Particle Equilibrium and Free Body Diagrams Study Guide

Free Body Diagrams and Assessment Requirements

  • Assessment Rules for Free Body Diagrams (FBDs):

    • Full credit on exam problems requiring a free body diagram will not be awarded if the free body diagram is omitted.
    • Online homework platforms (such as Connect) feature drawing mechanisms for free body diagrams that can be frustrating to navigate; submitting hard-copy paper solutions ensures full credit for free body diagrams.

  • Purpose and Representation of Free Body Diagrams:

    • A free body diagram isolates an object or particle to display all external forces acting upon it.
    • Objects small enough relative to system dimensions are treated as point mass particles of negligible size and are represented visually by a dot.
    • Realistic or semi-realistic pictorial sketches of objects are also acceptable representations on free body diagrams.

Non-Contact and Contact Forces

  • Non-Contact Forces:

    • Definition: Forces that act across a distance without requiring direct physical touch or contact between objects.
    • Gravity:
    • The primary non-contact force encountered in static analysis, directed vertically downward toward the Earth.
    • Negligible Mass Condition: When an object has an extremely small mass relative to external forces, the force of gravity is negligible and does not alter system equilibrium. In such cases, gravitational force is omitted.
    • Diagnostic Check: Always evaluate whether gravity is present and significant in every system.
    • Magnetism:
    • A non-contact force exerted by magnetic fields.
    • Practical learning materials include acquiring two cow magnets and a box of paper clips (such as from Wayne Farm and Flea Dump Field).
    • Electric Forces:
    • Non-Contact forces resulting from electrostatic charges.
    • Low Humidity / Cold Weather Behavior: Decreased ambient humidity prevents static charge dissipation, causing clothing removed from a dryer without dryer sheets to exhibit static cling. Separating statically clung garments in darkness reveals visible electrical sparks.
    • High Humidity / Summer Behavior: Atmospheric moisture transports electrical charges, preventing charge accumulation and causing electrostatic forces to decay rapidly.

  • Contact Forces:

    • Definition: Forces that result directly from physical contact between bodies.
    • Diagnostic Identification Method: Identify contact forces by asking, "What is touching the object?" Every contact force must be directly linked to a physical object touching the target body.

Principles of 2D Particle Equilibrium

  • Fundamental Equations of Static Equilibrium:

    • For a particle to remain in static equilibrium, the vector sum of all acting forces must equal zero:     ∑F=0\sum \mathbf{F} = 0
    • In a two-dimensional rectangular coordinate system, this vector equation breaks down into two independent scalar equations:     ∑Fx=0\sum F_x = 0∑Fy=0\sum F_y = 0

  • Resolution Methods:

    • Rectangular Components Method:
    • Resolves each force vector into orthogonal xx and yy components using right-triangle trigonometry.
    • Standard method required when four or more forces act on a single particle.
    • Force Triangle Method:
    • Formed by arranging force vectors head-to-tail to form a closed vector polygon.
    • Highly efficient method specifically suited for equilibrium systems acted upon by exactly three concurrent forces.
    • Employs geometric angle properties and the Law of Sines to solve unknown force magnitudes.

Analysis of Concurrent Cables Supporting a Suspended Weight (Problem 43)

  • System Configuration and Free Body Diagram Setup:

    • Target particle: Connection knot located at point CC.
    • Particle mass: Negligible mass; gravitational force acting on knot CC itself is ignored.
    • Applied downward force: Suspended load of weight W=400 lbW = 400\,\text{lb} (the pound is a direct unit of force).
    • Contact elements touching point CC:
    • Cable ACAC exerting tension force TACT_{AC} (simplified as AA) directed from point CC toward point AA.
    • Cable BCBC exerting tension force TBCT_{BC} (simplified as BB) directed from point CC toward point BB.
    • Cable tension behavior: Tension forces always pull along the direction of the cable away from the point of application.
    • Spatial orientation angles:
    • Cable ACAC forms an angle of 50∘50^\circ relative to the horizontal (derived using alternate interior angles for parallel lines).
    • Cable BCBC forms an angle of 30∘30^\circ relative to the horizontal.

