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FCTC Mechanical Reasoning Principles 1 Flashcards

6 cards from real FCTC practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 6 FCTC Mechanical Reasoning Principles 1 flashcards as text

  1. A movable pulley system uses 2 rope segments to support a load. If the load weighs 400 lbs, how much effort force is required (ignoring friction)?

    Answer: 200 lbs

    A movable pulley provides a mechanical advantage equal to the number of rope segments supporting the load. With 2 supporting segments, the mechanical advantage is 2, so the required effort is 400 ÷ 2 = 200 lbs.

  2. A wheel with a radius of 12 inches is attached to an axle with a radius of 3 inches. If a force of 50 lbs is applied to the rim of the wheel, how much load can the axle lift?

    Answer: 200 lbs

    The mechanical advantage of a wheel and axle equals the wheel radius divided by the axle radius: 12 ÷ 3 = 4. Multiply the applied force by the MA: 50 × 4 = 200 lbs.

  3. A hydraulic system has a small piston with an area of 2 square inches and a large piston with an area of 10 square inches. If 50 lbs of force is applied to the small piston, what output force is produced at the large piston?

    Answer: 250 lbs

    By Pascal's Law, pressure is transmitted equally throughout the fluid. Pressure = 50 ÷ 2 = 25 psi. Output force = 25 × 10 = 250 lbs. The force scales with the ratio of the piston areas.

  4. An inclined plane is 10 feet long and 2 feet high. What is the mechanical advantage of this ramp?

    Answer: 5

    The mechanical advantage of an inclined plane is calculated by dividing the length of the slope by its vertical height: 10 ÷ 2 = 5. This means you only need 1/5 of the load's weight to push it up the ramp.

  5. Which class of lever has the load (resistance) positioned between the fulcrum and the effort force?

    Answer: Second class

    In a second-class lever, the load sits between the fulcrum and the effort. A wheelbarrow is the classic example — the wheel is the fulcrum, the load sits in the middle, and you lift the handles at the far end.

  6. A spring with a spring constant of 20 lbs per inch is compressed by applying 80 lbs of force. How far does the spring compress?

    Answer: 4 inches

    Using Hooke's Law, compression distance = Force ÷ Spring constant = 80 ÷ 20 = 4 inches. The spring constant describes how many pounds of force are needed to compress the spring by one inch.