Mechanical Equilibrium, Forces & Torque – Comprehensive Notes

Free Body Diagrams (FBD)

• Purpose
  • Visual tool for isolating and representing all forces on a single object.
  • Essential first step on MCAT force problems; draw before doing any math.
• How to construct
  • Represent object as a point or box; draw vectors emanating outward.
  • Label each with magnitude (N) & direction; include normal force, weight, tension, friction, applied, etc.
  • Choose a convenient coordinate system (often align an axis with motion or an inclined plane).
  • Indicate angles between forces and axes; list known trig values.
• Example: 3 ropes pulling a tire
  • Given forces: $F<em>1 = 100\,\text{N at }30^\circ\text{ above }+x$, $F</em>2 = 125\,\text{N at }37^\circ\text{ above }-x$, $F_3 = 125\,\text{N along }-y$.
  • Resolve into components (use provided trig):
  • $F_{1x} = 100\cos30^\circ = 86.6\,\text{N}$
  • $F_{1y} = 100\sin30^\circ = 50\,\text{N}$
  • $F_{2x} = 125\cos37^\circ = 100\,\text{N}$ (to the left)
  • $F_{2y} = 125\sin37^\circ = 75\,\text{N}$ (up)
  • $F<em>{3x}=0$, $F</em>{3y}=-125\,\text{N}$
  • Sum of components:
  • $F_{\text{net},x} = +86.6 - 100 = -13.4\,\text{N}$ (left)
  • $F_{\text{net},y} = 50 + 75 -125 = 0$
  • Resultant: $|\vec F_{\text{net}}| = 13.4\,\text{N left}$ (purely horizontal).

Translational Equilibrium

• Definition
  • First condition of equilibrium: $\sum \vec F = 0$.
  • No linear acceleration ( $a = 0$ ); velocity is constant (can be 0 or non-zero).
• Newton’s Laws connection
  • Newton I: Object maintains state of motion if $\sum \vec F = 0$.
  • Newton II: If $\sum \vec F = 0$, then $\vec a = \frac{\sum \vec F}{m} = 0$.
  • Newton III: Forces come in action–reaction pairs, helpful when ropes, pulleys, contact involved.
• Analyzing problems
  1. Draw FBD for each interacting object.
  2. Choose axes; break forces into perpendicular components.
  3. Write $\sum F<em>x = 0$ & $\sum F</em>y = 0$; solve simultaneously.
• Example: Two-block static system
  • Block A (mass $m_A = 15\,\text{kg}$) on horizontal surface, connected by rope over pulley to hanging Block B.
  • Coefficient of static friction at A–surface: $\mu_s = 0.2$.
  • Max $m_B$ before motion when static friction is at maximum.
  • Forces on A: $T$ (right), $f<em>s^{\max} = \mu</em>s N = \mu<em>s m</em>A g$ (left), $N = m<em>A g$ (up), $m</em>A g$ (down).
  • Forces on B: $m_B g$ (down), $T$ (up).
  • Equilibrium → $T = m<em>B g = f</em>s^{\max} = \mu<em>s m</em>A g$ ⇒ $m<em>B = \mu</em>s m_A = 0.2(15\,\text{kg}) = 3.0\,\text{kg}$.

Rotational Equilibrium & Torque

• Torque (moment of force)
  • Vector, causes angular acceleration.
  • Magnitude: $\tau = rF\sin\theta$ where
  • $r$ = lever arm (distance from pivot to line of action of force).
  • $F$ = applied force.
  • $\theta$ = angle between $\vec r$ and $\vec F$.
  • Right-hand rule gives direction (into/out of page).
• Second condition of equilibrium: $\sum \tau = 0$ about any pivot.
  • Convention: Counter-clockwise positive, clockwise negative (MCAT default).
  • If satisfied and $\sum \vec F = 0$, both translational & rotational equilibrium hold.
• Possible motions while in rotational equilibrium
  • Not rotating at all (MCAT’s usual intent).
  • Rotating with constant $\omega$ (rarely tested).
• Problem-solving steps
  1. Pick a pivot (often the fulcrum) to eliminate unknown forces with zero lever arm.
  2. Write $\sum \tau = 0$, assigning signs.
  3. If needed, also apply $\sum \vec F = 0$.
• Example: Balanced seesaw
  • Seesaw mass $m<em>s = 5\,\text{kg}$ (centered), Block 1: $m</em>1 = 10\,\text{kg}$ located $r<em>1 = 2\,\text{m}$ left of fulcrum; Block 2 at $r</em>2 = 0.5\,\text{m}$ right.
  • Rotational equilibrium: $r<em>1 m</em>1 g = r<em>2 m</em>2 g$ ⇒ $m<em>2 = \frac{r</em>1 m<em>1}{r</em>2} = \frac{2\times10}{0.5} = 40\,\text{kg}$.
  • Translational equilibrium for vertical forces: $N = (m<em>s + m</em>1 + m_2)g$ ⇒ $N = (5 + 10 + 40)\,\text{kg}\times 10\,\text{m/s}^2 = 550\,\text{N}$ (exact calc $\approx 539\,\text{N}$ if $g = 9.8\,\text{m/s}^2$).

Key Equations & Constants

• Force: $\vec F = m\vec a$.
• Static friction (max): $f<em>s^{\max} = \mu</em>s N$.
• Kinetic friction: $f<em>k = \mu</em>k N$ (acts during sliding, independent of speed under MCAT assumptions).
• Torque: $\vec \tau = \vec r \times \vec F$, $|\tau| = rF\sin\theta$.
• Translational Equilibrium: $\sum \vec F = 0$.
• Rotational Equilibrium: $\sum \tau = 0$.
• Common trig values: $\sin30^\circ = 0.5$, $\cos30^\circ = 0.866$, $\sin37^\circ = 0.6$, $\cos37^\circ = 0.8$, $\sin90^\circ = 1$.
• Standard gravity: $g \approx 9.8\,\text{m/s}^2$ (MCAT often rounds to $10\,\text{m/s}^2$).

Strategy & Real-World Connections

• Always start with an FBD—reduces errors and clarifies unknowns.
• Choose pivot wisely to cancel unknown support forces when applying $\sum \tau = 0$ (e.g., pick the fulcrum, hinge, or point of contact).
• Remember: Equilibrium ≠ Rest; constant velocity (linear or angular) still counts.
• Medicine tie-ins
  • Lever arms mirror musculoskeletal system (bones as levers, joints as fulcrums, muscles applying force).
  • Translational & rotational equilibrium analyze posture, orthopedic supports, prosthetic design.
• Ethical/Professional point: Solid grasp of mechanics underlies medical technology (wheelchairs, surgical robots, imaging gantries); building competency now will aid patient safety and innovation.

Recap of Chapter Take-aways

• Kinematics gives motion; dynamics (forces & torques) explain causes.
• Four constant-acceleration equations allow time-position-velocity linkage.
• When $\sum \vec F = 0$ → no linear acceleration; when $\sum \tau = 0$ → no angular acceleration.
• Newton’s three laws remain the backbone; all MCAT mechanics scenarios—linear, inclined, projectile, circular—reduce to their application.
• Practice combining translational & rotational conditions; MCAT favors systems (pulleys, beams, seesaws, elevators) where both apply simultaneously.