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Chapter 4A. Translational Chapter 4A. Translational
EquilibriumEquilibrium
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
A PowerPoint Presentation byA PowerPoint Presentation by
Paul E. Tippens, Professor of PhysicsPaul E. Tippens, Professor
of Physics
Southern Polytechnic State UniversitySouthern Polytechnic State
University
© 2007
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A MOUNTAIN CLIMBER exerts action forces on crevices and ledges,
which produce reaction forces on the climber, allowing him to scale
the cliffs.
Photo by Photo Disk Vol. 1/Getty
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Objectives: After completing this Objectives: After completing
this module, you should be able to:module, you should be able
to:
•• State and describe with examples NewtonState and describe
with examples Newton’’s s three laws of motion.three laws of
motion.
•• State and describe with examples your State and describe with
examples your understanding of the understanding of the first
condition for first condition for equilibriumequilibrium..
•• Draw Draw freefree--body diagramsbody diagrams for objects in
for objects in translational equilibrium.translational
equilibrium.
•• Write and apply the Write and apply the first condition for
first condition for equilibriumequilibrium to the solution of
problems to the solution of problems similar to those in this
module. similar to those in this module.
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NewtonNewton’’s First Laws First Law
Newton’s First Law: An object at rest or an object in motion at
constant speed will remain at rest or at constant speed in the
absence of a resultant force.
NewtonNewton’’s First Law:s First Law: An object at rest or an
An object at rest or an object in motion at constant speed will
remain object in motion at constant speed will remain at rest or at
constant speed in the absence of a at rest or at constant speed in
the absence of a resultant force.resultant force.
A glass is placed on a board and the board is jerked quickly to
the right. The glass tends to remain at rest while the board is
removed.
A glass is placed on a board and the board is jerked quickly to
the right. The glass tends to remain at rest while the board is
removed.
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NewtonNewton’’s First Law (Cont.)s First Law (Cont.)
Newton’s First Law: An object at rest or an object in motion at
constant speed will remain at rest or at constant speed in the
absence of a resultant force.
NewtonNewton’’s First Law:s First Law: An object at rest or an
An object at rest or an object in motion at constant speed will
remain object in motion at constant speed will remain at rest or at
constant speed in the absence of a at rest or at constant speed in
the absence of a resultant force.resultant force.
Assume glass and board move together at constant speed. If the
board stops suddenly, the glass tends to maintain its constant
speed.
Assume glass and board move together at constant speed. If the
board stops suddenly, the glass tends to maintain its constant
speed.
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Understanding the First Law:Understanding the First Law:
(a) The driver is forced to move forward. An object at rest
tends to remain at rest.
Discuss what the driver experiences when a car accelerates from
rest and then applies the brakes.
(b) Driver must resist the forward motion as brakes are applied.
A moving object tends to remain in motion.
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NewtonNewton’’s Second Laws Second Law
Newton’s second law of motion will be discussed quantitatively
in a later chapter, after we have covered acceleration.
Newton’s second law of motion will be discussed quantitatively
in a later chapter, after we have covered acceleration.
Acceleration is the rate at which the speed of an object
changes. An object with an acceleration of 2 m/s2, for example, is
an object whose speed increases by 2 m/s every second it
travels.
Acceleration is the rate at which the speed of an object
changes. An object with an acceleration of 2 m/s2, for example, is
an object whose speed increases by 2 m/s every second it
travels.
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NewtonNewton’’s Second Law:s Second Law:
• Second Law: Whenever a resultant force acts on an object, it
produces an acceleration - an acceleration that is directly
proportional to the force and inversely proportional to the
mass.
•• Second Law:Second Law: Whenever a resultant force Whenever a
resultant force acts on an object, it produces an acts on an
object, it produces an acceleration acceleration -- an acceleration
that is an acceleration that is directly proportional to the force
and directly proportional to the force and inversely proportional
to the mass.inversely proportional to the mass.
Fam
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Acceleration and Force With Acceleration and Force With Zero
Friction ForcesZero Friction Forces
Pushing the cart with twice the force produces twice the
acceleration. Three times the force triples the acceleration.
