11.2 Work, Energy, Power & Simple Machines (Mechanical Advantage)

11.2 Work, Energy, Power & Simple Machines (Mechanical Advantage)

Mechanical devices across aircraft systems—such as flight control push-pull rods, bellcranks, landing gear actuators, cargo winches, and flap screw jacks—apply classical mechanics to transmit forces and perform work. An Aviation Maintenance Technician must be able to calculate work, energy, horsepower, mechanical advantage, and efficiency across all classes of simple machines in accordance with FAA-H-8083-30B.

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1. Mechanical Work and Energy Transformations

Definition of Mechanical Work

In physics, work ($W$) is performed only when an applied force ($F$) causes physical displacement ($d$) of an object in the direction of the applied force: $W = F \times d$

• If force is applied at an angle $\theta$ to the direction of motion: $W = F \times d \times \cos(\theta)$.
• No Displacement = No Work: If a technician pushes with $200\text{ lbs}$ of force against a stationary hangar door that does not move ($d = 0$), the physical work performed is strictly zero foot-pounds.
• Units of Work:
  • English Engineering Unit: Foot-Pound ($\text{ft-lb}$) — The work done when a constant force of 1 pound moves an object through a distance of 1 foot.
  • SI Metric Unit: Joule (J) — The work done when a force of 1 Newton displaces an object by 1 meter ($1\text{ J} = 1\text{ N}\cdot\text{m}$).
  • Conversion Equivalence: $1\text{ ft-lb} = 1.3558\text{ Joules} \approx 1.356\text{ J}$; $1\text{ Joule} = 0.7376\text{ ft-lb}$.

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Forms of Energy: Potential vs. Kinetic Energy

Energy is defined as the capacity or ability to perform physical work. Energy cannot be created or destroyed, only transformed from one state to another (Law of Conservation of Energy):

1. Potential Energy ($PE$): Energy stored within a physical body due to its spatial position, internal stress, or chemical configuration:
   • Gravitational Potential Energy: An aircraft of weight $W$ at an altitude $h$ above ground level possesses stored potential energy: $PE = m \times g \times h = W \times h \qquad [\text{ft-lb or Joules}]$
   • Elastic Potential Energy: Energy stored in compressed landing gear oleo strut springs or emergency blow-down accumulators ($PE = \frac{1}{2} k x^2$).
   • Chemical Potential Energy: Energy locked in hydrocarbon bonds of aviation fuel (Avgas $\approx 19,000\text{ BTU/lb}$, Jet-A $\approx 18,500\text{ BTU/lb}$).
2. Kinetic Energy ($KE$): Energy possessed by an object due to its mass and physical velocity: $KE = \frac{1}{2} m v^2 = \frac{W v^2}{2g} \qquad [\text{ft-lb or Joules}]$

Energy Interchange in Flight Maneuvers:
       [ High Altitude / Low Airspeed ]  =====> High PE, Low KE
                     │
                     ▼  Aircraft enters a dive (Gravity accelerates mass)
       [ Low Altitude / High Airspeed ]  =====> Low PE, High KE
                     │
                     ▼  Touchdown & Braking (Friction converts KE to thermal heat)
       [ Stationary Aircraft on Runway ] =====> Zero Flight KE (Brake Discs at 600°C)

Aircraft Brake Kinetic Energy Note: During an emergency rejected takeoff (RTO), an aircraft's total kinetic energy ($KE = \frac{1}{2} m v^2$) must be absorbed entirely by multi-disc carbon brakes and converted into thermal energy ($Q$). Because $KE$ is proportional to the square of velocity ($v^2$), doubling the takeoff abort speed quadruples the kinetic energy and thermal load on the brake assemblies.

2. Power and Mechanical Horsepower Ratings

Definition of Power

Power ($P$) is the time rate at which mechanical work is performed or energy is transformed: $P = \frac{\text{Work } (W)}{\text{Time } (t)} = \frac{\text{Force } (F) \times \text{Displacement } (d)}{\text{Time } (t)} = F \times v$

Mechanical Horsepower ($HP$)

Established by James Watt, standard mechanical horsepower equates the sustained pulling capability of a draft horse to mechanical units:

$1\text{ Mechanical Horsepower (HP)} = 550\text{ ft-lb/second} = 33,000\text{ ft-lb/minute} = 746\text{ Watts} = 0.746\text{ kW}$

$\text{Horsepower (HP)} = \frac{\text{Force (lbs)} \times \text{Distance (ft)}}{550 \times \text{Time (seconds)}} = \frac{\text{Force (lbs)} \times \text{Distance (ft)}}{33,000 \times \text{Time (minutes)}}$

3. Mechanical Advantage (TMA vs. AMA) and Efficiency

Simple machines do not create energy; they merely transform input force and displacement into output force and displacement. In every machine, mechanical friction and elastic deformation absorb a portion of input work as wasted thermal energy.

