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05B: STATICS05B: STATICS

San Juanico Bridge

Objectives: After completing Objectives: After completing this lecture, you should be this lecture, 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 lecture. to those in this lecture.

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?

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.

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

In this lecture, we will review the first condition for equilibrium (treated in Part 5A); 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 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: F = 0 Fx = Bx - Ax

Fy = By + Ay - W

350 550

A B

500 N

Bx - Ax = 0 Bcos θB – Acos θA = 0

By + Ay – W = 0Bsin θB + Asin θA – W = 0

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

B = 410 NB = 410 N

A = 287 NA = 287 N

AB

w

350 550

Bx

By

Ax

Ay

B cos θB – Acos θA = 0

B sin θB + Asin θA – W = 0

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


Center of GravityCenter of GravityThe center of gravity of an object is the point at which its weight is concentrated. It is a point where the object does not turn or rotate.

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.

Finding the center of Finding the center of gravitygravity

01. Plumb-line Method

02. Principle of Moments

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

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

m1 = 4 kg

m2 = 1 kgm3 = 6 kg

3 cm

2 cmSince the system is 2-D: Choose now an axis of rotation/reference

axis:Trace in the Cartesian plane in a case that your axis of rotation is at the origin

(0,0)

(0, 2cm)

(-3 cm, 2cm)

CG

Example 6:Example 6: 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 N4 m 6 m

xChoose axis at left, then sum

torques:

x = 2.00 m

x = 2.00 m

C.G.

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

Since the boom is uniform, it’s weight is located at its center of gravity (geometric

center)

Since the boom is uniform, it’s weight is located at its center of gravity (geometric

center)

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

w2w1

T

Fx

Fy

2 m3 m5 m

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

Choose axis of rotation at wall (least information)

(ccw):(ccw):

r

(cw):(cw):

r1 = 8 m

r2 = 5 m

r3 = 10 m

T = 2250 N

T = 2250 N

300

T

300

T

800 N

200 N

800 N

200 N

Fx

Fy

2 m3 m5 m


300

Ty

Tx

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

F(right) = F(right) = F(left):F(left):

F = 1953 N, θ = 356.3o

F = 1953 N, θ = 356.3o

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.)

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


First: First: (ccw)(ccw)

1 = Br3

2 = w1r1

Next: Next: (cw)(cw)3 = w2r2

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

B = 48 NB = 48 N

40 N 80 N

2 m 3 m7 m

A B

A B

w1 w2

r1 r2

r3

40 N 80 N

2 m 3 m7 m

A B


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

Translational Translational EquilibriumEquilibrium

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

Recall that B = 48 N

B

A

w2

w1

A = 72 NA = 72 N

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