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Acceleration and Mass Acceleration and Mass Again With Zero
FrictionAgain With Zero Friction
F F
aa/2
Pushing two carts with same force F produces one-half the
acceleration. The acceleration varies inversely with the amount of
material (the mass).
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NewtonNewton’’s Third Laws Third Law
• To every action force there must be an equal and opposite
reaction force.
•• To every action force there must be an To every action force
there must be an equal and opposite reaction force. equal and
opposite reaction force.
Force of Hands on Wall
Force of Wall on HandsForce
of Floor on Man
Force of Man on Floor
Force of Ceiling on Man
Force of Man on Ceiling
Action and reaction forces act on different objects.
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NewtonNewton’’s Third Laws Third LawTwo More Examples:Two More
Examples:Two More Examples:
Action and Reaction Forces Act on Different Objects. They Do Not
Cancel Each Other!
ActionReaction
Action
Reaction
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Translational EquilibriumTranslational Equilibrium
•• An object is said to be in An object is said to be in
Translational EquilibriumTranslational Equilibrium if if and only
if there is no and only if there is no resultant force. resultant
force.
•• This means that the sum This means that the sum of all acting
forces is zero.of all acting forces is zero.
In the example, the resultant of the three forces A, B, and C
acting on the ring must be zero.
In the example, the resultant of the three forces A, B, and C
acting on the ring must be zero.
A
C
B
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Visualization of ForcesVisualization of ForcesForce diagrams are
necessary for studying objects in equilibrium. Don’t confuse action
forces with reaction forces.
Equilibrium:
0F The action forces are each
ON the ring.
AB
C
• Force A: By ceiling on ring.
• Force B: By ceiling on ring.
• Force C: By weight on ring.
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Visualization of Forces (Cont.)Visualization of Forces
(Cont.)Now let’s look at the Reaction Forces for the same
arrangement. They will be equal, but opposite, and they act on
different objects.
Reaction forces:
Reaction forces are each exerted: BY the ring.
ArBr
Cr
• Force Ar : By ring on ceiling.
• Force Br : By ring on ceiling.
• Force Cr : By ring on weight.
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Vector Sum of ForcesVector Sum of Forces
•• An object is said to be in An object is said to be in
Translational EquilibriumTranslational Equilibrium if and only if
there is no if and only if there is no resultant force. resultant
force.
•• The vector sum of all The vector sum of all forces acting
forces acting onon the ring the ring is zero in this case.is zero
in this case.
W
400
AB
C
Vector sum: F = A + B + C = 0
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Vector Force DiagramVector Force Diagram
W
400
AB
C
W
400
A
B
C Ax
Ay
A free-body diagram is a force diagram
Ay
showing all the elements in this diagram: axes, vectors,
components, and angles.
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FreeFree--body Diagrams:body Diagrams:
• Read problem; draw and label sketch.
• Isolate a common point where all forces are acting.
• Construct force diagram at origin of x,y axes.
• Dot in rectangles and label x and y components opposite and
adjacent to angles.
• Label all given information and state what forces or angles
are to be found.
•• Read problem; draw and label sketch.Read problem; draw and
label sketch.
•• Isolate a common point where all forces are Isolate a common
point where all forces are acting.acting.
•• Construct force diagram at origin of x,y axes.Construct force
diagram at origin of x,y axes.
•• Dot in rectangles and label x and y components Dot in
rectangles and label x and y components opposite and adjacent to
angles.opposite and adjacent to angles.
•• Label all given information and state what Label all given
information and state what forces or angles are to be found.forces
or angles are to be found.
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Look Again at Previous ArrangementLook Again at Previous
Arrangement
W
400
AB
C
1. Isolate point.
2. Draw x,y axes.
3. Draw vectors.
4. Label components.
5. Show all given information.
A
400
W
AyB
C
Ay
Ax
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Example 1. Example 1. Draw a freeDraw a free--body diagram for
the body diagram for the arrangement shown on left. The pole is
light arrangement shown on left. The pole is light and of
negligible weight.and of negligible weight.