  • Component Resolution Method:

    • Vector component expressions:
    • Force BB:       Bx=B×cos⁡(30∘)B_x = B \times \cos(30^\circ)By=B×sin⁡(30∘)B_y = B \times \sin(30^\circ)
    • Force AA:       Ax=−A×cos⁡(50∘)A_x = -A \times \cos(50^\circ)Ay=A×sin⁡(50∘)A_y = A \times \sin(50^\circ)
    • Downward Load WW:       Wx=0W_x = 0Wy=−400 lbW_y = -400\,\text{lb}
    • Equilibrium equation along the xx-axis:     ∑Fx=B×cos⁡(30∘)−A×cos⁡(50∘)=0\sum F_x = B \times \cos(30^\circ) - A \times \cos(50^\circ) = 0
    • Equilibrium equation along the yy-axis:     ∑Fy=B×sin⁡(30∘)+A×sin⁡(50∘)−400 lb=0\sum F_y = B \times \sin(30^\circ) + A \times \sin(50^\circ) - 400\,\text{lb} = 0
    • Step-by-Step Algebraic Substitution:
    • Isolate BB in the xx-equilibrium equation:       B×cos⁡(30∘)=A×cos⁡(50∘)B \times \cos(30^\circ) = A \times \cos(50^\circ)B=A×cos⁡(50∘)cos⁡(30∘)B = A \times \frac{\cos(50^\circ)}{\cos(30^\circ)}cos⁡(50∘)cos⁡(30∘)≈0.7422\frac{\cos(50^\circ)}{\cos(30^\circ)} \approx 0.7422B=0.7422×AB = 0.7422 \times A
    • Substitute B=0.7422×AB = 0.7422 \times A into the yy-equilibrium equation:       (0.7422×A)×sin⁡(30∘)+A×sin⁡(50∘)−400 lb=0(0.7422 \times A) \times \sin(30^\circ) + A \times \sin(50^\circ) - 400\,\text{lb} = 0
    • Factor out force magnitude AA:       A×(0.7422×sin⁡(30∘)+sin⁡(50∘))=400 lbA \times \left(0.7422 \times \sin(30^\circ) + \sin(50^\circ)\right) = 400\,\text{lb}
    • Numerical evaluation of trigonometric terms:       sin⁡(30∘)=0.5\sin(30^\circ) = 0.5sin⁡(50∘)≈0.7660\sin(50^\circ) \approx 0.76600.7422×0.5=0.37110.7422 \times 0.5 = 0.37110.3711+0.7660=1.137140.3711 + 0.7660 = 1.13714A×1.13714=400 lbA \times 1.13714 = 400\,\text{lb}
    • Calculate magnitude AA:       A=400 lb1.13714≈351.8 lbA = \frac{400\,\text{lb}}{1.13714} \approx 351.8\,\text{lb}
    • Calculate magnitude BB:       B=0.7422×351.8 lb≈261.1 lbB = 0.7422 \times 351.8\,\text{lb} \approx 261.1\,\text{lb}
    • Rounded values for system entry:
    • Tension TAC=352 lbT_{AC} = 352\,\text{lb}
    • Tension TBC=261 lbT_{BC} = 261\,\text{lb}

  • Force Triangle Method:

    • Vector triangle construction:
    • Draw downward vertical force vector of magnitude 400 lb400\,\text{lb}.
    • Attach tail of vector BB (pointing upward and to the right) to tip of downward load vector.
    • Attach tail of vector AA (pointing upward and to the left) to tip of vector BB to close the triangle back at the origin.
    • Calculation of interior triangle angles:
    • Interior angle opposite vector AA: Cable BB makes 30∘30^\circ with horizontal, yielding an angle of 90∘−30∘=60∘90^\circ - 30^\circ = 60^\circ with the vertical.
    • Interior angle opposite vector BB: Cable AA makes 50∘50^\circ with horizontal, yielding an angle of 90∘−50∘=40∘90^\circ - 50^\circ = 40^\circ with the vertical.
    • Interior angle opposite 400 lb400\,\text{lb} load: Combined angle between cables relative to horizontal is 50∘+30∘=80∘50^\circ + 30^\circ = 80^\circ (or 180∘−40∘−60∘=80∘180^\circ - 40^\circ - 60^\circ = 80^\circ).
    • Application of the Law of Sines:
    • Solving for magnitude AA:       Asin⁡(60∘)=400 lbsin⁡(80∘)\frac{A}{\sin(60^\circ)} = \frac{400\,\text{lb}}{\sin(80^\circ)}A=400 lb×sin⁡(60∘)sin⁡(80∘)≈352 lbA = \frac{400\,\text{lb} \times \sin(60^\circ)}{\sin(80^\circ)} \approx 352\,\text{lb}
    • Solving for magnitude BB:       Bsin⁡(40∘)=400 lbsin⁡(80∘)\frac{B}{\sin(40^\circ)} = \frac{400\,\text{lb}}{\sin(80^\circ)}B=400 lb×sin⁡(40∘)sin⁡(80∘)≈261 lbB = \frac{400\,\text{lb} \times \sin(40^\circ)}{\sin(80^\circ)} \approx 261\,\text{lb}

Analysis of System with Applied Horizontal Force (Problem 44)

  • System Configuration and Free Body Diagram Setup:

    • Target particle: Point CC.
    • Applied external load: Horizontal force of 6 kN6\,\text{kN} acting to the right.
    • Cable angles specified relative to the vertical axis:
    • Cable ACAC forms an angle of 55∘55^\circ relative to the vertical.
    • Cable BCBC forms an angle of 30∘30^\circ relative to the vertical.
    • Conversion to reference angles relative to the horizontal axis:
    • Cable ACAC horizontal reference angle: 90∘−55∘=35∘90^\circ - 55^\circ = 35^\circ.
    • Cable BCBC horizontal reference angle: 90∘−30∘=60∘90^\circ - 30^\circ = 60^\circ.

