Chapter 5B Rotational Equilibrium A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University A PowerPoint.

Chapter 5B Chapter 5B Rotational Rotational EquilibriumEquilibrium

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007

TheThe Golden Golden Gate Bridge Gate Bridge provides an provides an excellent excellent example of example of balanced balanced forces and forces and torques. torques. Engineers Engineers must design must design such structures such structures so that so that rotational and rotational and translational translational equilibrium is equilibrium is maintained.maintained.

Photo © EP 101 Photodisk/Getty

Objectives: After completing Objectives: After completing this module, you should be this module, you should be able to:able to:• State and describe with examples State and describe with examples

your understanding of the your understanding of the first and first and second conditions for equilibriumsecond conditions for equilibrium..

• Write and apply the Write and apply the first and second first and second conditions for equilibriumconditions for equilibrium to the to the solution of physical problems similar solution of physical problems similar to those in this module. to those in this module.

Translational EquilibriumTranslational Equilibrium

The linear speed is The linear speed is notnot changing with changing with time. There is no resultant force and time. There is no resultant force and therefore zero acceleration. therefore zero acceleration. Translational equilibrium exists.Translational equilibrium exists.

Car at rest Constant speed

0; F 0; No change in a v

Rotational EquilibriumRotational Equilibrium

The angular speed is The angular speed is notnot changing changing with time. There is no resultant with time. There is no resultant torque and, therefore, zero change torque and, therefore, zero change in rotational velocity. Rotational in rotational velocity. Rotational equilibrium exists.equilibrium exists.

Wheel at rest

Constant rotation

0; . No change in rotation

EquilibriumEquilibrium

• An object is said to be in An object is said to be in equilibriumequilibrium if if and only if there is no resultant force and only if there is no resultant force and no resultant torque.and no resultant torque.

0; 0x yF F 0; 0x yF F First First Condition:Condition:

0 0 Second Second Condition:Condition:

Does Equilibrium Exist?Does Equilibrium Exist?

A sky diver who reaches terminal speed?

A fixed pulley rotating at constant speed?

A sky diver moments after the jump?

Yes or No? Yes

Yes

300

TIs the system at left in

equilibrium both translationally and

rotationally?

YES! Observation shows that no part of

the system is changing its state of

motion.

No

Statics or Total Statics or Total EquilibriumEquilibrium

StaticsStatics is the physics that treats is the physics that treats objects at rest or objects in constant objects at rest or objects in constant

motion.motion.In this module, we will review the first condition for equilibrium (treated in Part 5A of these modules); then we will extend our treatment by working with the second condition for equilibrium. Both conditions must be satisfied for true equilibrium.

In this module, we will review the first condition for equilibrium (treated in Part 5A of these modules); then we will extend our treatment by working with the second condition for equilibrium. Both conditions must be satisfied for true equilibrium.

Translational Equilibrium Translational Equilibrium OnlyOnly

If all forces act at the same point, then there is no torque to consider and one need only apply the first condition for equilibrium:

Fx= 0; Fy= 0

• Solve for unknowns.

• Construct free-body diagram.

• Sum forces and set to zero:

Review: Free-body Review: Free-body DiagramsDiagrams

• Read problem; draw and label sketch.Read problem; draw and label sketch.

• Construct force diagram for each Construct force diagram for each object, vectors at origin of x,y axes.object, vectors at origin of x,y axes.

• Dot in rectangles and label x and y Dot in rectangles and label x and y compo-nents opposite and adjacent to compo-nents opposite and adjacent to angles.angles.

• Label all components; choose positive Label all components; choose positive direction.direction.

• Read problem; draw and label sketch.Read problem; draw and label sketch.

• Construct force diagram for each Construct force diagram for each object, vectors at origin of x,y axes.object, vectors at origin of x,y axes.

• Dot in rectangles and label x and y Dot in rectangles and label x and y compo-nents opposite and adjacent to compo-nents opposite and adjacent to angles.angles.

• Label all components; choose positive Label all components; choose positive direction.direction.

Example 1.Example 1. Find the tension in ropes A Find the tension in ropes A and B.and B.

80 N

A B600

• Read problem; draw sketch; construct a Read problem; draw sketch; construct a free-body diagram, indicating components.free-body diagram, indicating components.

80 N

AB

600

Free-body Diagram:

By

Bx

• Choose x-axis horizontal and choose Choose x-axis horizontal and choose right direction as positive (+). There is right direction as positive (+). There is no motion.no motion.

