Spring Basics and Extended Body Equilibrium Notes

Springs: basics and Hooke's law

• Springs are the final topic of particle equilibrium; then extended body equilibrium follows.
• Everyday relevance: springs are everywhere (e.g., pencils, absorption systems).
• Two key quantities to find the force from a spring:
  • Spring constant, denoted by $k$ (units: $\text{N/m}$).
  • Displacement of the spring, measured from its rest state.
• Rest length (unstretched length) is LO; this is the reference length of the spring when no force is applied.
• Change in length (displacement) when the spring is deformed:
  • For stretching: $\Delta = L<em>S - L</em>O\,$ where $L_S$ is the final length after stretch.
  • For compression: define a different delta: $\Delta<em>c = L</em>O - L<em>C$, where $L</em>C$ is the compressed length.
• Hooke's law relates force to displacement:$F = k\Delta$ for stretching (pull force) and, for compression, the magnitude is $F = k\Delta_c$ (the spring pushes back).
• Direction and action of the spring:
  • When stretched, the spring pulls on the attached object toward its rest position.
  • When compressed, the spring pushes away from the attached object (toward its rest position).
• Geometry tips for the lengths:
  • Original (unstretched) length LO is the distance between the spring’s endpoints before loading.
  • Final (stretched) length LF is the distance after loading.
• In calculations, signs matter when summing forces in equilibrium. We often present magnitudes first, then assign signs based on the chosen system.
• In the stretch case: LO is the baseline; LS is the position after deformation; Delta = LS - LO; Force magnitude = $F = k\Delta$.
• In the compression case: define Delta2 = LO - LC; Force magnitude = $F = k\Delta_2$.
• Summary of the two cases:
  • Stretch: force magnitude $F = k(LS - LO)$, direction is along the spring toward shortening.
  • Compression: force magnitude $F = k(LO - LC)$, direction is along the spring away from the shortened end.
• Notes on sign conventions:
  • When solving equilibrium problems, you’ll assign positive/negative signs to forces depending on your axis. The magnitudes are often written first, and signs are applied in the vector sum equations.
• Example intuition: to stretch a spring by a certain amount, you must apply a force proportional to that amount, and the spring reacts with an equal and opposite force on the object it’s attached to.

Example: two symmetric springs attached to rings A and B connected by a cable

• Setup:
  • Two springs, each with spring constant $k = 100\ \text{N/m}$.
  • The springs are connected to O-rings at A and B via a cable, with the system initially symmetric.
  • Original geometry: each spring length is 2 m; there is 1 m between the rings; the depth (vertical distance) is 1.5 m.
  • After attaching two masses at rings A and B, the rings sag by $s = 0.5\ \text{m}$.
• Quantities given/derived:
  • Sag after loading: the vertical distance from the original plane increases by 0.5 m for each side.
  • Original length LO of each spring (before loading): use the initial geometry. With horizontal span 2 m and vertical rise 1.5 m, the length LO is
    $LO = \sqrt{2^2 + (1.5)^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5\ \text{m}.$
  • Final length LF after sag: horizontal distance remains 2 m, vertical distance becomes 1.5 m + 0.5 m = 2.0 m, so
    $LF = \sqrt{2^2 + 2^2} = \sqrt{8} \approx 2.828\ \text{m}.$
  • Spring extension: $\Delta = LF - LO = \sqrt{8} - 2.5 \approx 2.828 - 2.5 = 0.328\ \text{m}.$
  • Spring force: $F_s = k\Delta = 100\times 0.328 \approx 32.8\ \text{N}.$
• Force decomposition and free-body considerations: ring A (one of the rings) as the body of interest.
  • Forces on ring A:
  • Weight of the weight attached at ring A: denote as the tension in the cable AB, $T_{AB}$, which supports the mass at A (for a stationary ring, weight is balanced by vertical components).
  • Tension in the cable AB pulling to the right from A toward B: $T_{AB}$.
  • Force from the spring on A: the spring pulls toward C (the other end of the spring). This force is of magnitude $F_s$ and points along the spring toward C (i.e., away from A toward the spring’s other end).
  • Components of the spring force: use the geometry of the spring to find vertical and horizontal components.
  • In the given final geometry, the line from A to C forms a right triangle with legs of 2 m (horizontal) and 2 m (vertical), so the final spring segment has length $LF = \sqrt{8}$ m.
  • The angle relative to the horizontal has sin(theta) = opposite/hypotenuse = 2/\sqrt{8} = 1/\sqrt{2}.
  • Therefore, the vertical component of the spring force is
    $F<em>{s,y} = F</em>s \sin\theta = F<em>s \left(\frac{2}{\sqrt{8}}\right) = \frac{F</em>s}{\sqrt{2}}.$
  • Free-body equations for ring A (assuming vertical equilibrium, no horizontal acceleration):
  • Vertical: $F<em>{s,y} - w</em>A = 0$ where $w_A = m g$ is the weight of the attached mass at A.
  • Horizontal: the horizontal components balance with cable tension: not needed to solve for mass here if symmetry is assumed and vertical balance is primary.
  • Symmetry simplification: since the two masses are assumed equal due to symmetry and equal spring constants, ring A and ring B have the same mass: $m<em>1 = m</em>2 = m.$
• Solve for mass using vertical equilibrium at A:
  • From vertical balance, $m g = F<em>{s,y} = \frac{F</em>s}{\sqrt{2}}.$
  • Therefore, the mass is
    $m = \frac{F_s}{g\sqrt{2}} = \frac{32.843}{9.81 \cdot \sqrt{2}} \approx 2.37\ \text{kg}.$
• Numerical note: the exact numeric for the spring force might be printed as $32.843\ \text{N}$ due to rounding in intermediate steps; the final mass estimate is ~$2.37\ \text{kg}$ for each weight.
• Why the two masses are equal: due to perfect symmetry in spring constants and geometry; if the springs or geometry were not symmetric, the masses could differ and additional information would be required.
• Takeaway: solving this example combines Hooke's law, geometry to resolve force components, and vertical force balance to extract the mass.

