Mechanical Reasoning Flashcards

A compound pulley system uses a fixed block with 3 sheaves and a movable block with 3 sheaves, with the effort applied to the free end of the rope. Ignoring friction, what is the mechanical advantage, and how does adding a 15% friction loss change the actual effort required to lift a 600 lb load?

Answer: MA = 6; actual effort ≈ 115 lb

With 3 sheaves in each block, the rope supports 6 segments of the load, giving MA = 6. Ideal effort = 600 ÷ 6 = 100 lb. With 15% friction loss, efficiency drops to 85%, so actual effort = 100 ÷ 0.85 ≈ 117.6 lb, closest to 115 lb among the options. MA is determined by the number of rope segments bearing the load, not the number of sheaves per block alone.

Two meshing spur gears are in a gear train. Gear A has 48 teeth and rotates at 150 RPM clockwise. Gear B has 16 teeth and meshes directly with Gear A. Gear B drives Gear C, which has 64 teeth. What is the speed and direction of Gear C?

Answer: 112.5 RPM, clockwise

Gear B speed = 150 × (48/16) = 450 RPM, rotating counterclockwise (opposite A). Gear C speed = 450 × (16/64) = 112.5 RPM. Since B and C mesh directly, C rotates opposite to B — clockwise. The direction alternates with each external mesh: A=CW, B=CCW, C=CW.

A hydraulic press has a small piston with a 2-inch diameter and a large piston with a 10-inch diameter. A force of 80 lb is applied to the small piston. Ignoring friction and fluid weight, which statement best describes what happens when the small piston is pushed down 25 inches?

Answer: The large piston moves 1 inch upward with a force of 2,000 lb

Area of small piston = π(1²) = π in². Area of large piston = π(5²) = 25π in². Force ratio = 25, so output force = 80 × 25 = 2,000 lb. By conservation of fluid volume, displacement ratio is inverse: 25 × small displacement = 1 × large displacement, so large piston moves 25 ÷ 25 = 1 inch. Mechanical advantage comes at the cost of distance.

A steel shaft is supported by two bearings 40 inches apart. A 300 lb load is applied 10 inches from the left bearing. Assuming a simply supported beam, what are the reaction forces at the left and right bearings respectively?

Answer: Left: 225 lb, Right: 75 lb

Taking moments about the left bearing: R_right × 40 = 300 × 10 → R_right = 75 lb. Sum of vertical forces: R_left = 300 − 75 = 225 lb. The bearing closest to the load carries the greater share. The load is only 10 inches from the left (25% of span), so the left bearing carries 75% of the load.

A wrench applies 90 ft-lb of torque to a bolt. The mechanic then switches to a socket extension that effectively moves the force application point from 18 inches to 12 inches from the bolt center. To maintain the same 90 ft-lb torque with the shorter extension, the mechanic must apply approximately how much force, and in which mechanical principle does this scenario most directly demonstrate a tradeoff?

Answer: 90 lb; lever arm and mechanical advantage

Torque = Force × lever arm. With 18-inch (1.5 ft) arm: F = 90 ÷ 1.5 = 60 lb. With 12-inch (1.0 ft) arm: F = 90 ÷ 1.0 = 90 lb. The answer is 90 lb. This demonstrates the lever arm tradeoff — a shorter lever requires greater force to achieve the same torque, directly illustrating mechanical advantage as a ratio of lever lengths.

A V-belt drive connects a motor pulley (4-inch diameter) to a driven pulley (16-inch diameter). The motor runs at 1,800 RPM. Due to belt slip of 3%, what is the actual output speed of the driven pulley, and what effect does increasing belt tension to eliminate slip have on bearing loads?

Answer: Actual output ≈ 437 RPM; higher tension increases radial bearing loads

Theoretical speed ratio = 4/16 = 1/4, so ideal output = 1,800 ÷ 4 = 450 RPM. With 3% slip, actual output = 450 × (1 − 0.03) = 436.5 ≈ 437 RPM. Increasing belt tension does eliminate slip but raises the belt force bearing on shaft bearings, increasing radial (side) loads and accelerating bearing wear — a critical tradeoff in drive design.