Engineering Engineering Mechanics:Mechanics:
STATICSSTATICS
Anthony Bedford and Wallace Fowler
SI Edition
Teaching SlidesChapter 5:
Objects in Equilibrium
(C) 2005 Pearson Education South Asia Pte Ltd 2
Chapter OutlineChapter Outline
The Equilibrium Equations 2-Dimensional Applications
(C) 2005 Pearson Education South Asia Pte Ltd 3
5.1 5.1 The Equilibrium EquationsThe Equilibrium Equations
When an object acted upon by a system of forces & moments is in equilibrium, the following conditions are satisfied:
1. The sum of the forces is zero:
Σ F = 0 (5.1)2. The sum of the moments about any point is
zero:
Σ Many point = 0 (5.2)
(C) 2005 Pearson Education South Asia Pte Ltd 4
When the loads and reactions on an object in equilibrium form a two-dimensional system of forces and moments, they are related by three scalar equilibrium equations:
Σ Fx = 0 (5.3)
Σ Fy = 0 (5.4)
Σ Many point = 0 (5.5) More than three independent equilibrium equations
cannot be obtained from a two-dimensional free-body diagram, which means we can solve for at most three unknown forces or couples.
5.1 5.1 The Scalar Equilibrium EquationsThe Scalar Equilibrium Equations
(C) 2005 Pearson Education South Asia Pte Ltd 5
5.1 5.1 The Equilibrium EquationsThe Equilibrium Equations
Eqs. (5.1) & (5.2) imply that the system of forces & moments acting on an object in equilibrium is equivalent to a system consisting no forces & no couples
If the sum of the forces on an object is zero & the sum of the moments about 1 point is zero, then the sum of the moments about every point is zero
(C) 2005 Pearson Education South Asia Pte Ltd 6
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
Supports: Forces & couples exerted on an object by its
supports are called reactions, expressing the fact that the supports “react” to the other forces & couples or loads acting on the object
20 KN-m
(C) 2005 Pearson Education South Asia Pte Ltd 7
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
Supports: Some very common kinds of supports are
represented by stylized models called support conventions if the actual supports exert the same reactions as the models
(C) 2005 Pearson Education South Asia Pte Ltd 8
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
The Pin Support: Figure a: a pin support
a bracket to which an object (such as a beam) is attached by a smooth pin that passes through the bracket & the object
Figure b: side view
(C) 2005 Pearson Education South Asia Pte Ltd 9
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
To understand the reactions that a pin support can exert:
Imagine holding the bar
attached to the pin support If you try to move the bar without rotating it
(i.e. translate the bar), the support exerts a reactive force that prevents this movement
However, you can rotate the bar about the axis of the pin
The support cannot exert a couple about the pin axis to prevent rotation
(C) 2005 Pearson Education South Asia Pte Ltd 10
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
The arrows indicate the directions of the reactions if Ax & Ay are positive
If you determine Ax or Ay to be negative, the reaction is in the direction opposite to that of the arrow
Thus, a pin support can’t exert a couple about the pin axis but it can exert a force on the object in any direction, which is usually expressed by representing the force in terms of components
(C) 2005 Pearson Education South Asia Pte Ltd 11
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
The pin support is used to represent any real support capable of exerting a force in any direction but not exerting a couple
Used in many common devices, particularly those designed to allow connected parts to rotate relative to each other
(C) 2005 Pearson Education South Asia Pte Ltd 12
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
The Roller Support: A pin support mounted on wheels Like a pin support, it cannot exert a couple
about the axis of the pin Since it can move freely in the direction
parallel to the surface on which it rolls, it can’t exert a force parallel to the surface but can exert a force normal (perpendicular) to this surface
(C) 2005 Pearson Education South Asia Pte Ltd 13
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
Other commonly used conventions equivalent to the roller support:
The wheels of vehicles & wheels supporting parts of machines are roller supports if the friction forces exerted on them are negligible in comparison to the normal forces
(C) 2005 Pearson Education South Asia Pte Ltd 14
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
A plane smooth surface can also be modeled by a roller support:
Beams & bridges are sometimes supported in this way so that they will be free to undergo thermal expansion & contraction
(C) 2005 Pearson Education South Asia Pte Ltd 15
