Chapter 5B Chapter 5B Rotational EquilibriumRotational EquilibriumA 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

TheThe Golden Gate Golden Gate Bridge Bridge provides provides an excellent an excellent example of example of balanced forces balanced forces and torques. and torques. Engineers must Engineers must design such design such structures so that structures so that rotational and rotational and translational translational equilibrium is equilibrium is maintained.maintained. Photo © EP 101 Photodisk/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 your State and describe with examples your

understanding of the understanding of the first and second first and second conditions for equilibriumconditions for equilibrium..

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

Translational EquilibriumTranslational Equilibrium

The linear speed is The linear speed is notnot changing with time. changing with time. There is no resultant force and therefore zero There is no resultant force and therefore zero acceleration. Translational equilibrium exists.acceleration. 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 with changing with time. There is no resultant torque and, time. There is no resultant torque and, therefore, zero change in rotational therefore, zero change in rotational velocity. Rotational equilibrium exists.velocity. Rotational 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 First First Condition:Condition:

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 EquilibriumStatics or Total Equilibrium

StaticsStatics is the physics that treats objects at is the physics that treats objects at rest or objects in constant motion.rest or objects in constant 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 OnlyTranslational Equilibrium OnlyIf 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: FreeReview: Free--body Diagramsbody Diagrams

• Read problem; draw and label sketch.

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

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

• Label all components; choose positive direction.

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

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

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

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

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

80 N

A B600

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

80 N

AB

600

Free-body Diagram:

By

Bx

•• Choose xChoose x--axis horizontal and choose right axis horizontal and choose right direction as positive (+). There is no motion.direction as positive (+). There is 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 and B.Find tension in ropes A and B.

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

80 N

AB

600

Free-body Diagram:

By

Bx

0; 0; x yF F

80 N

A

B sin B sin 606000

B cos 60o

Bx

By

FF xx = 0= 0FF yy = 0= 0

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

AB

W

350 550

Bx

By

Ax

Ay

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

Fy = By + Ay – 500 N = 0W = 500 N

350 550

A B

500 N

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

350 550

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 exerted by Find the forces exerted by supports supports AA and and BB. Neglect the weight . Neglect the weight of the of the 1010--mm boom.boom.

40 N 80 N

2 m 3 m7 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 m7 mA B

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

40 N 80 N

2 m 3 m7 mA 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 each magnitudes of each torque.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: (Cont.)Example 3: (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 m7 m

A B

40 N 80 N

2 m 3 m7 mA B

Force A is ignored: Neither ccw nor cw

40 N 80 N

2 m 3 m7 m

A B

40 N 80 N

2 m 3 m7 mA B

Example 3 (Cont.)Example 3 (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 NB = 48.0 N

40 N 80 N

2 m 3 m7 m

A B

40 N 80 N

2 m 3 m7 mA B

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

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

A + 48 N = 120 N

A = 72.0 NA = 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 m7 m

A B

40 N 80 N

2 m 3 m7 mA B

Example 3 (Cont.)Example 3 (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 m7 m

A B

40 N 80 N

2 m 3 m7 mA B

Reminder on Signs:Reminder on 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 = 0

We wrote: A + B = 40 N + 90 N

Example 4: Example 4: Find the tension in Find the tension in the rope and the force by the the rope and the force by the wall on the boom. The wall on the boom. The 1010--mm boom weighing boom weighing 200 N200 N. Rope is . Rope is 2 m2 m 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 N200 N

300

800 N200 N

TFx

Fy

2 m3 m5 m

300

T

800 N200 N

300

800 N200 N

TFx

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 NT = 2250 N

300

T

300

T

800 N200 N 800 N200 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, 356.30F = 1954 N, 356.30or

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 GravityExamples of Center of Gravity

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

Example 5:Example 5: Find the center of gravity of the Find the center of gravity of the apparatus shown below. Neglect the weight apparatus shown below. Neglect the weight of the connecting rods.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 mx = 2.00 m

SummarySummary

0xF

0yF

0

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

An object is said to An object is said to be in be in equilibriumequilibrium if if and only if there is and only if there is no resultant force no resultant force and no resultant and no resultant torque.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 5B Rotational EquilibriumRotational Equilibrium