Physics 170 Week 3, Lecture 2
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http://www.phas.ubc.ca/∼gordonws/170
Physics 170 203 Week 3 Lecture 2
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Textbook Chapter 3: Section 3.3
Physics 170 203 Week 3 Lecture 2
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Learning Goals:
• A brief review of the conditions for equilibrium of a particle.
• Illustrate the properties of systems of forces with an example.
• Become familiar with techniques for solving systems of linear
equations.
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Review:
• The condition for equilibrium of a particle is
X
F~i = 0
i
• If some components of F~i are not known, the equilibrium
condition can be used to find them. It contains three equations
– therefore it can be used to find three unknown quantities.
• To solve a problem, one should begin by drawing a free body
diagram.
• Analysis includes solving linear equations.
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Example:
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The 100 kg crate is supported by three cords, one of which is
connected to a spring. Determine the tension in the cords AC and
AD and the stretch of the spring.
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Strategy for finding a solution:
• Obtain a mathematical description of Force vectors acting at
~.
A: F~AB , F~AC , F~AD , W
• Directions of all force vectors are known.
~ is known, magnitudes of F~AB , F~AC , F~AD are
• Magnitude of W
unkowns.
~ , we apply
• Once we have expressions for F~AB , F~AC , F~AD , W
P~
the equation
Fi = 0:
~ =0
F~AB + F~AC + F~AD + W
This contains three equations, to solve for the three unknowns
FAB , FAC , FAD .
• Spring constant is known, so we will be able to use FAB = ks
to determine stretch of spring, s, once we know F~AB .
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The first step is:
Draw a free body diagram
Physics 170 203 Week 3 Lecture 2
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The first step is:
Draw a free body diagram
Physics 170 203 Week 3 Lecture 2
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The first step is:
Draw a free body diagram
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Find a mathematical description of vectors:
F~AC
F~AB = FAB î
³
´
= FAC cos 120î + cos 135ĵ + cos 60k̂
³
1
F~AD = FAD p
(−1)2
+
(2)2
+
(2)2
´
−1î + 2ĵ + 2k̂
µ
¶
1
2
2
= FAD − î + ĵ + k̂
3
3
3
~ = −(100)(9.81)N k̂
W
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Apply the theory:
0=
X
~
F~i = F~AB + F~AC + F~AD + W
i
As an explicit vector identity, this is
³
´
FAB î + FAC cos 120î + cos 135ĵ + cos 60k̂ +
µ
¶
2
2
1
+FAD − î + ĵ + k̂ − (100)(9.81)N k̂ = 0
3
3
3
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Find equations for components:
Given the vector equation
³
´
FAB î + FAC cos 120î + cos 135ĵ + cos 60k̂ +
µ
¶
1
2
2
+FAD − î + ĵ + k̂ − (100)(9.81)N k̂ = 0
3
3
3
we can find the following three equations for the components
FAB + FAC cos 120 − FAD
FAC cos 135 + FAD
FAC cos 60 + FAD
Physics 170 203 Week 3 Lecture 2
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=0
3
2
=0
3
2
− (100)(9.81)N = 0
3
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Solve linear equations
FAB + cos 120FAC
cos 135FAC
cos 60FAC
1
− FAD
3
2
+ FAD
3
2
+ FAD
3
=
0
=
0
=
(100)(9.81)N
Techniques:
• algebra
• matrix inversion
• use Maple, or your caluclator
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Solve linear equations using algebra
1
FAB + cos 120FAC − FAD = 0
3
2
cos 135FAC + FAD = 0
3
2
cos 60FAC + FAD = (100)(9.81)N
3
Look for easiest identities first. Subtract second from third eqn.
(cos 60 − cos 135)FAC = (100)(9.81)N , FAC =
(100)(9.81)N
(cos 60 − cos 135)
Use second equation
FAD = −
3 cos 135
3 cos 135 (100)(9.81)N
FAC = −
2
2
(cos 60 − cos 135)
Use first equation
(100)(9.81)(− cos 120 − (1/2)(cos 135))N
1
FAB = − cos 120FAC + FAD =
3
(cos 60 − cos 135)
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FAC
=
FAD
=
FAB
=
FAB = 694N
(100)(9.81)N
(cos 60 − cos 135)
3 cos 135 (100)(9.81)N
−
2
(cos 60 − cos 135)
(100)(9.81)(− cos 120 − (1/2)(cos 135))N
(cos 60 − cos 135)
FAC = 813N
FAD = 862N
Spring stretches according to
FAB = ks → 693.7N = (1500N/m)s
s = .462m
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Solve linear equations by inverting a matrix
FAB + cos 120FAC
cos 135FAC
cos 60FAC
1
− FAD
3
2
+ FAD
3
2
+ FAD
3
=
0
=
0
=
(100)(9.81)N
Assemble the linear equations into one matrix equation:


