0AX20-0.86FBC
• AX20-1500
• AX130

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Solution Continued

•  

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

• ?Fx0
• 0AX-BX
• AXBX
• ?Fy0
• 0450cos45O-CY
• CY450(0.707)
• CY318.2N
• ?Fz0
• 0AZ-BZCZ-450sin45O
• ?Mx0
• 0-0.8BZ-450cos45O(0.4)-450sin45O(1.2)318.2(0.4)
  1.2CZ
• ?My0
• 00.6CZ-300
• CZ500N

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Solution Continued

• ?Mx0
• 0-0.8BZ-450cos45O(0.4) 450sin45O(1.2)318.2(0.4)
  1.2CZ
• 00.8BZ-450cos45O(0.4)-450sin45O(1.2)318.2(0.4)1
  .2(500)
• 0-0.8BZ-127.3-381.8127.3600
• BZ272.8N
• ?Fx0
• 0AX-BX
• AXBX238.7N
• ?Fz0
• 0AZ-BZCZ-450sin45O
• 0AZ-272.8500-318.2
• AZ91.0N
• AX 238.7N
• AZ 91.0N
• BX 238.7N
• BZ 272.8N

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FBD-1
6-106
This is an FBD of just the bucket
EY
EX
E
FAC
1200lb
1ft
60O
C
0.25ft
1.50ft
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Sum Forces and Moments about E
For this problem finding EX and EY is not needed

• ?Fx0
• 0FAC(cos60O)-EX
• ?Fy0
• 0-FAC(sin60O)EY-1200
• ?ME0
• 0FAC(cos60O)(1)-(1200)(1.5)FAC(sin60O)(0.25)
• 00.5FAC-18000.217FAC
• 18000.717FAC
• FAC2512.2lbs

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FBD-2
FBD for hydraulic cylinder
Y
FAB
45O
FAD
X
60O
FAC 2512.2lbs
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Solution
There are no distances now so we know there are
no moments

• ?Fx0
• 0FAD-FAB(sin45O)-2512.2(cos60O)
• ?Fy0
• 02512.2(sin60O)-FAB(cos45O)
• 0.707FAB 2175.6
• FAB 3077.2lbs
• ?Fx0
• 0FAD-FAB(sin45O)-2512.2(cos60O)
• FAD3077.2(0.707)1256.1
• FAD 3431.7lbs