Chapter 9

Static Equilibrium

Rigid Objects in Equilibrium

If a rigid body is in equilibrium, neither its linear motion nor its rotational motion changes.

0=∑ xF ∑ = 0yF ∑ = 0τ

0== yx aa 0=α

Rigid Objects in Equilibrium

EQUILIBRIUM OF A RIGID BODY A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium, the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero.

∑ = 0τ∑ = 0yF0=∑ xF

Rigid Objects in Equilibrium

Reasoning Strategy 1.  Select the object to which the equations for equilibrium are to be applied. 2. Draw a free-body diagram that shows all of the external forces acting on the object.

3.  Choose a convenient set of x, y axes and resolve all forces into components that lie along these axes. 4.  Apply the equations that specify the balance of forces at equilibrium. (Set the net force in the x and y directions equal to zero.)

5.  Select a convenient axis of rotation. Set the sum of the torques about this axis equal to zero. 6. Solve the equations for the desired unknown quantities.

Rigid Objects in Equilibrium

Example: A Diving Board A woman whose weight is 530 N is poised at the right end of a diving board with length 3.90 m. The board has negligible weight and is supported by a fulcrum 1.40 m away from the left end. Find the forces that the bolt and the fulcrum exert on the board.

Rigid Objects in Equilibrium

022 =−=∑ WWF τ

F2 =530 N( ) 3.90 m( )

1.40 m=1480 N

22

WWF =

Rigid Objects in Equilibrium

021 =−+−=∑ WFFFy

0N 530N 14801 =−+− F

N 9501 =F

Center of Gravity

DEFINITION OF CENTER OF GRAVITY The center of gravity of a rigid body is the point at which its weight can be considered to act when the torque due to the weight is being calculated.

Center of Gravity

When an object has a symmetrical shape and its weight is distributed uniformly, the center of gravity lies at its geometrical center.

Center of Gravity

++

++=

21

2211

WWxWxW

xcg

Center of Gravity

Example: The Center of Gravity of an Arm The horizontal arm is composed of three parts: the upper arm (17 N), the lower arm (11 N), and the hand (4.2 N). Find the center of gravity of the arm relative to the shoulder joint.

Center of Gravity

++

++=

21

2211

WWxWxW

xcg

xcg =17 N( ) 0.13 m( )+ 11 N( ) 0.38 m( )+ 4.2 N( ) 0.61 m( )

17 N+11 N+ 4.2 N= 0.28 m

17 N

11 N

4.2 N

Center of Gravity

Conceptual Example: Overloading a Cargo Plane This accident occurred because the plane was overloaded toward the rear. How did a shift in the center of gravity of the plane cause the accident?

Center of Gravity

Finding the center of gravity of an irregular shape.

Rigid Objects in Equilibrium

Example: The Massive Diving Board A woman whose weight is 530 N is poised at the right end of a diving board with length 3.90 m. The board now has a weight of 1400 N and is supported by a fulcrum 1.40 m away from the left end. Find the forces that the bolt and the fulcrum exert on the board. How do the results change from the “Massless Diving Board” example done earlier?

lW/2

WD

Rigid Objects in Equilibrium

lW/2

WD

τ∑ = F2ℓ2 −WℓW −WDℓW2

#

$%

&

'(= 0

F2 =WℓWℓ2

+WDℓW2ℓ2

!

"#

$

%&

F2 =530 N( ) 3.90 m( )

1.40 m+ (1400 N) 3.90 m

2 1.40 m( )

!

"##

$

%&&

=1480 N+1950 N = 3430 N

Rigid Objects in Equilibrium

lW/2

WD

Fy∑ = −F1 +F2 −W −WD = 0

−F1 +3430 N− 530 N−1400 N = 0

F1 =1500 N

Rigid Objects in Equilibrium

Example: Bodybuilding The arm is horizontal and weighs 32.2 N. The deltoid muscle can supply 1840 N of force. What is the weight of the heaviest dumbell he can hold?

Rigid Objects in Equilibrium

0=+−−=∑ Mddaa MWW τ

ℓM = 0.150 m( )sin13.0"

Rigid Objects in Equilibrium

Wd =−Waℓa +MℓM

ℓd

=− 32.2 N( ) 0.280 m( )+ 1840 N( ) 0.150 m( )sin13.0!

0.620 m= 85.6 N