10.2 Pulleys, Wheel and Axle Systems, and Inclined Planes

10.2 Pulleys, Wheel and Axle Systems, and Inclined Planes

Core Principle: Rotational and inclined simple machines transform linear forces into rotational torques and trade displacement distance for mechanical advantage. On the CAT-ASVAB Mechanical Comprehension subtest, you will encounter diagram-based questions requiring you to determine rope tensions, count load-supporting strands, calculate pulling forces, evaluate wheel-to-axle diameter ratios, and determine mechanical advantage for ramps, wedges, and screw jacks.

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Pulley Fundamentals: Fixed vs. Movable Pulleys

A pulley is a grooved wheel (called a sheave) mounted on an axle within a supporting frame that carries a flexible rope, cable, or chain. Pulleys fall into two primary categories:

+-----------------------------------------------------------------------------------------+
|                              FIXED PULLEY VS. MOVABLE PULLEY                            |
+--------------------------+------------------------------+-------------------------------+
| Mechanical Feature       | Fixed Pulley                 | Movable Pulley                |
+--------------------------+------------------------------+-------------------------------+
| Axle Mounting            | Bolted to fixed beam/ceiling | Attached directly to the load |
|                          | (stationary in space)        | (moves vertically with load)  |
+--------------------------+------------------------------+-------------------------------+
| Force Direction          | Reverses direction           | Same direction                |
|                          | (pull down ➔ lift up)        | (pull up ➔ lift up)           |
+--------------------------+------------------------------+-------------------------------+
| Ideal Mech Advantage     | IMA = 1                      | IMA = 2                       |
+--------------------------+------------------------------+-------------------------------+
| Effort Force (Ideal)     | F_effort = F_load            | F_effort = F_load / 2         |
+--------------------------+------------------------------+-------------------------------+
| Rope Distance Pulled     | d_effort = d_load            | d_effort = 2 × d_load         |
+--------------------------+------------------------------+-------------------------------+
| Lever Model Equivalent   | Class 1 Lever (Equal arms)   | Class 2 Lever (Fulcrum at rim)|
+--------------------------+------------------------------+-------------------------------+

1. The Fixed Pulley ($IMA = 1$)

• Mechanism: The axle is anchored to a stationary structure (ceiling, mast, beam). As you pull downward on one side with effort force $F_e$, the load rises upward on the opposite side with force $F_L$.
• Advantage: Provides zero force multiplication ($IMA = 1$). Its value lies entirely in directional convenience—allowing an operator to use body weight and gravity to pull downward rather than lifting against gravity.
• Equilibrium: $F_{effort} = F_{load}$ and $d_{effort} = d_{load}$.

      ╔══════════╗ (Ceiling Anchor)
           ║
        ┌──╨──┐
        │ (O) │ ➔ Fixed Sheave (Stationary Axle)
        └──┬──┘
   [Effort]│  │
     (Pull │  │ [Load]
     Down) ▼  ▲ (Lifts Up)

2. The Movable Pulley ($IMA = 2$)

• Mechanism: One end of the rope is anchored to an overhead support, the rope loops under the pulley sheave, and the load is suspended from the pulley's axle frame. The operator pulls upward on the free rope end.
• Advantage: The load's weight is shared equally between two vertical rope segments ($T_1 + T_2 = F_{load}$). Because both strands support the load, the operator only exerts half the load's weight: $F_{effort} = \frac{F_{load}}{2} \implies IMA = 2$
• Distance Trade-Off: To raise the movable pulley and load by vertical height $h$, both supporting rope strands must shorten by $h$, requiring the operator to pull $2h$ of rope ($d_e = 2 \cdot d_L$).

      ╔══════════╗ (Ceiling Anchor)
           │
           │  ▲ [Effort] (Pulls Upward)
        ┌──┴──┬──┐
        │   ( O )│ ➔ Movable Sheave (Rides with Load)
        └──┬─────┘
           ▼
        [Load: F_L]

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Compound Pulley Systems: Block and Tackle

A block and tackle combines multiple fixed and movable pulleys mounted inside rigid block housings to achieve high mechanical advantage for heavy lifting operations (engine cranes, naval rigging, recovery winches).

