Equilibrium)ataPoint - Memphis Equilibrium.pdfEquilibrium)ataPoint ... 41 Particle Equilibrium...

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Equilibrium at a Point

“Never slap a man who's chewing tobacco.” -‐Will Rogers

Wednesday, September 12, 2011 Particle Equilibrium 2

Objec3ves

¢  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

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Tools

¢  Basic Trigonometry ¢  Pythagorean Theorem ¢  Algebra ¢  Scalar analysis of forces


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.

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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


Equilibrium

¢  Stated mathema3cally, sta3c equilibrium would be

ΔvΔt

= 0

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Equilibrium

¢  If we express this as an instantaneous change in velocity, the expression becomes

dv

dt= 0


Equilibrium

¢  From physics, you may remember that instantaneous change in velocity is also known as accelera3on so

dv

dt= a= 0

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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

dt= a= 0

F= ma


Equilibrium

¢ Mass is a scalar quan3ty so there is no direc3on assigned to it. Force however will have the same direc3on as the accelera3on

dv

dt= a= 0

F= ma

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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

dt= a= 0 = F

mF= 0


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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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

F

∑ i= 0


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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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

0

0

0

x

y

z

F

F

F

=

=

=

∑∑∑


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

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¢  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



¢  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

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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.

Problem F3-‐4



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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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)


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

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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.

Problem F3-‐4



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

Problem F3-‐4



We can add a force representing the action of gravity on the block (the weight) acting straight down.

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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.

Problem F3-‐4



Finally, we add a force representing the reaction of the smooth surface on the box.

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Problem F3-‐4



We can consider the box as a point and make all the force coincident at a point.

Problem F3-‐4



We can label each of the forces acting on the system to make the work and representation a bit easier.

F1

F2

F3

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Problem F3-‐4



In order to make the math easier, we can set the x-axis along the 0.4m side and the y-axis parallel to the 0.3m side.

F1

F2

F3

x

y

Problem F3-‐4



Now we can write each of the forces in Cartesian form.

F1

F2

F3

x

y

F1x =45F1

F1y =35F1

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Problem F3-‐4




F1

F2

F3

x

y

F2x = 0F2y = F2

Problem F3-‐4




F1

F2

F3

x

y

F3x = −F3 cos 450( )

F3y = −F3 sin 450( )

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Problem F3-‐4



Summing the forces in the x and y directions we have.

F1

F2

F3

x

y

Fx∑ = 45F1 + 0 − F3 cos 45

0( )Fy∑ = 3

5F1 + F2 − F3 sin 45

0( )

Problem F3-‐4



Both of these sums must be equal to 0.

F1

F2

F3

x

y

Fx∑ = 45F1 + 0 − F3 cos 45

0( ) = 0

Fy∑ = 35F1 + F2 − F3 sin 45

0( ) = 0

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Problem F3-‐4



F3 is equal to the weight of the block and weight is equal to mass times gravity.

F1

F2

F3

x

y

Fx∑ = 45F1 + 0 −mgcos 45

0( ) = 0

Fy∑ = 35F1 + F2 −mgsin 45

0( ) = 0

Problem F3-‐4



Substituting what we know F1

F2

F3

x

y

Fx∑ = 45F1 + 0 − 5kg( ) 9.81m

s2⎛⎝⎜

⎞⎠⎟ cos 45

0( ) = 0

Fy∑ = 35F1 + F2 − 5kg( ) 9.81m

s2⎛⎝⎜

⎞⎠⎟ sin 45

0( ) = 0

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Problem F3-‐4



Using the first equation and isolating F1.

F1

F2

F3

x

y

45F1 + 0 − 5kg( ) 9.81m

s2⎛⎝⎜

⎞⎠⎟ cos 45

0( ) = 0

F1 =545kg( ) 9.81m

s2⎛⎝⎜

⎞⎠⎟ cos 45

0( )F1 = 43.35N

Problem F3-‐4



Using the expression for a spring.

F1

F2

F3

x

y

F1 = 43.35NF1 = k1 Δl( )

43.35N = 200 Nm

⎛⎝⎜

⎞⎠⎟ Δl( )

Δl = 0.22m

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Problem F3-‐4



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

F2

F3

x

y

F1 = 43.35NF1 = k1 Δl( )

43.35N = 200 Nm

⎛⎝⎜

⎞⎠⎟ Δl( )

Δl = 0.22m

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

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