Engineering Mechanics: STATICS

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Engineering MechanicsSTATICS

• Anthony Bedford and Wallace Fowler
• SI Edition

Teaching Slides Chapter 5 Objects in Equilibrium
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Chapter Outline

• The Equilibrium Equations
• 2-Dimensional Applications
• Statically Indeterminate Objects
• 3-Dimensional Applications
• 2-Force 3-Force Members
• Computational Mechanics

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5.1 The 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
• S F 0
  (5.1)
• 2. The sum of the moments about any point is
• zero
• S Many point 0
  (5.2)

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5.1 The 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

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5.1 The Equilibrium Equations

• For an object subjected to concurrent forces F1,
  F2, , FN no couples

• Moment about point P is zero
• The only condition imposed by equilibrium on a
  set of concurrent forces is that their sum is
  zero
• F1 F2 ??? FN 0 (5.3)

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5.1 The Equilibrium Equations

• To determine the sum of the moments about a line
  L due to a system of forces moments acting on
  an object

• Choose any point P on the line determine the
  sum of moments S MP about P
• The sum of the moments about the line is the
  component of S MP parallel to the line

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5.1 The Equilibrium Equations

• If the object is in equilibrium, S MP 0
• The sum of the moments about any line due to the
  forces couples acting on an object in
  equilibrium is zero

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5.2 2-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
• E.g. a bridge is held up by the reactions exerted
  by its supports the loads are the forces
  exerted by the weight of the bridge itself, the
  traffic crossing it the wind
• 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

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5.2 2-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

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5.2 2-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

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5.2 2-Dimensional Applications

• Thus, a pin support cant 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

• 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

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5.2 2-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

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5.2 2-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
  cant exert a force parallel to the surface but
  can exert a force normal (perpendicular) to this
  surface

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5.2 2-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

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5.2 2-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

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5.2 2-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)

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5.2 2-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

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5.2 2-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

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5.2 2-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

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5.2 2-Dimensional Applications

• The term MA is the couple exerted by the support
  the curved arrow indicates its direction
• Fence posts lampposts 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

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5.2 2-Dimensional Applications

• Table 5.1 summarizes the support conventions
  commonly used in 2-D applications

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5.2 2-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

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5.2 2-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

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5.2 2-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

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5.2 2-Dimensional Applications

• Dont 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
• 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

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5.2 2-Dimensional Applications

• The Scalar Equilibrium Equations
• When the loads reactions on an object in
  equilibrium form a 2-D system of forces
  moments, they are related by 3 scalar equilibrium
  equations
• S Fx 0
  (5.4)
• S Fy 0
  (5.5)
• S Many point 0
  (5.6)

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5.2 2-Dimensional Applications

• More than 1 equation can be obtained from Eq.
  (5.6) by evaluating the sum of the moments about
  more than 1 point
• But the additional equations will not be
  independent of Eqs. (5.4)?(5.6)
• In other words, more than 3 independent equations
  cannot be obtained from a 2-D free-body diagram,
  which means we can solve for at most 3 unknown
  forces or couples

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5.2 2-Dimensional Applications

• The seesaw found on playgrounds, consisting of a
  board with a pin support at the center that
  allows it to rotate, is a simple familiar
  example that illustrates the role of Eq. (5.6)
• If 2 people of unequal weight sit at the seesaws
  ends, the heavier person sinks to the ground

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5.2 2-Dimensional Applications

• To obtain equilibrium, that person must move
  closer to the center
• Draw the free-body diagram of the seesaw showing
  the weights of the people W1 W2 the reactions
  at the pin support

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5.2 2-Dimensional Applications

• Evaluating the sum of the moments about A, we
  find that the equilibrium equations are
• S Fx Ax 0
  (5.7)
• S Fy Ay ? W1 ? W2 0
  (5.8)
• S Mpoint A D1W1 ? D2W2 0
  (5.9)
• Thus, Ax 0, Ay W1 W2 D1W1 D2W2
• The last condition indicates that the relation
  between the positions of the 2 persons necessary
  for equilibrium

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5.2 2-Dimensional Applications

• To demonstrate that an additional independent
  equation is not obtained by evaluating the sum of
  the moments about a different point
• Sum the moments about the right end of the seesaw
• S Mright end (D1 D2)W1 ? D2Ay 0

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5.2 2-Dimensional Applications

• This equation is a linear combination of Eqs.
  (5.8) (5.9)
• (D1 D2)W1 ? D2Ay ?D2(Ay ? W1 ? W2)
• (D1W1
  ? D2W2) 0

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5.2 2-Dimensional Applications

• Until now we have assumed in examples problems
  that the tension in a rope or cable is the same
  on both sides of a pulley

• Consider this pulley
• In its free-body diagram, we do not assume that
  the tensions are equal
• Summing the moments about the center of the
  pulley
• S Mpoint A RT1 ? RT2 0

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5.2 2-Dimensional Applications

• The tensions must be equal if the pulley is in
  equilibrium
• However, notice that we have assumed that the
  pulleys support behaves like a pin support
  cannot exert a couple on the pulley
• When that is not true for example, due to
  friction between the pulley the support the
  tensions are not necessarily equal

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Example 5.1 Reactions at Pin Roller Supports

• The beam in Fig. 5.13 has a pin at A roller
  supports at B is subjected to a 2-kN force.
  What are the reactions at the supports?