+-----------------------------------------------------------------------------------------+
|                    GOLDEN RULE FOR FINDING IMA OF A BLOCK AND TACKLE                    |
+-----------------------------------------------------------------------------------------+
|  The Ideal Mechanical Advantage (IMA) equals the EXACT NUMBER OF ROPE STRANDS THAT      |
|  DIRECTLY SUPPORT THE MOVABLE LOAD BLOCK.                                               |
|                                                                                         |
|  CRITICAL ASVAB TEST RULES:                                                             |
|  1. Strands pulling UPWARD on the movable block ALWAYS COUNT.                           |
|  2. If the effort rope exits DOWNWARD from a FIXED overhead pulley, DO NOT count the    |
|     pull strand (it only redirects force, adding zero mechanical advantage).            |
|  3. If the effort rope exits UPWARD from a MOVABLE block, DO COUNT the pull strand      |
|     (it directly supports and lifts the load).                                          |
+-----------------------------------------------------------------------------------------+

Continuous Rope Tension Principle

In any single continuous rope woven through a frictionless pulley system, tension $T$ is identical throughout every segment of the rope: $T = F_{effort} = \frac{F_{load}}{IMA}$

Standard Block and Tackle Configurations Table

+-----------------------------------------------------------------------------------------+
|                        COMMON BLOCK AND TACKLE EXAM CONFIGURATIONS                      |
+-------------------+--------------+--------------+---------------+-----------------------+
| System Name       | Fixed Sheaves| Movable Sheav| Pull Direction| Resulting Advantage   |
+-------------------+--------------+--------------+---------------+-----------------------+
| Single-Movable    | 1 Fixed      | 1 Movable    | Downward      | IMA = 2 (2 strands)   |
| Luff Tackle       | 2 Fixed      | 1 Movable    | Downward      | IMA = 3 (3 strands)   |
| Double-Double     | 2 Fixed      | 2 Movable    | Downward      | IMA = 4 (4 strands)   |
| 5-Strand Tackle   | 2 Fixed      | 2 Movable    | Upward        | IMA = 5 (5 strands)   |
| Three-Fold        | 3 Fixed      | 3 Movable    | Downward      | IMA = 6 (6 strands)   |
+-------------------+--------------+--------------+---------------+-----------------------+

• Worked Example (4-Strand System): A double-sheave fixed block and double-sheave movable block lift a 1,200-lb engine block. The rope exits downward from the top fixed block.
  • Number of supporting strands pulling up on the movable block: $n = 4$.
  • Ideal Mechanical Advantage: $IMA = 4$.
  • Ideal effort force required: $F_e = \frac{1,200\text{ lbs}}{4} = 300\text{ lbs}$.
  • Cable pulled to lift the engine 3 feet: $d_e = 4 \times 3\text{ ft} = 12\text{ feet}$.
  • If the system has an efficiency of $\eta = 80%$ ($0.80$): $AMA = IMA \times \eta = 4 \times 0.80 = 3.2$ $F_{effort, actual} = \frac{1,200\text{ lbs}}{3.2} = 375\text{ lbs}$

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Wheel and Axle Mechanics

A wheel and axle consists of two rigidly joined concentric cylinders of different diameters that rotate together around a shared central rotational axis. When the wheel completes one rotation, the axle completes exactly one rotation.

                     ┌──────────────────────────┐
                     │        WHEEL (R)         │
                     │      ┌────────────┐      │
                     │      │  AXLE (r)  │      │
                     │      │     ●      │      │
                     │      └────────────┘      │
                     │                          │
                     └──────────────────────────┘

The Continuous Rotating Lever Analogy

A wheel and axle functions as a continuous $360^\circ$ rotating lever:

• The center of the shared shaft represents the fulcrum.
• The outer wheel radius ($R_{wheel}$) represents the effort arm.
• The inner axle radius ($r_{axle}$) represents the load arm.