 

1 cos 120 − 13
FAB
0
2 
 0 cos 135

FAC  = 
0
3
2
0 cos 60
FAD
(100)(9.81)N
3
Check that the rules of matrix multiplication
(matrix)(vector)=(vector) reproduce the linear equations when they
are applied.
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Now, we must

1
0
0
solve the matrix equation

 

0
cos 120 − 13
FAB
2 

FAC  = 
0
cos 135
3
2
FAD
(100)(9.81)N
cos 60
3
It is solved by finding the inverse of the matrix:

−1 
 
1 cos 120 − 13
1 cos 120 − 13
1
2 
2 
 0 cos 135
 0 cos 135
= 0
3
3
2
2
0 cos 60
0
cos
60
0
3
3
0
1
0

0
0
1
Multiply each side of the equation by the inverse matrix:

 
−1 

FAB
1 cos 120 − 13
0
2 
 FAC  =  0 cos 135


0
3
2
FAD
0 cos 60
(100)(9.81)N
3
This is an eqn for the forces. They can be found if the inverse is
known.
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How to find the inverse,

1
0
0
cos 120
cos 135
cos 60
− 13
2
3
2
3
−1

Sign on to your physics account. Go to maple(either in windows or
by typing maple as a unix command line). Use the maple
command:
f := linalg[matrix](3, 3, [1, cos(120), −1/3, 0, cos(135), 2/3, 0, cos(60), 2/3]);
This inputs the matrix. Then say
linalg[inverse](M );
and it will give the inverse:

cos(120)+cos(60)
1 − 12 2cos(135)−cos(60)
0
1

cos(135)−cos(60)
0
−3 cos(60)/3
cos(135)−cos(60)
Physics 170 203 Week 3 Lecture 2
1 2 cos 120+cos(135)
2 cos(135)−cos(60)
−1
cos(135)−cos(60)
3 cos(135)/2
cos(135)−cos(60)



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Then, the final answer is obtained by doing the matrix
multiplication

 
FAB
1
 FAC  =  0
FAD
0
or



1
FAB
 FAC  = 
0
FAD
0
cos 120
cos 135
cos 60
− 13
2
3
2
3
cos(120)+cos(60)
− 12 2cos(135)−cos(60)
1
cos(135)−cos(60)
−3 cos(60)/3
cos(135)−cos(60)
−1 


0


0
(100)(9.81)N
1 2 cos 120+cos(135)
2 cos(135)−cos(60)
−1
cos(135)−cos(60)
3 cos(135)/2
cos(135)−cos(60)


0

0 

981N
Matrix multiplication gives

  1 2 cos 120+cos(135) 981N 
FAB
2 cos(135)−cos(60)

−1
 FAC  = 
 cos(135)−cos(60) 981N 
3 cos(135)/2
FAD
981N
cos(135)−cos(60)
Calculator leads to same result as before.
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For the next lecture, please read
Textbook Chapter 4: Section 4.1-2
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