Chapter 9 Static Equilibrium
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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
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