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
These supports are similar to the roller support in that they cannot exert a couple & can only exert a force normal to a particular direction (friction is neglected)
(a) Pin in a slot (b) Slider in a slot (c) Slider on a shaft
(C) 2005 Pearson Education South Asia Pte Ltd 16
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
In these supports, the supported object is attached to a pin or slider that can move freely in 1 direction but is constrained in the perpendicular direction
Unlike the roller support, these supports can exert a normal force in either direction
(C) 2005 Pearson Education South Asia Pte Ltd 17
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
The Fixed Support: The fixed support shows the supported object
literally built into a wall (built-in)
To understand the reactions: Imagine holding a bar attached to the fixed
support
(C) 2005 Pearson Education South Asia Pte Ltd 18
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
If you try to translate the bar, the support exerts a reactive force that prevents translation
If you try to rotate the bar, the support exerts a reactive couple that prevents rotation
A fixed support can exert 2 components of force & a couple
(C) 2005 Pearson Education South Asia Pte Ltd 19
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
The term MA is the couple exerted by the support & the curved arrow indicates its direction
Fence posts have fixed supports The attachments of parts connected so that
they cannot move or rotate relative to each other, such as the head of a hammer & its handle, can be modeled as fixed supports
(C) 2005 Pearson Education South Asia Pte Ltd 20
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
Table 5.1 summarizes the support conventions commonly used in 2-D applications:
(C) 2005 Pearson Education South Asia Pte Ltd 21
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
Table 5.1
(C) 2005 Pearson Education South Asia Pte Ltd 22
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
Free-Body Diagrams: By using the support conventions, we can
model more elaborate objects & construct their free-body diagrams in a systematic way
Example: a beam with a pin support at the left end &
a roller support on at the right end & is loaded with a force F
The roller support rests on a surface inclined at 30°
(C) 2005 Pearson Education South Asia Pte Ltd 23
5.2 5.2 2-Dimensional Applications2-Dimensional Applications To obtain a free-body
diagram of the beam, isolate it from its supports
Complete the free-body diagram by showing the reactions that may be exerted on the beam by the supports
Notice that the reaction at B exerted by the roller support is normal to the surface on which the support rests
(C) 2005 Pearson Education South Asia Pte Ltd 24
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
Example: The object in this figure has a fixed support
at the left end A cable passing over a pulley is attached to
the object at 2 points
Isolate it from its supports & complete the free-body by showing the reactions at the fixed support & the forces exerted by the cable
(C) 2005 Pearson Education South Asia Pte Ltd 25
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
Don’t forget the couple at the fixed support Since we assume the tension in the cable is the
same on both sides of the pulley, the 2 forces exerted by the cable have the magnitude T
(C) 2005 Pearson Education South Asia Pte Ltd 26
5.2 5.2 2-Dimensional Applications2-Dimensional Applications
Once you have obtained the free-body diagram of an object in equilibrium to identify the loads & reactions acting on it, you can apply the equilibrium equations
(C) 2005 Pearson Education South Asia Pte Ltd 27
Example 5.1 Reactions at Pin & Roller Example 5.1 Reactions at Pin & Roller SupportsSupports
The beam in Fig. 5.1 has a pin at A & roller supports at B & is subjected to a 2-kN force. What are the reactions at the supports?
Fig. 5.1
(C) 2005 Pearson Education South Asia Pte Ltd 28
Example 5.1 Reactions at Pin & Roller Example 5.1 Reactions at Pin & Roller SupportsSupports
StrategyStrategy To determine the reactions exerted on the beam by
its supports, draw a free-body diagram of the beam isolated from the supports. The free-body diagram must show all external forces & couples acting on the beam, including the reactions exerted by the supports. Then determine the unknown reactions by applying equilibrium equations
(C) 2005 Pearson Education South Asia Pte Ltd 29
Example 5.1 Reactions at Pin & Roller Example 5.1 Reactions at Pin & Roller SupportsSupports
SolutionSolutionDraw the Free-Body Diagram:Isolate the beam from its supports & show the loads & the reactions that may be exerted by the pin & roller supports.