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Example 5.1 Reactions at Pin Roller Supports

• Strategy
• 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

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Example 5.1 Reactions at Pin Roller Supports

• Solution
• Draw 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
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Example 5.1 Reactions at Pin Roller Supports

• Solution
• Apply the Equilibrium Equations
• Summing the moments about point A
• S Fx Ax ? Bsin 30 0
• S Fy Ay Bcos 30 ? 2 kN 0
  
• S 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

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Example 5.1 Reactions at Pin Roller Supports

• Solution
• Confirm that the equilibrium equations are
  satisfied
• S Fx 0.69 kN ? (1.39 kN)sin 30 0
• S Fy 0.80 kN (1.39 kN)cos 30 ? 2 kN
  0
• S Mpoint A (5 m)(1.39 kN)cos 30 ? (3 m)(2 kN)
  0
• Critical 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

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Example 5.1 Reactions at Pin Roller Supports

• Critical 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

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Example 5.1 Reactions at Pin Roller Supports

• Critical Thinking
• Equilibrium equations
• S Fx Ax ? Bsin 30 0
• S Fy ?Ay Bcos 30 ? 2 kN 0
  
• S 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

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Example 5.2 Reactions at a Fixed Support

• The object in Fig. 5.14 has a fixed support at
  A is subjected to 2 forces a couple. What are
  the reactions at the support?

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Example 5.2 Reactions at a Fixed Support

• Strategy
• 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.

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Example 5.2 Reactions at a Fixed Support

• Solution
• 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.

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Example 5.2 Reactions at a Fixed Support

• Solution
• Draw the Free-Body Diagram

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Example 5.2 Reactions at a Fixed Support

• Solution
• Apply the Equilibrium Equation
• Summing the moments about point A
• S Fx Ax 100cos 30 N 0
• S Fy Ay ? 200 N 100sin 30 N 0
  
• S 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.

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Example 5.2 Reactions at a Fixed Support

• Critical Thinking
• Why dont 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

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Example 5.3 Reactions on a Cars Tires

• The 14 000-N car in Fig. 5.15 is stationary.
  Determine the normal forces exerted on the front
  rear tires by the road.

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Example 5.3 Reactions on a Cars Tires

• Strategy
• Draw the free-body diagram of the car, showing
  the forces exerted on its tires by the road at A
  B apply the equilibrium equations to
  determine the forces on the front rear tires.

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Example 5.3 Reactions on a Cars Tires

• Solution
• Draw the Free-Body Diagram
• Isolate the car show its weight the
  reactions exerted by the road.
• There are 2 unknown reactions the forces A
  B exerted on the front rear tires.

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Example 5.3 Reactions on a Cars Tires

• Solution
• Apply the Equilibrium Equations
• The forces have no x components.
• Summing the moments about point B
• S Fy A B ? 14000 N 0
  
• S Mpoint B ? (2 m)(14000 N) (3 m)A
  0
• Solving these equations, the reactions are A
  9333 N
• B 4667 N.

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Example 5.3 Reactions on a Cars Tires

• Critical Thinking
• This example doesnt fall within our definition
  of a 2-D system of forces moments because the
  forces acting on the car are not coplanar

• From the oblique view of the free-body diagram of
  the car, you can see the total forces acting on
  the individual tires

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Example 5.3 Reactions on a Cars Tires

• Critical Thinking
• Total normal force on the front tires
• AL AR A
• Total normal force on the rear tires
• BL BR B
• The sum of the forces in the y direction
• S Fy AL AR BL BR ? 14000 N 0
• A B ? 14000 N 0

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Example 5.3 Reactions on a Cars Tires

• Critical Thinking
• Since the sum of the moments about any line due
  to the forces couples acting on an object in
  equilibrium is zero, the sum of the moments about
  the z axis due to the forces acting on the car is
  zero
• S Mz axis (3 m)(AL AR) ? (2 m)(14000 N)
• (3 m)A ? (2 m)(14000 N) 0
• Thus we obtain the same equilibrium equations we
  did when we solved the problem using a 2-D
  analysis

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Example 5.4 Choosing the Point About Which to
Evaluate Moments

• The structure AB in Fig. 5.16 supports a
  suspended 2-Mg (megagram) mass. The structure is
  attached to a slider in a vertical slot at A
  has a pin support at B. What are the reactions at
  A B?