Example 1 (Continued).Example 1 (Continued). Find Find AA and and BB..

80 N

A B600

80 N

AB

600

Free-body Diagram:

By

Bx

Bx = B cos 600; By = B sin 600Bx = B cos 600; By = B sin 600

Note: The components Bx and By can be found from right triangle trigonometry:

Example 1 (Cont.).Example 1 (Cont.). Find tension in ropes A Find tension in ropes A and B.and B.

• Apply the first condition for Apply the first condition for equilibrium.equilibrium.

80 N

AB

600

Free-body Diagram:

By

Bx

0; 0; x yF F 0; 0; x yF F

80 N

A

B sin B sin 606000

B cos 60o

Bx

By

FFxx = 0 = 0 FFyy = 0 = 0

Example 2.Example 2. Find tension in ropes A Find tension in ropes A and B.and B.

AB

W

350 550

Bx

By

Ax

Ay

Recall: Fx = Fy = 0 Fx = Bx - Ax = 0

Fy = By + Ay – 500 N = 0

W = 500 N

350 550

A B

500 N

Example 2 (Cont.)Example 2 (Cont.) Simplify by rotating Simplify by rotating axes:axes:

Recall that W = 500 N

Fx = B - Wx = 0

Fy = A - Wy = 0

B = Wx = (500 N) cos 350

B = 410 NB = 410 N

A = Wx = (500 N) sin 350

A = 287 NA = 287 N

350550

AB

W

Wx

Wy

xy

Total EquilibriumTotal EquilibriumIn general, there are six degrees of

freedom (right, left, up, down, ccw, and cw):

Fx= 0 Right = left

Fx= 0 Up = down

ccw (+) cw (-)

(ccw)= (ccw)

General Procedure:General Procedure:

• Draw free-body diagram and label.

• Choose axis of rotation at point where least information is given.

• Extend line of action for forces, find moment arms, and sum torques about chosen axis:

• Sum forces and set to zero: Fx= 0; Fy= 0

• Solve for unknowns.

Example 3:Example 3: Find the forces Find the forces exerted by supports exerted by supports AA and and BB. . Neglect the weight of the Neglect the weight of the 10-m10-m boom.boom.

40 N 80 N

2 m

3 m

7 m

A BDraw free-body

diagram

Rotational Equilibrium:

Choose axis at point of unknown

force.At A for

example.

40 N 80 N

2 m

3 m

7 m

A B

Example 3 Example 3 (Cont.)(Cont.)

40 N 80 N

2 m

3 m

7 m

A B

Torques about axis ccw are equal to

those cw.

ccw (+) cw (-)

(ccw) = (ccw) = (cw)(cw)

Note: When applying

we need only the we need only the absoluteabsolute (positive) (positive) magnitudes of magnitudes of each torque.each torque.

(+) = (-)

Essentially, we are saying that the torques are balanced about a chosen

axis.

Essentially, we are saying that the torques are balanced about a chosen

axis.

Example 3: Example 3: (Cont.)(Cont.)

Rotational Equilibrium:

or

(ccw) = (ccw) = (cw)(cw) With respect to Axis

A:CCW Torques:Forces B and 40 N.

CW Torques: 80 N force.

40 N 80 N

2 m

3 m

7 m

A B

40 N 80 N

2 m

3 m

7 m

A B

Force A is ignored: Neither ccw nor cw

40 N 80 N

2 m

3 m

7 m

A B

40 N 80 N

2 m

3 m

7 m

A B

Example 3 Example 3 (Cont.)(Cont.)

First: First: (ccw)(ccw)

1 = B (10 m)

2 = (40 N) (2 m) = 80 Nm

Next: Next: (cw)(cw)

3 = (80 N) (7 m) = 560 Nm

B(10 m) + 80 Nm = 560 Nm

(ccw) = (ccw) = (cw)(cw)

B = 48.0 N

B = 48.0 N

40 N 80 N

2 m

3 m

7 m

A B

40 N 80 N

2 m

3 m

7 m

A B

Example 3 Example 3 (Cont.)(Cont.)