W
300A
BC
700 N
Careful: The pole can only push or pull since it
has no weight.
The force B is the force exerted on the rope by the pole. Don’t
confuse it with the reaction force exerted by the rope on the
pole.
The force B is the force exerted on the rope by the pole. Don’t
confuse it with the reaction force exerted by the rope on the
pole.
B 300
A
C
700 N
Ay
Ax
Isolate the rope at the end of the boom. All forces must act ON
the rope!
On On roperopeB
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Translational EquilibriumTranslational Equilibrium
•• The The First Condition for First Condition for
EquilibriumEquilibrium is that there be is that there be no
resultant force. no resultant force.
•• This means that the sum of This means that the sum of all
acting forces is zero.all acting forces is zero.
0xF 0yF
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Example 2.Example 2. Find the tensions in ropes A Find the
tensions in ropes A and B for the arrangement shown.and B for the
arrangement shown.
200 N
400
AB
C
The Resultant Force on the ring is zero:
R = F = 0
Rx = Ax + Bx + Cx = 0
Ry = Ay + By + Cy = 0
200 N
400
A
B
C Ax
Ay Ay
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Example 2. (Cont.) Finding components.Example 2. (Cont.) Finding
components.
Recall trigonometry
to find components:
The components of the vectors are
found from the free- body diagram.
200 N
400
A
B
C Ax
Ay
BxCy Cx = 0
Cy = -200 N
Opp = Hyp x sin
Adj = Hyp x cosAx = A cos 400
Ay = A sin 400A
By = 0
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Example 2. Continued . . .Example 2. Continued . . .
W
400
A
B
C
A free-body diagram must represent all forces as components
along x and y-axes. It must also show all given information.
A free-body diagram must represent all forces as components
along x and y-axes. It must also show all given information.
Components
Ax = A cos 400
Ay = A sin 400
Bx = B; By = 0
Cx = 0; Cy = W
Ax
Ay Ay
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Example 2. Continued . . .Example 2. Continued . . .
200 N
400
AB
C
200 N
400
A
BC Ax
Ay Ay
Fx = 0 Fy = 0
Components
Ax = A cos 400
Ay = A sin 400
Bx = B; By = 0
Cx = 0; Cy = W
0 0cos40 0; or cos40x B AF A B 0 0sin 40 200 N 0; sin 40 200 Nor
yF AA
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Example 2. Continued . . .Example 2. Continued . . .
200 N
400
A
BC Ax
Ay Ay
Solve first for A 0
200 N 311 Nsin 40
A
0 0cos40 (311 N)cos40 ; B =238 NB A
Solve Next for B
The tensions in A and B are A = 311 N; B = 238 N
Two equations;
two unknowns
0cos 40B A
0sin 40 200 NA
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Problem Solving StrategyProblem Solving Strategy
1. Draw a sketch and label all information.
2. Draw a free-body diagram.
3. Find components of all forces (+ and -).
4. Apply First Condition for Equilibrium:
Fx = 0 ; Fy = 0
5. Solve for unknown forces or angles.
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Example 3.Example 3. Find Tension in Ropes A and B.Find Tension
in Ropes A and B.
300 600
A B
400 N
AB
400 N
1. Draw free1. Draw free--body diagram.body diagram.
2. Determine angles.2. Determine angles.
300 600 300 600Ay
By
Ax Bx
3. Draw/label components.3. Draw/label components.
Next we will find components of each vector.
Next we will find components of each vector.
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Example 3.Example 3. Find the tension in ropes A and B.Find the
tension in ropes A and B.
Fx = Bx - Ax = 0
Fy = By + Ay - W = 0
Bx = AxBy + Ay = W
AB
W 400 N
300 600Ay
By
Ax Bx
4. Apply 1st Condition for Equilibrium:
First Condition for Equilibrium:
Fx = 0 ; Fy = 0
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Example 3.Example 3. Find the tension in ropes A and B.Find the
tension in ropes A and B.
