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

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

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.

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.

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.

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.

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.

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

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.

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).

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.

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

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

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.

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.

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

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.

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.

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

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

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

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

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

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

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

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

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.

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.

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

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

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.

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

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

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

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

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

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.

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.

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.

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

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.

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

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.

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