F(up) = F(up) = F(down) F(down)

A + 48 N = 120 N

A = 72.0 N

A = 72.0 N

Translational Translational EquilibriumEquilibrium

Fx= 0; Fy= 0Fx= 0; Fy= 0

A + B = 40 N + 80 N

A + B = 120 N

Recall that B = 48.0 N

40 N 80 N

2 m

3 m

7 m

A B

40 N 80 N

2 m

3 m

7 m

A B

Example 3 Example 3 (Cont.)(Cont.)Check answer

by summing torques about right end to verify A = 72.0 N(ccw) = (ccw) = (cw)(cw)

(40 N)(12 m) + (80 N)(3 m) = A (10 m)480 Nm + 240 Nm = A (10

m)A = 72.0

NA = 72.0

N

40 N 80 N

2 m

3 m

7 m

A B

40 N 80 N

2 m

3 m

7 m

A B

Reminder on Reminder on Signs:Signs:

F(up) = F(up) = F(down) F(down)

Absolute values Absolute values apply for:apply for:

We used absolute (+) values for both

UP and DOWN terms.

Instead of: Fy = A + B – 40 N - 80 N = 0We wrote: A + B = 40 N + 90 N

Example 4: Example 4: Find the Find the tension in the rope and the tension in the rope and the force by the wall on the force by the wall on the boom. The boom. The 10-m10-m boom boom weighing weighing 200 N200 N. Rope is . Rope is 2 2 mm from right end. from right end.

300

T

800 N

For purposes of summing torques, we consider entire weight to act at center of

board.

For purposes of summing torques, we consider entire weight to act at center of

board.

300

T

800 N

200 N

300

800 N

200 N

T

Fx

Fy

2 m3 m5 m

300

T

800 N

200 N

300

800 N

200 N

T

Fx

Fy

2 m3 m5 m

Example 4 Example 4 (Cont.)(Cont.)

Choose axis of rotation at wall (least information)

(ccw):(ccw):

r

Tr = T (8 m)sin 300 = (4 m)T

(cw):(cw): (200 N)(5 m) + (800 N)(10 m) = 9000 Nm

(4 m)T = 9000 Nm

T = 2250 N

T = 2250 N

300

T

300

T

800 N

200 N

800 N

200 N

Fx

Fy

2 m3 m5 m

Example 4 Example 4 (Cont.)(Cont.)

300

Ty

Tx

F(up) = F(up) = F(down):F(down): Ty + Fy = 200 N + 800 N

Fy = 1000 N - T sin 300 Fy = -125

N

F(right) = F(right) = F(left):F(left): Fx = Ty = (2250 N) cos 300

Fx = 1950 N

F = 1954 N, 356.30

F = 1954 N, 356.30

or

Fy = 200 N + 800 N - Ty ;

Fy = 1000 N - (2250 N)sin 300

Center of GravityCenter of GravityThe center of gravity of an object is the point at which all the weight of an object might be considered as acting for purposes of treating forces and torques that affect the object.

The single support force has line of action that passes through the c. g. in any orientation.

Examples of Center of Examples of Center of GravityGravity

Note: C. of G. is not always inside material.

Example 5:Example 5: Find the center of gravity of Find the center of gravity of the apparatus shown below. Neglect the apparatus shown below. Neglect the weight of the connecting rods.the weight of the connecting rods.

30 N 10 N 5 N

4 m 6 mC. of G. is point at which a single up- upward force F will balance the system.

x

Choose axis at left, then sum

torques:(ccw) = (ccw) = (cw)(cw)

F

Fx = (10 N)(4 m) + (5 N)(10 m)

Fx = 90.0 Nm

F(up) = F(up) = F(down):F(down):

F = 30 N + 10 N + 5 N

(45 N) x = 90 N

x = 2.00 m

x = 2.00 m

SummarySummary

0xF 0xF

0yF 0yF

0 0

An object is said An object is said to be in to be in equilibriumequilibrium if if and only if there and only if there is no resultant is no resultant force and no force and no resultant torque.resultant torque.

An object is said An object is said to be in to be in equilibriumequilibrium if if and only if there and only if there is no resultant is no resultant force and no force and no resultant torque.resultant torque.

Conditions for Equilibrium:

Summary: ProcedureSummary: Procedure

• Draw free-body diagram and label.

• Choose axis of rotation at point where least information is given.

• Extend line of action for forces, find moment arms, and sum torques about chosen axis:

• Sum forces and set to zero: Fx= 0; Fy= 0

• Solve for unknowns.

CONCLUSION: Chapter 5BCONCLUSION: Chapter 5BRotational EquilibriumRotational Equilibrium