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10.2 Pulleys, Wheel and Axle Systems, and Inclined Planes
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10.2 Pulleys, Wheel and Axle Systems, and Inclined Planes
Key Takeaways Explain This Topic AI
A fixed pulley anchored to a static overhead support provides IMA = 1 and merely redirects the applied pulling force, whereas a movable pulley rides with the load to provide IMA = 2, halving the required input effort while doubling the rope pull distance.
In block and tackle compound systems, IMA is determined exclusively by counting the number of rope strands directly supporting the movable load block (strands pulling upward); downward-pulling effort strands exiting fixed top sheaves do not add mechanical advantage.
Under frictionless conditions, continuous rope tension T is uniform throughout every segment of a single-line pulley system (T = F_effort = F_load / IMA), while sheave bearing friction lowers actual mechanical advantage (AMA = IMA × η).
Wheel and axle mechanisms function as continuous 360° rotating levers where IMA = R_wheel / r_axle = D_wheel / d_axle when effort turns the outer wheel to multiply torque (winches, steering wheels, screwdrivers) or IMA < 1 when turning the axle to multiply rim speed (bicycle wheels, vehicle drive axles).
Inclined planes (IMA = L / h), wedges (IMA = L / w), and screw jacks (IMA = 2πr / p) belong to the inclined plane family, trading increased linear or rotational travel distance for amplified output lifting and splitting forces.
Last updated: August 2026
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.
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 ( I M A = 1 IMA = 1 I M A = 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 F_e F e , the load rises upward on the opposite side with force F L F_L F L .
Advantage: Provides zero force multiplication ( I M A = 1 IMA = 1 I M A = 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 e f f o r t = F l o a d F_{effort} = F_{load} F e ff or t = F l o a d and d e f f o r t = d l o a d d_{effort} = d_{load} d e ff or t = d l o a d . ╔══════════╗ (Ceiling Anchor)
- ║
- ┌──╨──┐
- │ (O) │ ➔ Fixed Sheave (Stationary Axle)
- └──┬──┘
- [Effort]│ │
- (Pull │ │ [Load]
- Down) ▼ ▲ (Lifts Up)
- 2. The Movable Pulley ( I M A = 2 IMA = 2 I M A = 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 l o a d T_1 + T_2 = F_{load} T 1 + T 2 = F l o a d ). Because both strands support the load, the operator only exerts half the load's weight: F e f f o r t = F l o a d 2 ⟹ I M A = 2 F_{effort} = \frac{F_{load}}{2} \implies IMA = 2 F e ff or t = 2 F l o a d ⟹ I M A = 2
Distance Trade-Off: To raise the movable pulley and load by vertical height h h h , both supporting rope strands must shorten by h h h , requiring the operator to pull 2 h 2h 2 h of rope ( d e = 2 ⋅ d L d_e = 2 \cdot d_L d e = 2 ⋅ d L ). ╔══════════╗ (Ceiling Anchor)
- │
- │ ▲ [Effort] (Pulls Upward)
- ┌──┴──┬──┐
- │ ( O )│ ➔ Movable Sheave (Rides with Load)
- └──┬─────┘
- ▼
- [Load: F_L]
- 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 T T is identical throughout every segment of the rope : T = F e f f o r t = F l o a d I M A T = F_{effort} = \frac{F_{load}}{IMA} T = F e ff or t = I M A F l o a d
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 n = 4 n = 4 .
Ideal Mechanical Advantage: I M A = 4 IMA = 4 I M A = 4 .
Ideal effort force required: F e = 1 , 200 lbs 4 = 300 lbs F_e = \frac{1,200\text{ lbs}}{4} = 300\text{ lbs} F e = 4 1 , 200 lbs = 300 lbs .
Cable pulled to lift the engine 3 feet: d e = 4 × 3 ft = 12 feet d_e = 4 \times 3\text{ ft} = 12\text{ feet} d e = 4 × 3 ft = 12 feet .
If the system has an efficiency of η = 80 % \eta = 80\% η = 80% ( 0.80 0.80 0.80 ): A M A = I M A × η = 4 × 0.80 = 3.2 AMA = IMA \times \eta = 4 \times 0.80 = 3.2 A M A = I M A × η = 4 × 0.80 = 3.2 F e f f o r t , a c t u a l = 1 , 200 lbs 3.2 = 375 lbs F_{effort, actual} = \frac{1,200\text{ lbs}}{3.2} = 375\text{ lbs} F e ff or t , a c t u a l = 3.2 1 , 200 lbs = 375 lbs
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 ∘ 360^\circ 36 0 ∘ rotating lever:
The center of the shared shaft represents the fulcrum .
The outer wheel radius ( R w h e e l R_{wheel} R w h ee l ) represents the effort arm .