Bx = Ax
By + Ay = W
AB
W 400 N
300 600Ay
By
Ax Bx
Using Trigonometry, the first condition yields:
B cos 600 = A cos 300
A sin 300 + B sin 600 = 400 N
Ax = A cos 300; Ay = A sin 300
Bx = B cos 600
By = B sin 600
Wx = 0; Wy = -400 N
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Example 3 (Cont.)Example 3 (Cont.) Find the tension in A and
B.Find the tension in A and B.
AB
W 400 N
300 600Ay
By
Ax Bx
0
0
cos30 1.73cos 60
AB A B = 1.732 AB = 1.732 A
We will first solve the horizontal equation for B in terms of
the unknown A:
B cos 600 = B cos 300
A sin 300 + B sin 600 = 400 N
We now solve for A and B: Two Equations and Two Unknowns.
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Example 3 (Cont.)Example 3 (Cont.) Find Tensions in A and B.
Find Tensions in A and B.
A sin 300 + B sin 600 = 400 NB = 1.732 A
A sin 300 + (1.732 A) sin 600 = 400 N
0.500 A + 1.50 A = 400 N A = 200 NA = 200 N
AB
400 N
300 600Ay
By
Ax Bx
B = B = 1.7321.732 AANow apply Trig to:
Ay + By = 400 N
A sin 600 + B sin 600 = 400 N
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Example 3 (Cont.)Example 3 (Cont.) Find B with A = 200 N.Find B
with A = 200 N.
Rope tensions are: Rope tensions are: A = A = 200 N200 N and and
B = B = 346 N346 N
This problem is made much simpler if you notice This problem is
made much simpler if you notice that the angle between vectors B
and A is 90that the angle between vectors B and A is 900 0 and and
rotate the x and y axes (Continued)rotate the x and y axes
(Continued)
B = 1.732 A
A = A = 200 N200 N
B = 1.732(400 N)
B = B = 346 N346 N
AB
W 400 N
300 600Ay
By
Ax Bx
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Example 4.Example 4. Rotate axes for same example.Rotate axes
for same example.
300 600A B
400 N
AB
400 N
300 600 300 600Ay
By
Ax Bx
We recognize that We recognize that AA and and BB are at right
angles, are at right angles, and choose the and choose the
xx--axisaxis along along BB –– not horizontally. not horizontally.
The The yy--axisaxis will then be along will then be along
AA——withwith W W offsetoffset..
xy
WW
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Since Since AA and and BB are perpendicular, we can are
perpendicular, we can find the new angle find the new angle
from geometry.from geometry.
You should show that the angle will be 300. We now only work
with components of W.
xy
AB
300 600
400 N
A B
W =400 N
xy
606000303000
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Recall Recall W = W = 400 N. Then we have:400 N. Then we
have:
Apply the first condition for Equilibrium, and . . . Apply the
first condition for Equilibrium, and . . .
AABB
xy
300WW xx
WW yy
WW xx = = (400 N)(400 N) coscos 303000
WW yy = = (400 N)(400 N) sin 30sin 3000
Thus, the components of Thus, the components of the weight
vector are:the weight vector are:
WW xx = = 346 N; 346 N; WW yy = = 200 N200 N
B – Wx = 0 and A – Wy = 0B – Wx = 0 and A – Wy = 0
400 N
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Example 4 (Cont.)Example 4 (Cont.) We Now Solve for A and B:We
Now Solve for A and B:
Fx = B - Wx = 0
Fy = A - Wy = 0
B = Wx = (400 N) cos 300
B = 346 NB = 346 N
A = Wy = (400 N) sin 300
A = 200 NA = 200 N
AABB
400 N400 N
xy
303000WWxx
WWyy
Before working a Before working a problem, you might problem,
you might see if rotation of the see if rotation of the
axes helps.axes helps.
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SummarySummary
• Newton’s First Law: An object at rest or an object in motion
at constant speed will remain at rest or at constant speed in the
absence of a resultant force.