+-----------------------------------------------------------------------------------------+
|                             WHEEL AND AXLE OPERATION MODES                              |
+------------------------------------+----------------------------------------------------+
| MODE A: Effort Applied to WHEEL    | MODE B: Effort Applied to AXLE                     |
+------------------------------------+----------------------------------------------------+
| • Multiplies FORCE (IMA > 1)       | • Multiplies SPEED & DISTANCE (IMA < 1)            |
| • Large wheel turns small axle     | • Small axle turns large wheel                     |
| • IMA = R_wheel / r_axle           | • IMA = r_axle / R_wheel                           |
| • Torque generated: τ = F_e × R    | • Outer rim velocity is magnified                  |
| • Examples: Steering wheel, door   | • Examples: Bicycle rear wheel, automobile drive   |
|   knob, screwdriver handle, winch, |   axle turning road tires, helicopter rotor hub,   |
|   water well windlass, pencil sharp|   marine propeller shaft turning blades            |
+------------------------------------+----------------------------------------------------+

Wheel and Axle Equations

1. Ideal Mechanical Advantage (Effort on Wheel): $IMA = \frac{R_{wheel}}{r_{axle}} = \frac{D_{wheel}}{d_{axle}} = \frac{\text{Circumference of Wheel}}{\text{Circumference of Axle}}$ (Because the ratio of radii equals the ratio of diameters, either dimension can be used directly).
2. Torque Equilibrium: $\tau_{wheel} = \tau_{axle} \implies F_{effort} \cdot R_{wheel} = F_{load} \cdot r_{axle}$ $F_{effort} = F_{load} \times \left(\frac{r_{axle}}{R_{wheel}}\right) = \frac{F_{load}}{IMA}$

The Winch / Windlass (Crank and Drum)

A manual recovery winch consists of a crank handle of radius $R$ turning a cable drum of radius $r$.

• Calculation Example: A field rescue winch has a crank handle radius $R = 16\text{ inches}$ and a cable drum radius $r = 2\text{ inches}$. To pull a $1,600\text{-lb}$ vehicle: $IMA = \frac{R}{r} = \frac{16\text{ in}}{2\text{ in}} = 8.0$ $F_{effort} = \frac{1,600\text{ lbs}}{8.0} = 200\text{ lbs}$

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The Inclined Plane Family: Ramps, Wedges & Screws

The inclined plane family consists of simple machines that use sloping surfaces to trade extended displacement distance for reduced applied force.

+-----------------------------------------------------------------------------------------+
|                               THE INCLINED PLANE FAMILY                                 |
+-------------------+--------------------+------------------------+-----------------------+
| Machine Type      | Geometric Form     | Mechanical Advantage   | Typical Applications  |
+-------------------+--------------------+------------------------+-----------------------+
| Inclined Plane    | Stationary sloping | IMA = Length / Height  | Loading ramps, ADA    |
| (Ramp)            | ramp surface       | IMA = L / h = 1/sin(θ) | ramps, switchbacks    |
+-------------------+--------------------+------------------------+-----------------------+
| Wedge             | Portable, moving   | IMA = Length / Width   | Axe, wood splitter,   |
|                   | double incline     | IMA = L / w            | chisel, knife, doorstop|
+-------------------+--------------------+------------------------+-----------------------+
| Screw             | Helical incline on | IMA = 2πr / Pitch      | Screw jack, C-clamp,  |
|                   | cylindrical shaft  | IMA = (2 · π · r) / p  | wood screws, bolts    |
+-------------------+--------------------+------------------------+-----------------------+

   1. Inclined Plane:            2. Wedge:                 3. Screw Jack:
           /|                       ┌─────────────┐             ┌────────┐
      L   / |                  L    │      w      │             │ Handle │ (Radius r)
         /  | h               ──────┼─────────────┼──►          └───┬────┘
        /   |                       │  ╱       ╲  │                 │ (Pitch p)
       /____|                       └─╱─────────╲─┘             ════╧═════

1. Inclined Planes (Ramps)

An inclined plane allows heavy loads to be raised to an elevation using a significantly smaller pushing force along its length than direct vertical lifting.