Theoretical vs. Actual Mechanical Advantage

1. Theoretical Mechanical Advantage (TMA / Ideal IMA): The frictionless geometric ratio of the distance moved by the effort ($d_E$) to the distance moved by the resistance load ($d_R$): $TMA = \frac{\text{Effort Distance } (d_E)}{\text{Resistance Distance } (d_R)} = \frac{\text{Effort Arm Length } (L_E)}{\text{Resistance Arm Length } (L_R)}$
2. Actual Mechanical Advantage (AMA): The real-world ratio of the output resistance force overcome ($F_R$) to the input effort force applied ($F_E$), incorporating frictional losses: $AMA = \frac{\text{Output Resistance Force } (F_R)}{\text{Input Effort Force } (F_E)}$
3. Mechanical Efficiency ($\eta$): The ratio of useful work output to total work input, expressed as a percentage: $\text{Efficiency } (\eta) = \frac{\text{Work Output}}{\text{Work Input}} \times 100\% = \frac{F_R \times d_R}{F_E \times d_E} \times 100\% = \frac{AMA}{TMA} \times 100\%$

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4. The Six Simple Machines in Aviation Systems

1. Levers

A lever is a rigid bar free to pivot around a fixed axis called the fulcrum. Levers are categorized into three distinct classes depending upon the relative positions of the fulcrum, effort, and resistance:

• Class 1 Lever: The fulcrum is positioned between the effort and the resistance ($E - F - R$).
  • Characteristics: Reverses the direction of motion. $TMA = L_E / L_R$ can be greater than 1, equal to 1, or less than 1.
  • Aviation Examples: Elevator trim tab pushrod bellcranks, crowbars, wire cutters, scissors, see-saws.
• Class 2 Lever: The resistance is positioned between the fulcrum and the effort ($F - R - E$).
  • Characteristics: Effort arm ($L_E$) is always longer than the resistance arm ($L_R$). $TMA$ is always strictly greater than 1.0. Acts as a continuous force multiplier (force advantage at the expense of travel distance).
  • Aviation Examples: Aircraft main landing gear mechanical down-lock drag braces, wheelbarrows, nutcrackers, foot brake pedals.
• Class 3 Lever: The effort is applied between the fulcrum and the resistance ($F - E - R$).
  • Characteristics: Resistance arm ($L_R$) is always longer than the effort arm ($L_E$). $TMA$ is always strictly less than 1.0. Acts as a force divider but a distance and speed multiplier (requires high effort force, but moves the load rapidly through a large distance).
  • Aviation Examples: Aircraft main landing gear hydraulic retraction actuators (where the hydraulic cylinder connects near the pivot hinge to swing the long gear strut through a $90^\circ$ arc with a compact cylinder stroke), trailing-edge flap drive pushrods, human arm flexing at the elbow.

Lever Arm Distance Layouts:
  Class 1:  [ Effort ]<==== L_E ====>[ Fulcrum ]<==== L_R ====>[ Resistance ]
  Class 2:  [ Fulcrum ]<==== L_R ====>[ Resistance ]<==== L_E ====>[ Effort ]
  Class 3:  [ Fulcrum ]<==== L_E ====>[ Effort ]<==== L_R ====>[ Resistance ]

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2. Pulleys (Cable Systems)

Pulleys consist of grooved wheels (sheaves) supported in a frame, utilized to route flexible steel aircraft flight control cables (e.g., $7\times19$ MIL-DTL-83420 carbon or stainless steel cables) or lift heavy cargo:

• Fixed Pulley ($TMA = 1$): Attached to stationary airframe structure. Does not amplify force; merely changes the direction of cable travel ($d_E = d_R$).
• Movable Pulley ($TMA = 2$): Attached directly to the moving load. The load is suspended by two cable strands; effort force required is halved, but effort cable must be pulled twice the distance ($d_E = 2 \times d_R$).
• Block and Tackle ($TMA = n$): A combination of multiple fixed and movable sheaves. The $TMA$ equals the number of rope or cable strands directly supporting the movable load block (the pull strand is excluded if pulled downward away from the load block).

3. Wheel and Axle

A wheel and axle consists of a large diameter wheel (or crank handle) rigidly attached to a smaller diameter shaft or axle: $TMA = \frac{\text{Radius of Wheel } (R)}{\text{Radius of Axle } (r)} = \frac{\text{Diameter of Wheel } (D)}{\text{Diameter of Axle } (d)}$

• Aviation Applications: Pilot flight control yoke (large wheel diameter provides roll mechanical advantage over the aileron drive sprocket), manual emergency landing gear extension hand cranks, elevator trim wheel.

4. Inclined Plane

An inclined plane is a flat sloping surface used to elevate heavy cargo by exerting a smaller force over a longer distance: $TMA = \frac{\text{Length of Incline } (L)}{\text{Vertical Height of Rise } (h)}$

• Aviation Applications: Aircraft cargo loading ramps, fuselage boarding ramps, wedge-locking fasteners.