The inner axle radius ( r a x l e r_{axle} r a x l e ) 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
Ideal Mechanical Advantage (Effort on Wheel): I M A = R w h e e l r a x l e = D w h e e l d a x l e = Circumference of Wheel Circumference of Axle IMA = \frac{R_{wheel}}{r_{axle}} = \frac{D_{wheel}}{d_{axle}} = \frac{\text{Circumference of Wheel}}{\text{Circumference of Axle}} I M A = r a x l e R w h ee l = d a x l e D w h ee l = Circumference of Axle Circumference of Wheel (Because the ratio of radii equals the ratio of diameters, either dimension can be used directly).
Torque Equilibrium: τ w h e e l = τ a x l e ⟹ F e f f o r t ⋅ R w h e e l = F l o a d ⋅ r a x l e \tau_{wheel} = \tau_{axle} \implies F_{effort} \cdot R_{wheel} = F_{load} \cdot r_{axle} τ w h ee l = τ a x l e ⟹ F e ff or t ⋅ R w h ee l = F l o a d ⋅ r a x l e F e f f o r t = F l o a d × ( r a x l e R w h e e l ) = F l o a d I M A F_{effort} = F_{load} \times \left(\frac{r_{axle}}{R_{wheel}}\right) = \frac{F_{load}}{IMA} F e ff or t = F l o a d × ( R w h ee l r a x l e ) = I M A F l o a d
The Winch / Windlass (Crank and Drum)
A manual recovery winch consists of a crank handle of radius R R R turning a cable drum of radius r r r .
Calculation Example: A field rescue winch has a crank handle radius R = 16 inches R = 16\text{ inches} R = 16 inches and a cable drum radius r = 2 inches r = 2\text{ inches} r = 2 inches . To pull a 1 , 600 -lb 1,600\text{-lb} 1 , 600 -lb vehicle: I M A = R r = 16 in 2 in = 8.0 IMA = \frac{R}{r} = \frac{16\text{ in}}{2\text{ in}} = 8.0 I M A = r R = 2 in 16 in = 8.0 F e f f o r t = 1 , 600 lbs 8.0 = 200 lbs F_{effort} = \frac{1,600\text{ lbs}}{8.0} = 200\text{ lbs} F e ff or t = 8.0 1 , 600 lbs = 200 lbs
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: I M A = Ramp Length ( L ) Vertical Height ( h ) = 1 sin ( θ ) IMA = \frac{\text{Ramp Length } (L)}{\text{Vertical Height } (h)} = \frac{1}{\sin(\theta)} I M A = Vertical Height ( h ) Ramp Length ( L ) = sin ( θ ) 1 F e f f o r t = F l o a d × ( h L ) = F l o a d I M A F_{effort} = F_{load} \times \left(\frac{h}{L}\right) = \frac{F_{load}}{IMA} F e ff or t = F l o a d × ( L h ) = I M A F l o a d
Rule: A longer, gentler ramp provides higher I M A IMA I M A 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: I M A = Length of Wedge ( L ) Width (Thickness) at Base ( w ) IMA = \frac{\text{Length of Wedge } (L)}{\text{Width (Thickness) at Base } (w)} I M A = Width (Thickness) at Base ( w ) Length of Wedge ( L )
Rule: A longer, thinner wedge yields a higher I M A IMA I M A , 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 p p ): The linear distance between adjacent screw threads (the axial distance advanced in one complete 360 ∘ 360^\circ 36 0 ∘ rotation).
Radius ( r r r ): The radius of the turning handle, lever, or wrench.
In one complete 360 ∘ 360^\circ 36 0 ∘ turn:
- Effort moves through the circumference: d e f f o r t = 2 π r d_{effort} = 2 \pi r d e ff or t = 2 π r
- Load moves linearly by the thread pitch: d l o a d = p d_{load} = p d l o a d = p
- Ideal Mechanical Advantage: I M A = 2 ⋅ π ⋅ r p IMA = \frac{2 \cdot \pi \cdot r}{p} I M A = p 2 ⋅ π ⋅ r
Screw Jack Example: A screw jack with a pitch of 0.25 inches 0.25\text{ inches} 0.25 inches ( 1 / 4 in 1/4\text{ in} 1/4 in ) is operated using a 14 -inch 14\text{-inch} 14 -inch handle ( r = 14 in r = 14\text{ in} r = 14 in , use π ≈ 22 / 7 \pi \approx 22/7 π ≈ 22/7 ): d e = 2 × ( 22 7 ) × 14 in = 88 inches d_e = 2 \times \left(\frac{22}{7}\right) \times 14\text{ in} = 88\text{ inches} d e = 2 × ( 7 22 ) × 14 in = 88 inches I M A = 88 in 0.25 in = 352 IMA = \frac{88\text{ in}}{0.25\text{ in}} = 352 I M A = 0.25 in 88 in = 352
An input force of 20 lbs 20\text{ lbs} 20 lbs produces an ideal lifting thrust of F L = 20 × 352 = 7 , 040 lbs F_L = 20 \times 352 = 7,040\text{ lbs} F L = 20 × 352 = 7 , 040 lbs !
- 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.
What is the I M A IMA I M A of the system?
What ideal pulling force is required?
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.
What is the I M A IMA I M A of the screwdriver?
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.
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. |
- +------------------------------------+----------------------------------------------------+