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Equilibrium at a Point “Never slap a man who's chewing tobacco.” -­‐Will Rogers Objec3ves ¢ 
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Understand the concept of sta3c equilibrium Understand the use of the free-­‐body diagram to isolate a system for analysis Understand the reac3on provided by the connec3on to a rope Understand the reac3on provided by the connec3on to a spring Particle Equilibrium
Wednesday, September 12, 2011
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Tools ¢  Basic Trigonometry ¢  Pythagorean Theorem ¢  Algebra ¢  Scalar analysis of forces 3
Particle Equilibrium
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Scalar Representa3on ¢  We noted that the magnitude of forces is always posi3ve. ¢  A nega3ve sign in front of a magnitude actually represents the direc3on of the force and not its absolute magnitude. ¢  A posi3ve sign is in the direc3on of the label. ¢  A nega3ve sign is in the direc3on away from the label. 4
Particle Equilibrium
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Equilibrium ¢  The unifying concept to this course is that of sta3c equilibrium ¢  Sta3c equilibrium in this course means that there is no change in velocity with 3me 5
Particle Equilibrium
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Equilibrium ¢  Stated mathema3cally, sta3c equilibrium would be 6

Δv
=0
Δt
Particle Equilibrium
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Equilibrium ¢  If we express this as an instantaneous change in velocity, the expression becomes 
dv
=0
dt
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Particle Equilibrium
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Equilibrium ¢  From physics, you may remember that instantaneous change in velocity is also known as accelera3on so  
dv
=a=0
dt
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Particle Equilibrium
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Equilibrium ¢  Now also from physics, you may remember that accelera3on is produced by the ac3on of a force on a mass ¢  This is Newton’s first law of mo3on  
dv
=a=0
dt


F = ma
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Particle Equilibrium
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Equilibrium ¢  Mass is a scalar quan3ty so there is no direc3on assigned to it. Force however will have the same direc3on as the accelera3on 10
 
dv
=a=0
dt


F = ma
Particle Equilibrium
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Equilibrium ¢  We can then remove some of the elements in our long expression to get to the heart of the maOer in this course 
 
