CHAPTER 8CHAPTER 8Rotational EquilibriumRotational Equilibrium

Torque:Torque: The tendency of a force to rotate a body about some axis.

d

F

= F d

= F · d (in this situation)

Lever arm

d

Θ

F

= F d

= F · d cosΘ (in this situation)

TorqueTorque is equal to the product of the Force applied and the “lever arm.”Lever ArmLever Arm is distance from the axis of the perpendicular rotation to a line drawn along the direction of the force.

Center of Gravity of an ObjectCenter of Gravity of an Object

1. For a symmetrical object, the Center of Gravity is usually intuitively obvious. It is at the intersection of the x and y lines of symmetry.

Center of Gravity

2. For non-symmetrical objects, the Center of Gravity is easily calculated (as in the next example) or it is provided to you.

3. Calculus is used to find the Center of Gravity for very non-symmetrical objects. You are not expected to know how to use calculus to find centers of gravity.

Center of Gravity of an ObjectCenter of Gravity of an ObjectTorque calculations often involve Fg.Calculations of Torque are simplified if all the mass in the object can be thought of as existing at one point in the object.Center of Gravity

ExampleExample

0 1 32 4

xcg = m1x1 + m2x2 + m3x 3+m4x4

m1 + m2 + m3 + m4xcg = weighted average lever arm about an end pivot pointxcg = m(.5) + 2m(1.5) + m(2.5) + m(3.5) 5m

xcg = 1.9

Notice that this approach to the problem used symmetry to find the center of mass of each individual block.

Torque and EquilibriumTorque and Equilibrium

Fx,Net = 100N–100N = 0 NFy,Net = 200N–200N = 0 N

Will the box begin to accelerate in any direction?Yes, because the third condition for equilibrium is not met.

Third Condition for equilibrium:

Net = 0 Net = CW - CCW

Strategy for determining Net

• Pick a convenient rotational point (pivot point)• A or B in this case• A convenient pivot point has a zero lever arm and zero torque

Calculate and Sum the Torques (A = pivot point) A = 0 N (100N · 0 m = 0 N·m) B = 100N · 1.0m = 100 N·m (CCW) Net = -100 N·m (CCW)

20kg

B F=100NFN=200N

Frictionless Surface

F=100NA

1.0m

Fg=200N

Using Torque to Solve “Statics” Using Torque to Solve “Statics” ProblemsProblems

Statics Problems:Statics Problems: System is in static equilibrium.

Fx,Net = 0 NFy,Net = 0 N Net = 0 N·m

Three equations make it possible to solve for up to three unknowns.

Example ProblemExample Problem (Seesaw)A 40kg child and a 33kg child are to be balanced (static equilibrium) on a 5.5m seesaw. Where must the pivot point be placed relative to the smaller child? What is the upward force at the fulcrum? Problem Solving Problem Solving Strategy:Strategy:Sketch the Problem A

33kg

B

40kgFP

5.5mx 5.5 - x

A

33kg

B

40kgFP

5.5mx 5.5 - x

Choose a convenient pivot point (the fulcrum in this case). Set up the three equations and solve for the unknowns

FNet,y = 0 N FP = FA + FB = (33kg)(9.8m/s2)+(40kg)(9.8m/s2)

x = 3.0 m

Net = 0 N·m (33kg)(9.8m/s2)(x) = (40kg)(9.8m/s2)(5.5 –

x)

FNet,x = 0 N No Data

FP = 720N

Example ProblemExample Problem (Forces on Table Supports)

A uniform 20.0m, 1500kg beam supports a 15,000kg printing press 5.0m from the right support column. Calculate the force on each support column.

Sketch the ProblemSketch the Problem

Notice how Centers of Gravity are used to simplify problem.Choose a Convenient Pivot PointEither left or right column could be used. Choose left.

1500kg

FC1 FBeam

15,000kg

FPress

FC2

10.0m

5.0m

20.0m

Set up three equations and solve for unknowns.FNet,x = 0 N No DataFNet,y = 0 N FC1 + FC2 = FBeam + FPress

FC1 + FC2 = (1500kg)(g) + (15,000kg)(g) FC1 + FC2 = 162,000 N

Net = 0 N·m FC1 (0M) + FBeam (10.0m) + FPress (15.0m) = FC2 (20.0m) (14,700N)(10.0m) + (147,000N)(15.0m) = FC2 (20.0m)

FC2 = 118,000 N FC1 = 44,000 N

5.0m

1500kg

FC1 FBeam

15,000kg

FPress

FC2

10.0m

20.0m

Example ProblemExample Problem (Beam and Wire)A uniform beam 2.20m long with mass of 25.0kg is mounted by a hinge on a wall and rests horizontally when supported by a wire connecting the outside end of the beam to the wall. The wire makes an angle of 30° to the horizontal. A 280.kg mass is suspended from the far end of the beam. Determine the components of force that the hinge exerts on the beam and the component of the tension force in the wire.Sketch the ProblemSketch the Problem

Show the Forces and their Show the Forces and their componentscomponents

m=280kg

2.20m

30°

FH,

y

FH

FH,x

Fg,Beam

Fg,Block

FT,y

FT,x

FT

Choose a Convenient Pivot Point (Hinge)Choose a Convenient Pivot Point (Hinge)Setup the Three Equations and solve for three Setup the Three Equations and solve for three unknownsunknownsFNet,x = 0 N FH,x = FT,x

FNet,y = 0 N FH,y + FT,y = Fg,Beam + Fg,Block

Net = 0 N·m

(Fg,Beam)(1.10m) + (Fg,Block)(2.20m) = (FT,y)(2.20m)

Plug and Solve (Hint: FT,y=FT,xtan30 is a 4th equation to use)

FT,y = 2870 N FH,y = 120N FT,x = 4970 N FH,x = 4970 N

m=280kg

2.20m

30°

FH,

y

FH

FH,x

Fg,Beam

Fg,Block

FT,y

FT,x

FT