  • Equations of Equilibrium:

    • Summation of forces along the xx-axis:     ∑Fx=6 kN−A×cos⁡(35∘)−B×cos⁡(60∘)=0\sum F_x = 6\,\text{kN} - A \times \cos(35^\circ) - B \times \cos(60^\circ) = 0
    • Summation of forces along the yy-axis:     ∑Fy=A×sin⁡(35∘)−B×sin⁡(60∘)=0\sum F_y = A \times \sin(35^\circ) - B \times \sin(60^\circ) = 0

  • Algebraic Solution Strategy:

    • Isolate variable AA in the yy-component equation first because it contains no constant terms:     A×sin⁡(35∘)=B×sin⁡(60∘)A \times \sin(35^\circ) = B \times \sin(60^\circ)A=B×sin⁡(60∘)sin⁡(35∘)A = B \times \frac{\sin(60^\circ)}{\sin(35^\circ)}
    • Substitute the expression for AA into the xx-component equation to solve directly for BB, then back-substitute to find AA

Analysis of Mass-to-Weight Conversion and Geometry (Problem 45)

  • Mass versus Weight Conversion:

    • Given suspended mass: m=50 kgm = 50\,\text{kg} (kilogram is a unit of mass, not force).
    • Acceleration due to gravity: g=9.81 m/s2g = 9.81\,\text{m/s}^2 (or 9.8 m/s29.8\,\text{m/s}^2).
    • Downward gravitational force (weight):     W=m×gW = m \times gW=50 kg×9.81 m/s2=490.5 NW = 50\,\text{kg} \times 9.81\,\text{m/s}^2 = 490.5\,\text{N}

  • System Geometry and Component Equations:

    • Cable ACAC angle relative to horizontal: 60∘60^\circ (directed upward and leftward).
    • Cable BCBC angle relative to horizontal: 20∘20^\circ (directed downward and rightward).
    • Summation of forces along the xx-axis:     ∑Fx=−A×cos⁡(60∘)+B×cos⁡(20∘)=0\sum F_x = -A \times \cos(60^\circ) + B \times \cos(20^\circ) = 0
    • Summation of forces along the yy-axis:     ∑Fy=A×sin⁡(60∘)−B×sin⁡(20∘)−490.5 N=0\sum F_y = A \times \sin(60^\circ) - B \times \sin(20^\circ) - 490.5\,\text{N} = 0

  • Alternative Force Triangle Geometry Setup:

    • Downward vertical load: 490.5 N490.5\,\text{N}.
    • Angle between vertical load line and cable AA: 90∘−60∘=30∘90^\circ - 60^\circ = 30^\circ.
    • Angle between vertical load line and cable BB: 90∘+20∘=110∘90^\circ + 20^\circ = 110^\circ.
    • Remaining interior angle of vector triangle:     180∘−(30∘+110∘)=40∘180^\circ - (30^\circ + 110^\circ) = 40^\circ

Engineering versus Physics Vector Conventions

  • Engineering Mechanics Approach:

    • Employs acute reference angles (ranging from 0∘0^\circ to 90∘90^\circ) measured relative to the nearest horizontal or vertical coordinate axis.
    • Directional signs (+ or −+\text{ or }-) are assigned manually based on visual inspection of arrow directions in the free body diagram.
    • Example for a force AA directed upward and leftward at 50∘50^\circ to the horizontal:     Ax=−A×cos⁡(50∘)A_x = -A \times \cos(50^\circ)Ay=+A×sin⁡(50∘)A_y = +A \times \sin(50^\circ)

  • Standard Physics Approach:

    • Measures directional angles continuously counterclockwise starting from the positive xx-axis (0∘0^\circ to 360∘360^\circ).
    • Algebraic directional signs are determined automatically by the trigonometric functions.
    • Example for a force AA directed upward and leftward at 50∘50^\circ to the negative xx-axis:     θ=180∘−50∘=130∘\theta = 180^\circ - 50^\circ = 130^\circAx=A×cos⁡(130∘)A_x = A \times \cos(130^\circ)Ay=A×sin⁡(130∘)A_y = A \times \sin(130^\circ)
    • Mathematical Equivalence: Because cos⁡(130∘)=−cos⁡(50∘)\cos(130^\circ) = -\cos(50^\circ) and sin⁡(130∘)=sin⁡(50∘)\sin(130^\circ) = \sin(50^\circ), both conventions yield identical final component equations.