Phys 170 Lecture 14 1 Physics 170 Lecture 14 Chapter 5 - “Equilibium of a Rigid Body” Still More Example Problems
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Phys 170 Lecture 14 1
Physics 170 Lecture 14
Chapter 5 - “Equilibium of a Rigid Body”
Still More Example Problems
Phys 170 Lecture 14 2
PROBLEM 5-82 (page 257, 12th
edition)
Member AB is supported at B by a cable and at A by a smooth
fixed square rod which fits loosely through the square hole of the
collar.
• Determine the x, y, z components of reaction at A and the tension
in the cable when
!F = (20 i
!! 40!j ! 75k
!) lb .
PROBLEM 5-82 (page 257, 12th
edition)
Member AB is supported at B by a cable and at A by a smooth
fixed square rod which fits loosely through the square hole of the
collar.
• Determine the x, y, z components of reaction at A and the tension
in the cable when
!F = (20 i
!! 40!j ! 75k
!) lb .
F = 20i − 40 j − 75k( )
Phys 170 Lecture 14 3 F = 20i − 40 j − 75k( )
AX
AY
MZ
MY
MX
FBC
RA = 0, 0, 0( )
RB = 12, 4, 0( )
RC = 0, 8, 6( )
RF = 12, 0, 0( )
RCB = −12, +4, +6( ) LCB = 122 + 42 + 62 = 14FCB = −12, +4, +6( ) ⋅ FCB
LCB F = 20, −40, −75( )
MBC =
RB ×
FCB =
i j k12 4 0−12 4 6
⋅FCBLCB
=24−7296
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥⋅FCBLCB
Phys 170 Lecture 14 4 F = 20i − 40 j − 75k( )
AX
AY
MZ
MY
MX
FBC
RA = 0, 0, 0( )
RB = 12, 4, 0( )
RC = 0, 8, 6( )
RF = 12, 0, 0( )
RCB = −12, +4, +6( ) LCB = 122 + 42 + 62 = 14FCB = −12, +4, +6( ) ⋅ FCB
LCB F = 20, −40, −75( )
MF =
RF ×
F =
i j k12 0 020 −40 −75
=0900−480
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
Phys 170 Lecture 14 5 F = 20i − 40 j − 75k( )
AX
AY
MZ
MY
MX
FBC
FCB = −12, +4, +6( ) ⋅ FCB
LCB F = 20, −40, −75( )
FX : AX −12FCBLCB
+ 20 = 0
FY : AY + 4FCBLCB
− 40 = 0
FZ : 6 FCBLCB
− 75 = 0
Lucky day!Solve FZ for FCB/LCBPlug into the others
→FCBLCB
= 12.5
→ FCB = 175 lb
→ AX = 130 lb
→ AY = −10 lb
Phys 170 Lecture 14 6 F = 20i − 40 j − 75k( )
AX
AY
MZ
MY
MX
FBC
MX : MX + 24FCBLCB
+ 0 = 0
MY : MY − 72FCBLCB
+ 900 = 0
MZ : MZ + 96FCBLCB
− 480 = 0
Since we know FCB/LCBjust plug into the others
→ MZ = −720 ft-lb
MBC =
24−7296
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥⋅FCBLCB
MF =
0900−480
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
→ MY = 0 ft-lb
→ MX = −300 ft-lb
Phys 170 Lecture 14 7
Find the reactions at wheels A, B, and C
FB
FC
F1
F2
F3
x y
z
FA
Phys 170 Lecture 14 8
Find the reactions at wheels A, B, and C
FB
FC
F1
F2
F3
x y
z
FA
RA = 0, 0, 0( )
RB = +12, 35, 0( )
RC = −12, 35, 0( )
R1 = +12, 20, 0( )
R2 = −12, 8, 0( )
R3 = 0, 30, 0( )
F1 = 0, 0, −380( )
F2 = 0, 0, −500( )
F3 = 0, 0, −800( )
R1 ×
F1 =
i j k+12 20 00 0 −380
=−760045600
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
R2 ×
F2 =
i j k−12 8 00 0 −500
=−4000−60000
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
R3 ×
F3 =
i j k0 30 00 0 −800
=−2400000
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
Phys 170 Lecture 14 9
FB
FC
F1
F2
F3
x y
z
FA
RA = 0, 0, 0( )
RB = +12, 35, 0( )
RC = −12, 35, 0( )
F1 = 0, 0, −380( )
F2 = 0, 0, −500( )
F3 = 0, 0, −800( )
RB ×
FB =
i j k12 35 00 0 1
⋅FB =35−120
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥⋅FB
RC ×
FC =
i j k−12 35 00 0 1
⋅FC =35120
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥⋅FC
Phys 170 Lecture 14 10
FB
FC
F1
F2
F3
x y
z
FA
F1 = 0, 0, −380( )
F2 = 0, 0, −500( )
F3 = 0, 0, −800( )
RB ×
FB =
35−120
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥⋅FB
RC ×
FC =
35120
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥⋅FC
R1 ×
F1 =
−760045600
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
R2 ×
F2 =
−4000−60000
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
R3 ×
F3 =
−2400000
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
MX : −7600 − 4000 − 24000 + 35FB + 35FC = 0MY : 4560 − 6000 + 0 −12FB +12FC = 0FZ : FA +VB + FC − 380 − 500 − 800 = 0
FA FB FCMX 0 35 35 35600MY 0 −12 12 1440FZ 1 1 1 1680
FA = 662.8 lbFΒ = 448.6 lbFC = 568.6 lb
Phys 170 Lecture 14 11
For Next Time
Read Chapter 8 (Friction), Sections 1-3
Homework assignment 4 due tonight at 10 PM.
Homework Assignment 5 will appear at 6 PM
Practice problems & tutorials available (optional, no points)
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