05B: Equilibrium05B: Equilibrium

San Juanico Bridge

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:

Total EquilibriumTotal EquilibriumIn general, there are eight degrees of

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

Fx= 0 Right = left

Fy= 0 Up = down

ccw (+) cw (-)

(ccw)= (cw)

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.

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 = 4.00 kg

m2 = 1.00 kg

m3 = 6.00 kg

3.00 cm

2.00 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, 2.00 cm)

(-3.00 cm, 2.00 cm)

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 mx = 2 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 10.00-m10.00-m boom boom weighing weighing 200.0 N200.0 N. Rope is . Rope is 2.00 m2.00 m from right end. from right end.

300

T

800.0 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

800.0 N

200.0 N

300

800.0 N

200.0 N

T

Fx

Fy

2.00 m

3.00 m5.00 m

300

T

800.0 N

200.0 N

300

w2w1

T

Fx

Fy

2.00 m3.00 m5.00 m

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

Choose axis of rotation at wall (least information)

(ccw):(ccw):

r

(cw):(cw):

r1 = 8.00 m

r2 = 5.00 mr3 = 10.00 m

T = 2250 N

T = 2250 N

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

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

F = 1950 N, 356.3oF = 1950 N, 356.3o

Example 3:Example 3: Find the forces Find the forces exerted by supports exerted by supports AA and and BB. The . The weight of the weight of the 12-m12-m boom is 200 N. boom is 200 N.

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

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.