• Formulas: $IMA = \frac{\text{Ramp Length } (L)}{\text{Vertical Height } (h)} = \frac{1}{\sin(\theta)}$ $F_{effort} = F_{load} \times \left(\frac{h}{L}\right) = \frac{F_{load}}{IMA}$
• Rule: A longer, gentler ramp provides higher $IMA$ and requires less pushing force, but the load must travel a greater distance.

2. Wedges

A wedge is a portable inclined plane driven into or under an object to split, cut, or secure it.

• Formula: $IMA = \frac{\text{Length of Wedge } (L)}{\text{Width (Thickness) at Base } (w)}$
• Rule: A longer, thinner wedge yields a higher $IMA$, generating massive lateral splitting forces from modest driving hammer blows.

3. Screws (Helical Inclined Planes)

A screw is an inclined plane wrapped helically around a central cylindrical core.

• Pitch ($p$): The linear distance between adjacent screw threads (the axial distance advanced in one complete $360^\circ$ rotation).
• Radius ($r$): The radius of the turning handle, lever, or wrench.
• In one complete $360^\circ$ turn:
  • Effort moves through the circumference: $d_{effort} = 2 \pi r$
  • Load moves linearly by the thread pitch: $d_{load} = p$
• Ideal Mechanical Advantage: $IMA = \frac{2 \cdot \pi \cdot r}{p}$
• Screw Jack Example: A screw jack with a pitch of $0.25\text{ inches}$ ($1/4\text{ in}$) is operated using a $14\text{-inch}$ handle ($r = 14\text{ in}$, use $\pi \approx 22/7$): $d_e = 2 \times \left(\frac{22}{7}\right) \times 14\text{ in} = 88\text{ inches}$ $IMA = \frac{88\text{ in}}{0.25\text{ in}} = 352$
  • An input force of $20\text{ lbs}$ produces an ideal lifting thrust of $F_L = 20 \times 352 = 7,040\text{ lbs}$!

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Step-by-Step Problem Derivations

Problem 1: Block and Tackle System

A 600-lb crate is lifted using a block and tackle system with 2 fixed sheaves and 2 movable sheaves. The pull rope exits downward from the top fixed block.

1. What is the $IMA$ of the system?
2. What ideal pulling force is required?
3. If the crate is raised 4 feet, how much rope must be pulled?

Solution Steps:
Step 1: Count strands supporting the movable block. With 2 movable sheaves, 4 strands pull upward. The downward exit rope does not count. IMA = 4.
Step 2: Ideal effort force F_e = F_L / IMA = 600 lbs / 4 = 150 lbs.
Step 3: Distance pulled d_e = IMA × d_L = 4 × 4 ft = 16 feet.

Problem 2: Screwdriver Wheel-and-Axle

A mechanic uses a screwdriver with a handle diameter of 1.5 inches to drive a screw with a shank diameter of 0.25 inches.

1. What is the $IMA$ of the screwdriver?
2. If the mechanic exerts 12 lb-in of torque on the handle, what torque is transferred to the screw shank?

Solution Steps:
Step 1: IMA = D_handle / d_shank = 1.5 in / 0.25 in = 6.0.
Step 2: In a direct rigid shaft, torque is transmitted directly: Torque_shank = Torque_handle = 12 lb-in.
        The tangential force at the screw edge is multiplied: F_edge = IMA × F_handle.

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ASVAB Exam Pitfalls & Traps

+-----------------------------------------------------------------------------------------+
|                            PULLEYS & INCLINES COMMON TRAPS                              |
+------------------------------------+----------------------------------------------------+
| Exam Trap                          | Correct Mechanical Rule                            |
+------------------------------------+----------------------------------------------------+
| Counting downward pull strand in   | Downward pull rope from top fixed sheave does not  |
| block and tackle                   | support load weight; DO NOT count it in IMA.       |
+------------------------------------+----------------------------------------------------+
| Mixing radius and diameter in      | Use R/r OR D/d; never divide wheel diameter by     |
| wheel and axle ratios              | axle radius.                                       |
+------------------------------------+----------------------------------------------------+
| Omitting 2π in screw jack formulas | Handle travels in a full circle (2πr); do not use  |
|                                    | radius alone in IMA = 2πr / pitch.                 |
+------------------------------------+----------------------------------------------------+