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Engineering 36. Chp 5: 2D Equil Special Cases. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer 2D Equil → Special Cases. PARTICLE: Size & Shape of the Object can be neglected as long as all applied Forces have a Point of Concurrency

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Engineering 36

Chp5: 2D EquilSpecial Cases

Bruce Mayer, PE

Licensed Electrical & Mechanical

2d equil special cases
2D Equil → Special Cases
  • PARTICLE: Size & Shape of the Object can be neglected as long as all applied Forces have a Point of Concurrency
    • Covered in Detail in Chp03
  • TWO-FORCE MEMBER: A Structural Element of negligible Wt with only 2Forces acting on it
2d equil special cases1
2D Equil → Special Cases
  • THREE-FORCE MEMBER: A structural Element of negligible Wt with only 3Forces acting on it
    • The forces must be either concurrent or parallel.
      • In the PARALLEL Case the PoC is located at Infinity
      • The NONparallel Case canbe Very Useful in LoadAnalysis
2d equil special cases2
2D Equil → Special Cases
  • FRICTIONLESS PULLEY: For a frictionless pulley in static equilibrium, the tension in the cable is the same on both sides of the pulley
    • Discussed Briefly in Chp03
      • Will Prove the T1 = T2 = T Behavior Today
2d planar system equilibrium
2D Planar System Equilibrium
  • In 2D systems it is assumed that
    • The System Geometry resides completely the XY Plane
    • There is NO Tendency to
      • Translate in the Z-Direction
      • Rotate about the X or Y Axes
  • These Conditions Simplify The Equilibrium Equations
2d planar system
2D Planar System:
  • No Z-Translation → NO Z-Directed Force:
  • No X or Y Rotation → NO X or Y Applied Moments
special case 2 force member
Special Case: 2-Force Member
  • A 2-Force Member/Element is a Body with negligible Weight and Only two applied Forces.
  • Some Special Properties of 2-Frc Ele’s
    • the LoA’s of the Two Forces MUST Cross and thus Produce a PoC
      • Treat as a PARTICLE
    • The Crossed LoA’s Define a PLANE
      • Treat as PLANAR System
2 force element equilibrium
2-Force Element Equilibrium
  • Consider a L-Bracket plate subjected to two forces F1 and F2
  • For static equilibrium, the sum of moments about Pt-A must be zero. Thus the moment of F2 About Pt-A must be zero. It follows that the line of action of F2 must pass through Pt-A
  • Similarly, the line of action of F1 must pass through Pt-B for the sum of moments about Pt-B to be zero.
  • Requiring that the sum of forces in any direction be zero leads to the conclusion that F1 and F2 must have equal magnitude but opposite sense.
special case 2 force element
Special Case: 2-Force Element
  • Mathematically
    • Since the Two Forces Must be Concurrent
  • Since the System is in Equilibrium ΣF’s =0.
  • Thus the two force are Equal and Opposite; that is, the forces CANCEL
special case 3 force member
Special Case: 3-Force Member
  • A 3-Force Element is a PLANAR Body with negligible Weight with Exactly 3 applied Forces (No applied Moments).
  • Claim: If a Planar 3-Force Element is in Equilibrium, Then the LoA’s for the 3-Forces must be CONCURRENT
    • If the Claim is TRUE, then the 3-Force Element can be treated as a PARTICLE
3 force 2d body equilibrium
3-Force 2D Body Equilibrium
  • Consider a Planar rigid body subjected to forces acting at only 3 points.
  • The lines of action of intersect F1 & F2, at Pt-D. The moment of F1 and F2 about this point of intersection is zero.
  • Since the rigid body is in equilibrium, the sum of the moments of F1, F2, and F3 about ANY Pivot-Pt must be zero. It follows that the moment of F3 about D must be zero as well and that the line of action of F3 must pass through D.
  • The lines of action of the three forces must be Concurrent OR Parallel.
3 force 2d body parallel forces
3-Force 2D Body: Parallel Forces
  • If 3 Parallel Forces Maintain a Rigid Body in Static Equilibrium, The following Conditions MUST be Satisfied
    • For Translation Equilibrium







  • For Rotation Equilibrium



special case 3 force element
Special Case: 3-Force Element
  • Mathematically for ||-Forces
    • Since a Body in Equil. Has NO Net Moment
  • Since the System is in Equilibrium ΣF’s =0.
  • In Summary: The dmFm products and, 3 Forces, Sum to Zero
special case 3 force element1
Special Case: 3-Force Element
  • A Graphical Summary