Extended body equilibrium: coplanar/concurrent vs extended bodies

• When forces act on a rigid body at a single point, the system is coplanar and concurrent; equilibrium can be solved with two force balance equations (Fx = 0, Fy = 0).
• Extended body: forces are applied at different points on the body, so moments (torques) must be considered in addition to force balances.
• In extended-body problems, you generally need three independent equilibrium equations:
  • Sum of forces in x: $\sum F_x = 0$
  • Sum of forces in y: $\sum F_y = 0$
  • Sum of moments about an axis (usually z, coming out of the plane): $\sum \tau_z = 0$
• The third equation arises because forces have nonzero moment arms about the chosen point when applied away from it.
• Three-dimensional cases would need more equations (moment about multiple axes), but we focus on planar problems here.
• Method for rigid-body/extended-body problems (step-by-step): 1) Draw a free-body diagram (FBD) of the body of interest, isolating it from its surroundings. 2) Replace supports with reaction forces (instead of drawing the support shapes). For a two-dimensional case:
  • Pin support (A): provides two perpendicular reactions Ax and Ay (and a reaction moment if the pin also resists rotation; typically a pin does not provide a moment, but a fixed support would).
  • Roller support (B): provides a single reaction perpendicular to the contact surface; commonly B_y for a horizontal beam resting on a roller.
    3) Identify external loads acting on the body (point loads, distributed loads, weight, etc.).
    4) Apply equilibrium equations:
  • $\sum F<em>x = 0$, $\sum F</em>y = 0$, and $\sum \tau_z = 0$ about a chosen point.
    5) Choose the point for moments that minimizes the number of unknowns (often a point where a few forces pass through the point, so they do not contribute to the moment).
• Note about the pin and the roller (support modeling):
  • A pin at A restrains movement in both x and y directions. The reaction forces are Ax and Ay. The pin does not inherently add a moment in the simple model; if a moment reaction is needed, the support type would have to be different.
  • A roller at B resists motion perpendicular to the surface (vertical for a horizontal beam). The roller contributes a single reaction, usually B_y, oriented perpendicular to the surface.
• Free-body-diagram convention for supports (practical tip): replace the physical support with its reaction forces in the FBD (don’t draw the pin/roller itself; draw the forces they exert on the body).
• Center of gravity and distributed weight notes:
  • For a uniform beam, the self-weight acts at the center of gravity (midpoint of the beam).
  • In many problems, the self-weight of components is neglected if small compared to external loads; include it if the problem requires it.
• Example outline: simply supported beam with a pin at A and a roller at B, carrying a load W located at a distance L/3 from A.
  • Steps:
  • Draw the beam as a rectangle (to emphasize dimensions).
  • Identify supports: Pin at A (Ax, Ay) and roller at B (B_y).
  • External load W located at x = L/3 from A.
  • If included, the beam’s own weight W_beam acts at its center (x = L/2).
  • Write equilibrium equations:
    • Horizontal: $\sum F<em>x = 0 = A</em>x.$ (Often A_x = 0 if no horizontal loads are present.)
    • Vertical: $\sum F<em>y = 0 = A</em>y + B<em>y - W - W</em>{beam}.$
    • Moments about A: choose A to simplify; a common equation is
      $-W\left(\frac{L}{3}\right) - W<em>{beam}\left(\frac{L}{2}\right) + B</em>y\,L = 0.$
  • Solve the three equations for the unknown reactions (Ay, By, and potentially A_x if needed).
  • Important practical tip: choose the moment center at a point where reactions pass through the point to avoid including their moment arms (e.g., about point A, Ax and Ay do not contribute to the moment).
• Practical note: a common exam pitfall is forgetting to specify the moment point or incorrectly accounting for which forces produce moments about that point.
• Resource reference mentioned in class: there is a two-dimensional support reaction document (often found in the course portal) that shows how to model pin and roller reactions for problems like these.
• Connection to the course sequence:
  • Particles: solved with two equations (Fx and Fy).
  • Extended bodies: add the moment equation to handle rotations and moments due to forces with nonzero lever arms.
• Quick recap of the core ideas:
  • Springs: k, LO, LS, LF, Δ, F = kΔ, direction depends on stretch vs compression.
  • Extended bodies require three equilibrium equations: sum of forces in x, sum of forces in y, and sum of moments about z.
  • Use free-body diagrams to model supports with appropriate reaction forces, and include external loads and weights where applicable.