There are 3 unknown reactions: 2 components of force Ax & Ay at the pin support & a force B at the roller support
(C) 2005 Pearson Education South Asia Pte Ltd 30
Example 5.1 Reactions at Pin & Roller Example 5.1 Reactions at Pin & Roller SupportsSupportsSolutionSolutionApply the Equilibrium Equations:Summing the moments about point A:
Σ Fx = Ax Bsin 30° = 0
Σ Fy = Ay + Bcos 30° 2 kN = 0
Σ Mpoint A = (5 m)(Bcos 30°) (3 m)(2 kN) = 0
Solving these equations, the reactions are:Ax = 0.69 kN, Ay = 0.80 kN & B = 1.39 kN
(C) 2005 Pearson Education South Asia Pte Ltd 31
Example 5.1 Reactions at Pin & Roller Example 5.1 Reactions at Pin & Roller SupportsSupports
SolutionSolutionConfirm that the equilibrium equations are satisfied:
Σ Fx = 0.69 kN (1.39 kN)sin 30° = 0
Σ Fy = 0.80 kN + (1.39 kN)cos 30° 2 kN = 0
Σ Mpoint A = (5 m)(1.39 kN)cos 30° (3 m)(2 kN) = 0
Critical ThinkingCritical Thinking In drawing free-body diagrams, you should try to
choose the correct directions of the reactions because it helps to develop your physical intuition
(C) 2005 Pearson Education South Asia Pte Ltd 32
Example 5.1 Reactions at Pin & Roller Example 5.1 Reactions at Pin & Roller SupportsSupportsCritical ThinkingCritical Thinking However, if you choose an incorrect direction for
a reaction in drawing the free-body diagram of a single object, the value you obtain from the equilibrium equations for that reaction will be negative, which indicates that its actual direction is opposite to the direction you chose E.g. if we draw the free-body diagram of the
beam with the component Ay pointed downward:
(C) 2005 Pearson Education South Asia Pte Ltd 33
Example 5.1 Reactions at Pin & Roller Example 5.1 Reactions at Pin & Roller SupportsSupports
Critical ThinkingCritical Thinking Equilibrium equations:
Σ Fx = Ax Bsin 30° = 0
Σ Fy = Ay + Bcos 30° 2 kN = 0
Σ Mpoint A = (5 m)(Bcos 30°) (3 m)(2 kN) = 0
Solving, we obtain: Ax = 0.69 kN, Ay = 0.80 kN & B = 1.39 kN
The negative value of Ay indicates that the vertical force exerted on the beam by the pin support at A is in the direction opposite to the arrow, i.e. the force is 0.80 kN upward
(C) 2005 Pearson Education South Asia Pte Ltd 34
Example 5.2 Reactions at a Fixed SupportExample 5.2 Reactions at a Fixed Support
The object in Fig. 5.2 has a fixed support at A & is subjected to 2 forces & a couple. What are the reactions at the support?
Fig. 5.2
(C) 2005 Pearson Education South Asia Pte Ltd 35
Example 5.2 Reactions at a Fixed SupportExample 5.2 Reactions at a Fixed Support
StrategyStrategy Obtain a free-body diagram by isolating the object
from the fixed support at A & showing the reactions exerted at A, including the couple that may be exerted by a fixed support. Then determine the unknown reactions by applying the equilibrium equations.
(C) 2005 Pearson Education South Asia Pte Ltd 36
Example 5.2 Reactions at a Fixed SupportExample 5.2 Reactions at a Fixed Support
SolutionSolution Draw the Free-Body Diagram:
Isolate the object from its support & show the reactions at the fixed support.
There are 3 unknown reactions: 2 force components Ax & Ay & a couple MA. (Remember that we can choose the directions of these arrows arbitrarily)
Also resolve the 100-N force into its components.
(C) 2005 Pearson Education South Asia Pte Ltd 37
Example 5.2 Reactions at a Fixed SupportExample 5.2 Reactions at a Fixed Support
SolutionSolutionDraw the Free-Body Diagram:
(C) 2005 Pearson Education South Asia Pte Ltd 38
Example 5.2 Reactions at a Fixed SupportExample 5.2 Reactions at a Fixed Support
SolutionSolutionApply the Equilibrium Equation:Summing the moments about point A:
Σ Fx = Ax + 100cos 30° N = 0
Σ Fy = Ay 200 N + 100sin 30° N = 0
Σ Mpoint A = MA + 300 N-m (2 m)(200 N) (2 m)(100cos 30° N) + (4 m)(100sin 30° N)= 0
Solving these equations, Ax = 8.86 N, Ay = 150.0 N & MA = 73.2 N-m.