5. Screw Jack

A screw jack is an inclined plane wrapped spirally around a central cylinder (helix), combined with a lever arm (crank handle). For each complete revolution ($360^\circ$) of the lever arm ($L$), the effort travels the circumference of a circle ($2\pi L$), advancing the load by the pitch ($P$) (distance between adjacent screw threads): $TMA = \frac{2 \pi L}{\text{Pitch } (P)}$

• Aviation Applications: Transport aircraft horizontal stabilizer trim actuators (acme screw jacks), trailing edge flap ballscrew actuators, maintenance tripod wing jacks.

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5. Worked Calculation Examples

Example 1: Class 3 Lever Landing Gear Actuator Calculation

Problem: An aircraft main landing gear strut pivots at a hinge point (fulcrum). A hydraulic retraction cylinder connects to the strut at a distance of $6.0\text{ inches}$ from the pivot ($L_E = 6.0\text{ in}$). The landing gear center of gravity and wheel assembly is located $36.0\text{ inches}$ from the pivot ($L_R = 36.0\text{ in}$), presenting a total resistance force of $450\text{ lbs}$.

• Determine the lever class, the theoretical mechanical advantage ($TMA$), the required hydraulic cylinder effort force ($F_E$), and the cylinder stroke ($d_E$) needed to move the wheel through an arc distance of $24.0\text{ inches}$.

Solution:

1. Identify lever class: Effort is between fulcrum and resistance $\rightarrow$ Class 3 Lever.
2. Calculate Theoretical Mechanical Advantage: $TMA = \frac{L_E}{L_R} = \frac{6.0\text{ in}}{36.0\text{ in}} = \frac{1}{6} = 0.1667$
3. Calculate required cylinder effort force ($F_E$): $F_E = \frac{F_R}{TMA} = \frac{450\text{ lbs}}{0.1667} = 450 \times 6 = 2,700.0\text{ lbs}$
4. Calculate required cylinder stroke displacement ($d_E$): $d_E = d_R \times TMA = 24.0\text{ in} \times \left(\frac{1}{6}\right) = 4.0\text{ inches}$ (Analysis: The hydraulic actuator must exert a massive $2,700\text{ lbs}$ of force, but a stroke of only $4.0\text{ inches}$ retracts the wheel a full $24.0\text{ inches}$).

Example 2: Engine Hoist Block and Tackle with Frictional Efficiency

Problem: A maintenance technician utilizes a 4-strand block and tackle hoist to lift a $600\text{-lb}$ aircraft engine a vertical height of $6.0\text{ feet}$. Due to bearing friction, the hoist has an efficiency rating of $80%$ ($\eta = 0.80$).

• Calculate the theoretical mechanical advantage ($TMA$), actual mechanical advantage ($AMA$), effort force ($F_E$), length of rope pulled ($d_E$), and total work input.

Solution:

1. Calculate TMA based on supporting strands: $TMA = n = 4$
2. Calculate Actual Mechanical Advantage ($AMA$): $AMA = TMA \times \eta = 4 \times 0.80 = 3.20$
3. Calculate required effort pull force ($F_E$): $F_E = \frac{F_R}{AMA} = \frac{600\text{ lbs}}{3.20} = 187.5\text{ lbs}$
4. Calculate length of effort rope pulled ($d_E$): $d_E = d_R \times TMA = 6.0\text{ ft} \times 4 = 24.0\text{ feet}$
5. Verify work and efficiency: $W_{\text{out}} = F_R \times d_R = 600\text{ lbs} \times 6.0\text{ ft} = 3,600\text{ ft-lb}$ $W_{\text{in}} = F_E \times d_E = 187.5\text{ lbs} \times 24.0\text{ ft} = 4,500\text{ ft-lb}$ $\eta = \frac{3,600}{4,500} \times 100\% = 80.0\%$

Example 3: Horizontal Stabilizer Screw Jack Mechanical Advantage

Problem: A transport aircraft trim actuator uses a screw jack having a pitch of $0.25\text{ inch}$ ($4\text{ threads per inch}$), driven by an emergency manual trim wheel with a radius of $10.0\text{ inches}$.

• Compute the theoretical mechanical advantage ($TMA$) of the screw jack mechanism.

Solution:

1. Calculate the circumference traced by the handle: $C = 2 \pi L = 2 \times 3.14159 \times 10.0\text{ in} = 62.8318\text{ inches}$
2. Calculate Theoretical Mechanical Advantage: $TMA = \frac{2 \pi L}{\text{Pitch } (P)} = \frac{62.8318\text{ in}}{0.25\text{ in}} = 251.33$ (Conclusion: A manual effort force of only $10\text{ lbs}$ on the trim wheel can exert an axial thrust of $2,513\text{ lbs}$ on the stabilizer in frictionless conditions).