dv
F
=a=0=
dt
m

F =0
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Particle Equilibrium
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Equilibrium ¢  This means that the net resultant force ac3ng on a body must be equal to zero for the body to be in equilibrium 
F =0
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Particle Equilibrium
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Equilibrium ¢  If we have a number of forces ac3ng on a body, then the vector sum of those forces (the resultant) must be equal to 0 
∑ Fi = 0
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Particle Equilibrium
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Equilibrium ¢  Since the resultant is equal to 0, the coefficient of each of the components of the resultant must be equal to 0 
∑ Fx i= 0
∑ Fy j = 0
∑ Fz k = 0
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Particle Equilibrium
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Equilibrium ¢  If we take the sign from the direc3on of the vector components 3mes the magnitude for each of the components we can then write 15
∑F
∑F
∑F
x
=0
y
=0
z
=0
Particle Equilibrium
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Free-­‐Body Diagram ¢  The second important concept that we introduce is the idea of a free-­‐body diagram ¢  The free-­‐body diagram is an isola3on of an element and the iden3fica3on of all forces which are ac3ng on that element ¢  By an element, we mean a part of a mechanical system 16
Particle Equilibrium
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Free-­‐Body Diagram ¢  Forces can act on a system as either an applied force from some external source or ¢  Forces can act due to the connec3on of the system we have isolated to some other system in response to the applied forces – these forces are known as reac3ons 17
Particle Equilibrium
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Free-­‐Body Diagram ¢  In order to be able to draw a correct free-­‐body diagram (I will use FBD for the free-­‐body diagram) we have to understand what types of forces/reac3ons are generated through different types of connec3ons 18
Particle Equilibrium
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Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
spring.
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Particle Equilibrium
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Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
spring.
Nothing is the system is
accelerating. Since the block is
at rest, nothing is moving at all.
What forces are acting on the
block?
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Particle Equilibrium
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Springs ¢  Springs are systems that exert forces that are propor3onal to the condi3on of the spring ¢  Springs may be in compression (pushed together) or in tension (stretched apart) 21
Particle Equilibrium
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Springs ¢  Springs that are in compression tend to push on the objects that they are connected to ¢  Springs that are in tension (stretched) tend to pull on the objects that they are connected to ¢  Springs always generate a force that is along the axis of the spring 22
Particle Equilibrium
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Springs The force that they exert will be propor3onal to the amount they are compressed or extended ¢  The magnitude of the force will be given by ¢ 
F = k ( Δl )
k is the spring constant and has dimensions of force/length
Δl is the amount by which the spring changes from its original unstretched length.
If Δl is positive, the spring is stretched and it pulls on what it is connected to.
If Δl is negative, the spring is compressed and it pushes on what it is connected
to.
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Particle Equilibrium
Wednesday, September 12, 2011
Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
spring.
We will assume that the spring
is being stretched and draw a
force pulling on the block along
the line of the spring.
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Weight ¢  Weight always acts toward the center of the earth ¢  In typical problems, that will be towards the boOom of the page ¢  If the weight or mass of an element is not given, you may assume that it is negligible in the analysis that is being done 25
Particle Equilibrium
Wednesday, September 12, 2011
Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
spring.
We can add a force
representing the action of
gravity on the block (the weight)
acting straight down.
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Particle Equilibrium
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Smooth Surface ¢  A smooth or fric3onless surface provides a force that is perpendicular to the surface. ¢  It only stops something from moving through it. 27
Particle Equilibrium
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Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
spring.
Finally, we add a force
representing the reaction of the
smooth surface on the box.
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Particle Equilibrium
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Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
spring.
We can consider the box as a
point and make all the force
coincident at a point.
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Particle Equilibrium
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Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
spring.
F1
We can label each of the forces
acting on the system to make
the work and representation a
bit easier.
F2
F3
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Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
y
spring.
x
F1
In order to make the math
easier, we can set the x-axis
along the 0.4m side and the yaxis parallel to the 0.3m side.
F2
F3
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Particle Equilibrium
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Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
y
spring.
x
F1
Now we can write each of the
forces in Cartesian form.
4
F1
5
3
F1y = F1
5
F1x =
32
F2
F3
Particle Equilibrium
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Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
y
spring.
x
F1
Now we can write each of the
forces in Cartesian form.
F2
F2 x = 0
F2 y = F2
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F3
Particle Equilibrium
Wednesday, September 12, 2011
Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
y
spring.
x
F1
Now we can write each of the
forces in Cartesian form.
F3x = −F3 cos ( 45 0 )
F2
F3y = −F3 sin ( 45 0 )
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F3
Particle Equilibrium
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Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
y
spring.
x
F1
Summing the forces in the x
and y directions we have.
4
F1 + 0 − F3 cos ( 45 0 )
5
3
∑ Fy = 5 F1 + F2 − F3 sin ( 45 0 )
∑F
x
=
35
F2
F3
Particle Equilibrium
Wednesday, September 12, 2011
Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
y
spring.
x
F1
Both of these sums must be
equal to 0.
4
F1 + 0 − F3 cos ( 45 0 ) = 0
5
3
∑ Fy = 5 F1 + F2 − F3 sin ( 45 0 ) = 0
∑F
x
36
=
F2
Particle Equilibrium
F3
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Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
y
spring.
x
F1
F3 is equal to the weight of the
block and weight is equal to
mass times gravity.
4
F1 + 0 − mg cos ( 45 0 ) = 0
5
3
∑ Fy = 5 F1 + F2 − mgsin ( 45 0 ) = 0
∑F
x
=
37
F2
F3
Particle Equilibrium
Wednesday, September 12, 2011
Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
y
spring.
x
F1
Substituting what we know
∑F
=
4
m⎞
⎛
F1 + 0 − ( 5kg ) ⎜ 9.81 2 ⎟ cos ( 45 0 ) = 0 F
2
⎝
5
s ⎠
∑F
=
3
m⎞
⎛
F1 + F2 − ( 5kg ) ⎜ 9.81 2 ⎟ sin ( 45 0 ) = 0
⎝
5
s ⎠
x
y
F3
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Particle Equilibrium
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Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
y
spring.
x
F1
Using the first equation and
isolating F1.
4
m⎞
⎛
F1 + 0 − ( 5kg ) ⎜ 9.81 2 ⎟ cos ( 45 0 ) = 0 F2
⎝
5
s ⎠
5
m
( 5kg ) ⎛⎜⎝ 9.81 2 ⎞⎟⎠ cos ( 45 0 )
4
s
F1 = 43.35N
F1 =
39
F3
Particle Equilibrium
Wednesday, September 12, 2011
Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
y
spring.
x
F1
Using the expression for a
spring.
F1 = 43.35N
F1 = k1 ( Δl )
F2
N⎞
⎛
43.35N = ⎜ 200 ⎟ ( Δl )
⎝
m⎠
F3
Δl = 0.22m
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Particle Equilibrium
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Problem F3-­‐4 The block has a mass of 5 kg
and rests on the smooth
plane. Determine the
unstretched length of the
y
spring.
So the spring is increased by
0.22m. From the diagram the
stretched length of the spring is
0.5m, so the original length of
the spring was 0.28m.
F1 = 43.35N
x
F1
F2
F1 = k1 ( Δl )
N⎞
⎛
43.35N = ⎜ 200 ⎟ ( Δl )
⎝
m⎠
41
Δl = 0.22m
F3
Particle Equilibrium
Wednesday, September 12, 2011
Problem 3-­‐27 The 10-lb weight A is
supported by the cord AC
and roller C, and by spring
AB. If the spring has an
unstretched length of 8 in,
and the weight is in
equilibrium when d = 4 in.,
determine the k of the spring.
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Homework ¢  Problem 3-­‐2 ¢  Problem 3-­‐6 ¢  Problem 3-­‐21 43
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