AB is 3F Member(BC is 2F Member)

example pole raising
Example  Pole Raising
  • Solution Plan
    • Create a free-body diagram of the joist.
      • Note that the joist is a 3 force body acted upon by the ROPE, its WEIGHT, and the REACTION at A
    • The three forces must be concurrent for static equilibrium. Therefore, the reaction R must pass through the intersection of the lines of action of the weight and rope forces. Determine the direction of the reaction force R.
  • A man Raises a 10 kg Joist, of Length 4 m, by pulling on a rope.
  • Find the TENSION in the rope and the REACTION at A.
example pole raising1
Example  Pole Raising
  • Create a free-body diagram of the joist
  • Use LoA’s & Trigonometry to Determine the direction of the reaction force R

A LARGE, SCALED Diagram is REALLY Useful in this Problem

example pole raising2
Example  Pole Raising
  • Draw the Force Triangle to Scale
  • Use the Law of the Sines to Find the Reaction Force R
  • Solving find
special case frictionless pulley
Special Case: Frictionless Pulley
  • A FrictionLess Pulley is Typically used to change the Direction of a Cable or Rope in Tension

Pulley with PERFECT Axel (FrictionLess)

special case frictionless pulley1
A Perfect Axel Generates NO moment to Resist Turning.

Consider the FBD for a Perfect Pulley

Special Case: FrictionLess Pulley
  • Since the LoA’s for FAx & FAy Pass Thru the Axel-Axis Pt-A they Generate No moment about this point .
  • T1 and T2 have Exactly the SAME Lever arm, i.e., the Radius, R, of the Pulley
special case frictionless pulley2
Since the Pulley is in Equilibrium ΣMA = 0

Writing the Moment Eqn

Special Case: FrictionLess Pulley
  • Thus for the NO-Friction Perfect Pulley
fritionfilled pulley
FritionFilled Pulley
  • Consider the case where we have a pulley that is NOT Free Wheeling; i.e., the pulley resists rotation
  • Example: Automobile alternator changes thermal-mechanical energy into electrical energy
frictionfilled pulley
In Alternator Operation the generation of electricity produces a resisting moment that counters the direction of spin; The FBD in this case → FrictionFilled Pulley


  • The ΣMA = 0
frictionless pulley 3f mem
FrictionLess Pulley; 3F Mem
  • In the System at Right Member ABC, which is a FOUR-Force System, can be reduced to a 3-Force System using and Equivalent Resultant-Couple System at the Pulley
frictionless pulley 3f mem1
FrictionLess Pulley; 3F Mem
  • Recall that Forces Can be MOVED to a new point on a Body as long as the Rotation Tendency caused by the move is accounted for by the Addition of a COUPLE-Moment at the new Point
frictionless pulley 3f mem2
FrictionLess Pulley; 3F Mem
  • Apply the Equivalent Loading Method to a FrictionLess Pulley
  • From the Previous Discussion the MOMENT about the Axle (Pin) of a Frictionless pulley produced by the Tensions is ZERO
  • Thus Can Move the T’s to the Pin with a Couple of ZERO
frictionless pulley 3f mem3
FrictionLess Pulley; 3F Mem
  • The Equivalent Systems by MA= 0
frictionless pulley 3f mem4
FrictionLess Pulley; 3F Mem
  • Moving the FrictionLess Pulley Force-Resultant to the Pin at Pt-A produces the FBD Shown At Right
    • Now can Draw the Force Triangle
frictionless pulley important
FrictionLess Pulley - Important
  • For a FrictionLess Pulley the Tension Forces and be to the Pulley Axel (Pin) WithOUT the Addition of a Couple


special cases summarized
Special Cases Summarized
  • Particle:
  • 2-Force Element:
  • 3-Force Planar Element:
  • FrictionLess Pulley:
whiteboard work
WhiteBoard Work

Lets WorkThese NiceProblems


Engineering 36


Bruce Mayer, PE

Registered Electrical & Mechanical

jib problem
Jib Problem
  • The upper portion of the crane boom consists of the jib AB, which is supported by the pin at A, the guy line BC, and the backstay CD, each cable being separately attached to the mast at C. If the 5-kN load is supported by the hoist line, which passes over the pulley at B, determine the magnitude of the resultant force the pin exerts on the jib at A for equilibrium, the tension in the guy line BC, and the tension T in the hoist line. Neglect the weight of the jib. The pulley at B has a radius of 0.1 m.
disk problem
Disk Problem
  • The smooth disksD andE have a weight of 200 lb and 100 lb, respectively. Determine the largest horizontal force P that can be applied to the center of disk E without causing the disk D to move up the incline.