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Example 5.4 Choosing the Point About Which to
Evaluate Moments

• Strategy
• Draw the free-body diagram of the structure
  the suspended mass by removing the supports at A
  B. Notice that the support at A can exert only
  a horizontal reaction. Then use the equilibrium
  equations to determine the reactions at A B.

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Example 5.4 Choosing the Point About Which to
Evaluate Moments

• Solution
• Draw the Free-Body Diagram

Isolate the structure mass from the supports
show the reactions at the supports the force
exerted by the weight of the 2000-kg mass. The
slot at A can exert only a horizontal force on
the slider.
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Example 5.4 Choosing the Point About Which to
Evaluate Moments

• Solution
• Apply the Equilibrium Equations
• Summing the moments about point B
• S Fx Ax Bx 0
• S Fy By ? (2000)(9.81) N 0
  
• S Mpoint B (3 m)A (2 m)(200 N )(9.81)
  N 0
• The reactions are
• A ?13.1 kN, Bx 13.1 kN By 19.6
  kN.

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Example 5.4 Choosing the Point About Which to
Evaluate Moments

• Critical Thinking
• Although the point about which moments are
  evaluated in writing equilibrium equations can be
  chosen arbitrarily, a careful choice can often
  simplify your solution
• In this example, point B lies on the lines of
  action of the 2 unknown reactions Bx By
• By evaluating moments about B, we obtained an
  equation containing only 1 unknown, the reaction
  at A

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Design Example 5.5 Design for Human Factors

• Fig. 5.17 shows an airport luggage carrier
  its free-body diagram when it is held in
  equilibrium in the tilted position. If the
  luggage carrier supports a weight W 250 N, the
  angle ? 30, a 0.2 m, b 0.4 m d 1.2 m,
  what force F must the user exert?

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Design Example 5.5 Design for Human Factors

• Strategy
• The unknown reactions on the free-body diagram
  are the force F the normal force N exerted by
  the floor. If we sum moments about the center of
  the wheel C, we obtain an equation in which F is
  the only unknown reaction.

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Design Example 5.5 Design for Human Factors

• Solution
• Summing moments about C
• S M(point C) d(F cos ?) a(W sin ?) ? b(W cos
  ?) 0
• Solving for F
• 
  
  (1)
• Substituting the values of W, ?, a, b d yields
  the
• solution F 59.3 N.

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Design Example 5.5 Design for Human Factors

• Design Issues
• Design that accounts for human physical
  dimensions, capabilities characteristics is a
  special challenge
• This art is called design for human factors
• Here, we consider a simple device, the airport
  luggage carrier in Fig. 5.17 show how
  consideration of its potential users the
  constraints imposed by equilibrium equations
  affect its design

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Design Example 5.5 Design for Human Factors

• Design Issues
• The user moves the carrier by grasping the bar at
  the top, tilting it walking while pulling the
  carrier
• The height of the handle (the dimension h) needs
  to be comfortable
• Since h R d sin ?, if we choose values of h
  the wheel radius R, we obtain a relation between
  the length of the carriers handle d the tilt
  angle ?
• 
  (2)

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Design Example 5.5 Design for Human Factors

• Design Issues
• Substituting the expression for d into Eq. (1)
• 
  
  (3)
• Suppose that based on statistical data on human
  dimensions, we decide to design the carrier for
  convenient use by persons up to 1.9 m, which
  corresponds to a dimension h of approximately 0.9
  m
• Let R 0.075 m, a 0.15 m b 0.3 m

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Design Example 5.5 Design for Human Factors

• Design Issues
• The resulting value of F/W as a function of ? is
  shown in Fig. 5.18

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Design Example 5.5 Design for Human Factors

• Design Issues
• At ? 63, the force the user must exert is
  zero, which means the weight of the luggage acts
  at a point directly above the wheels
• This would be the optimum solution if the user
  could maintain exactly that value of ?
• However, ? inevitably varies the resulting
  changes in F make it difficult to control the
  carrier
• In addition, the relatively steep angle would
  make the carrier awkward to pull

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Design Example 5.5 Design for Human Factors

• Design Issues
• From this point of view, it is desirable to
  choose a design within the range of values of ?
  in which F varies slowly, say 30 ? ? ? 45
  (even though the force the user must exert is
  large in this range of ? in comparison with
  larger values of ?, it is only about 13 of the
  weight)
• Over this range of ?, the dimension of d varies
  from 1.65 m to 1.17 m
• A smaller carrier is desirable for lightness
  ease of storage, so we choose d 1.2 m for our
  preliminary design

69
Design Example 5.5 Design for Human Factors

• Design Issues
• We have chosen the dimension d based on
  particular values of the dimensions R, a b
• In an actual design study, we should carry out
  the analysis for expected ranges of values of
  these parameters
• Our final design would also reflect decisions
  based on safety (e.g. there must be adequate
  means to secure luggage no sharp projections),
  reliability (the frame must be sufficiently
  strong the wheels must have adequate reliable
  bearings) the cost of manufacture