(C) 2005 Pearson Education South Asia Pte Ltd 39
Example 5.2 Reactions at a Fixed SupportExample 5.2 Reactions at a Fixed Support
Critical ThinkingCritical Thinking Why don’t the 300 N-m couple & the couple MA
exerted by the fixed support appear in the first 2 equilibrium equations? A couple exerts no net force Also, because the moment due to a couple is
the same about any point, the moment about A due to the 300 N-m counterclockwise couple is 300 N-m counterclockwise
(C) 2005 Pearson Education South Asia Pte Ltd 40
ExExerciseercise 5. 5.66 Diving BoardDiving Board
The masses of the person and the diving board are 54 kg and 36 kg, respectively. Assume that they are in equilibrium.
(a) Draw the free-body diagram of the diving board.
(b) Determine the reactions at the supports A and B.
AnswersAx=0; Ay=-1.85 KN;By=2.74 KN
(C) 2005 Pearson Education South Asia Pte Ltd 41
ExExerciseercise 5. 5.77 Ironing BoardIroning Board
The ironing board has supports at A and B that can be modeled as roller supports.
(a) Draw the free-body diagram of the ironing board.
(b) Determine the reactions at A and B.
Answers
Ay=79.2 N
By=144.2 N
(C) 2005 Pearson Education South Asia Pte Ltd 42
ExExerciseercise 5. 5.1616 A Person Doing Push-upsA Person Doing Push-ups
A person doing push-ups pauses in the position shown. His mass is 80 kg. Assume that his weight W acts at the point shown. The dimensions shown are a = 250 mm, b = 740 mm, and c = 300 mm. Determine the normal force exerted by the floor (a) on each hand, (b) on each foot.
Answers
Force on each hand=FH=293.3 N
Force on each feet=FF=99.1 N
(C) 2005 Pearson Education South Asia Pte Ltd 43
ExExerciseercise 5. 5.19 Beam with cable passing 19 Beam with cable passing through pulleythrough pulley
(a) Draw the free-body diagram of the beam.
(b) Determine the tension in the cable and the reactions at A.
Answers
AX=554 N
AY=-160 N
T=640 N
(C) 2005 Pearson Education South Asia Pte Ltd 44
Exercise 5.26 WheelbarrowExercise 5.26 Wheelbarrow
The total weight of the wheelbarrow and its load is W = 100 lb.(a) If F = 0, what are the vertical reactions at A and B?(b) What force F is necessary to lift the support at A off the ground?
Answers
AX= 0 N
AY= 269.2 N
FB= 230.8 NF= 106.1 N
(C) 2005 Pearson Education South Asia Pte Ltd 45
Exercise 5.34Exercise 5.34
The forklift is stationary. The sign’s weight WS = 160 N acts at the point shown. The 50-N weight of the bar AD acts at the midpoint of the bar. Determine the tension in the cable AE and the reactions at D.
Answers
TAE= 165.2 N, DX= -155.2 N
DY= -153.5 N
(C) 2005 Pearson Education South Asia Pte Ltd 46
Exercise 5.36 TrussExercise 5.36 Truss
This structure, called a truss, has a pin support at A and a roller support at B and is loaded by two forces. Determine the reactions at the supports.Answers
AX= -1.828 KN
AY= 2.10 KN
BY= 2.46 KN
(C) 2005 Pearson Education South Asia Pte Ltd 47
Exercise 5.61Exercise 5.61
The dimensions a = 2 m and b = 1 m. The couple M = 2400 N-m. The spring constant is k = 6000 N/m, and the spring would be unstretched if h = 0.The system is in equilibrium when h = 2 m and the beam is horizontal. Determine the force F and the reactions at A.This structure, called a truss, has a pin support at A and a roller support at B and is loaded by two forces. Determine the reactions at the supports.
AnswersAX= 3045 NAY= -185 N F= 1845 N