•• NewtonNewton’’s First Law:s First Law: An object at rest or
an An object at rest or an object in motion at constant speed will
object in motion at constant speed will remain at rest or at
constant speed in the remain at rest or at constant speed in the
absence of a resultant force.absence of a resultant force.
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SummarySummary
• Second Law: Whenever a resultant force acts on an object, it
produces an acceleration, an acceleration that is directly
proportional to the force and inversely proportional to the
mass.
•• Second Law: Whenever a resultant force Second Law: Whenever a
resultant force acts on an object, it produces an acts on an
object, it produces an acceleration, an acceleration that is
acceleration, an acceleration that is directly proportional to the
force and directly proportional to the force and inversely
proportional to the mass.inversely proportional to the mass.
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SummarySummary• Third Law: To every action force there must
be
an equal and opposite reaction force. •• Third Law: To every
action force there must be Third Law: To every action force there
must be
an equal and opposite reaction force. an equal and opposite
reaction force.
ActionReaction
Action
Reaction
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FreeFree--body Diagrams:body Diagrams:
• Read problem; draw and label sketch.
• Isolate a common point where all forces are acting.
• Construct force diagram; At origin of x,y axes.
• Dot in rectangles and label x and y components opposite and
adjacent to angles.
• Label all given information and state what forces or angles
are to be found.
•• Read problem; draw and label sketch.Read problem; draw and
label sketch.
•• Isolate a common point where all forces are Isolate a common
point where all forces are acting.acting.
•• Construct force diagram; At origin of x,y axes.Construct
force diagram; At origin of x,y axes.
•• Dot in rectangles and label x and y components Dot in
rectangles and label x and y components opposite and adjacent to
angles.opposite and adjacent to angles.
•• Label all given information and state what Label all given
information and state what forces or angles are to be found.forces
or angles are to be found.
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Translational EquilibriumTranslational Equilibrium
•• The The First Condition for First Condition for
EquilibriumEquilibrium is that there be is that there be no
resultant force. no resultant force.
•• This means that the sum This means that the sum of all acting
forces is zero.of all acting forces is zero.
0xF 0yF
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Problem Solving StrategyProblem Solving Strategy
1. Draw a sketch and label all information.
2. Draw a free-body diagram.
3. Find components of all forces (+ and -).
4. Apply First Condition for Equilibrium:
Fx = 0 ; Fy = 0
5. Solve for unknown forces or angles.
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Conclusion: Chapter 4AConclusion: Chapter 4A Translational
EquilibriumTranslational Equilibrium
Chapter 4A. Translational EquilibriumSlide Number 2Objectives:
After completing this module, you should be able to:Newton’s First
LawNewton’s First Law (Cont.)Understanding the First Law:Newton’s
Second LawNewton’s Second Law:Acceleration and Force With Zero
Friction ForcesAcceleration and Mass Again With Zero
FrictionNewton’s Third LawNewton’s Third LawTranslational
EquilibriumVisualization of ForcesVisualization of Forces
(Cont.)Vector Sum of ForcesVector Force DiagramFree-body
Diagrams:Look Again at Previous ArrangementExample 1. Draw a
free-body diagram for the arrangement shown on left. The pole is
light and of negligible weight.Translational EquilibriumExample 2.
Find the tensions in ropes A and B for the arrangement
shown.Example 2. (Cont.) Finding components.Example 2. Continued .
. .Example 2. Continued . . .Example 2. Continued . . .Problem
Solving StrategyExample 3. Find Tension in Ropes A and B.Example 3.
Find the tension in ropes A and B.Example 3. Find the tension in
ropes A and B.Example 3 (Cont.) Find the tension in A and B.Example
3 (Cont.) Find Tensions in A and B. Example 3 (Cont.) Find B with A
= 200 N.Example 4. Rotate axes for same example.Since A and B are
perpendicular, we can find the new angle f from geometry.Recall W =
400 N. Then we have:Example 4 (Cont.) We Now Solve for A and
B:SummarySummarySummaryFree-body Diagrams:Translational
EquilibriumProblem Solving StrategyConclusion: Chapter
4A�Translational Equilibrium