Mechanical Reasoning Guide
1. Gear Direction and Rotation
How to read: Number each pair of meshing gears in sequence. Identify idler gears (gears that do not change ratio).
Gears In Sequence: Odd number of flips = same direction; even = opposite direction.
Gears In Sequence With Idler: With an idler gear in the train, an odd number of driven gears will produce an output that spins opposite the input, while an even number of driven gears will produce an output that spins the same as the input.
Gears transmit motion and torque. Each meshing reverses direction. Idler gears only reverse direction and do not affect speed.
Formulas:
Gear ratio = driven teeth ÷ driver teeth
driven = teeth on the output gear; driver = teeth on the input gearOutput speed = Input speed ÷ Gear ratio
speeds in rpmOutput torque = Input torque × Gear ratio
torque in lb·ft or N·m
Short Answer Practice:
- Question: A gear train has three meshes and the first gear rotates clockwise. In which direction will the final gear rotate?
Answer: Counterclockwise (opposite).
- Question: If the driver gear has 10 teeth and the driven gear has 20 teeth, what is the gear ratio?
Answer: 20 ÷ 10 = 2.
- Question: An input speed is 120 rpm with a gear ratio of 2. What is the output speed?
Answer: 120 ÷ 2 = 60 rpm.
2. Pulley Systems and Load Distribution
How to read: Identify each rope segment under tension supporting the load.
Then react: Use F = W ÷ n to find effort or n = W ÷ F to find rope count.
Pulleys redirect and distribute load across multiple rope segments, reducing the required effort to lift a weight.
Formula: Effort (F) = Load (W) ÷ Number of supporting ropes (n)
Short Answer Practice:
- Question: A 300 lb load is supported by 3 rope segments. What effort is required?
Answer: 300 ÷ 3 = 100 lb.
- Question: To lift a 400 lb load with 80 lb of effort, how many rope segments are needed?
Answer: 400 ÷ 80 = 5 ropes.
- Question: A load of 200 lb uses 4 ropes. What effort is needed?
Answer: 200 ÷ 4 = 50 lb.
3. Levers and Mechanical Advantage
How to read: Label the fulcrum (pivot), the effort point, and the load point, then measure distances d_E and d_L from the fulcrum.
Then react: Apply E × d_E = L × d_L or MA = d_E ÷ d_L.
Levers amplify force or distance by using different lever-arm lengths from the pivot (fulcrum).
Formulas:
Effort (E) × Effort Arm (d_E) = Load (L) × Load Arm (d_L)Mechanical Advantage (MA) = d_E ÷ d_L
Short Answer Practice:
- Question: If d_E = 4 ft and d_L = 1 ft, what is the mechanical advantage?
Answer: 4 ÷ 1 = 4.
- Question: To lift a 200 lb load with a lever that has MA = 4, what effort is required?
Answer: 200 ÷ 4 = 50 lb.
- Question: A lever with fulcrum between effort and load is classified as which class?
Answer: First-class lever.
4. Torque and Tool Use
How to read: Identify the applied force (F) and the distance (r) from the pivot to the line of action of F.
Then react: Compute τ = F × r.
Torque is the turning effect of a force applied at a distance from a pivot point.
Formula: Torque (τ) = Force (F) × Lever Arm (r)
Short Answer Practice:
- Question: A force of 30 lb is applied at 2 ft from the pivot. What is the torque?
Answer: 30 × 2 = 60 lb·ft.
- Question: To achieve 80 lb·ft of torque with a lever arm of 4 ft, how much force is needed?
Answer: 80 ÷ 4 = 20 lb.
- Question: What is the effect on torque when the lever arm increases while force remains the same?
Answer: Torque increases proportionally.
5. Real-World Elevator Systems
How to read: Note the car weight (W_car), the maximum load (W_load), and count support ropes (n).
Then react: Calculate tension, verify brake torque, and estimate governor activation speed.
Elevators use a balance of car weight, counterweight, and rope tension, with safety brakes and governors to prevent overspeed.
Formulas:
Tension (T) = (W_car + ½ W_load) ÷ nBrake torque (τ_brake) ≥ T × r_sheaveGovernor trip speed = (120 × RPM_sheave) ÷ grooves
Short Answer Practice:
- Question: If W_car = 5000 lb, W_load = 2000 lb, and n = 4, what is the tension per rope?
Answer: (5000 + 1000) ÷ 4 = 1500 lb.
- Question: For T = 1500 lb and r_sheave = 1 ft, what minimum brake torque is required?
Answer: ≥1500 lb·ft.
- Question: If RPM_sheave = 600 and grooves = 6, at what speed does the governor activate?
Answer: (120 × 600) ÷ 6 = 12000 rpm.
Quick “Dummies” Cheat Sheet
| Topic | Read, then React | Key Formula |
|---|
| Gears | Number meshes; mark idlers | Ratio = driven ÷ driver |
| Pulleys | Count tension ropes | F = W ÷ n |
| Levers | Label E, dE, L, dL | E·dE = L·dL |
| Torque | Identify F & r | τ = F·r |
| Elevator | Note W_car, W_load, n | T = (W_car + ½W_load) ÷ n |
| Units | Check units in question | lb → N: ×4.448 |