ASVAB Mechanical Comprehension Practice Test 9

Question 1 easy

A belt-driven pulley system consists of four pulleys connected by a continuous belt. Pulley A has a diameter of 12 inches, Pulley B has a diameter of 4 inches, Pulley C has a diameter of 8 inches, and Pulley D has a diameter of 6 inches. Which pulley will rotate at the fastest speed?

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Correct answer: B — Pulley B

In a belt-driven pulley system, all pulleys move at the same linear belt speed. Smaller pulleys must complete more rotations to cover the same linear distance, meaning the pulley with the smallest diameter (Pulley B, 4 inches) spins the fastest.

Question 2 easy

In a block and tackle pulley system, a worker pulls 12 meters of rope to raise a heavy crate by a height of 3 meters. What is the mechanical advantage of the pulley system?

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Correct answer: A — 4

Mechanical Advantage (MA) is calculated as the ratio of effort distance to load distance: MA = Effort Distance / Load Distance = 12 meters / 3 meters = 4.

Question 3 easy

A simple lever provides a mechanical advantage of 5. If the input arm (effort arm) is 100 cm long, what is the length of the output arm (load arm)?

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Correct answer: B — 20 cm

Mechanical Advantage for a lever is given by MA = Input Arm Length / Output Arm Length. Rearranging the formula gives Output Arm Length = Input Arm Length / MA = $$100\text{ cm} / 5 = 20$$ cm.

Question 4 medium

A crane mechanic utilizes a block and tackle system with a mechanical advantage of 5 to hoist a steel beam weighing 1,200 newtons. How much input force must be applied to lift the beam?

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Correct answer: C — 240 newtons

Mechanical Advantage (MA) is defined as Output Force (Load) / Input Force (Effort). Therefore, Input Force = Output Force / MA = $$\frac{1{,}200\text{ N}}{5} = 240$$ newtons.

Question 5 medium

A drive gear (Gear X) with 40 teeth is meshed with a middle idler gear, which in turn meshes with a driven gear (Gear Y) with 20 teeth. If Gear X turns clockwise for 6 full rotations, how will Gear Y move?

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Correct answer: B — Clockwise 12 rotations

1) Direction: An intermediate idler gear changes direction twice, so Gear Y turns in the SAME direction as Gear X (Clockwise). 2) Speed/Rotations: Gear Ratio = Teeth on Gear X / Teeth on Gear Y = $$\frac{40}{20} = 2$$. Rotations of Gear Y $$= 6 \times 2 = 12$$ rotations.

Question 6 medium

Which of the following mechanical components is installed between a rotating shaft and its support housing primarily to reduce rolling friction and support radial loads?

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Correct answer: B — Ball bearings

Ball bearings use smooth rolling spheres to replace sliding friction with rolling friction between a moving axle/shaft and a stationary housing.

Question 7 medium

Which real-world mechanical device functions on the exact same fundamental principle of cables, counterweights, and overhead pulleys as an elevator?

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Correct answer: B — A construction crane hoist

Both elevators and construction cranes utilize a system of pulleys and wire cables driven by motors to lift heavy loads vertically.

Question 8 medium

In a belt-and-pulley drive, what is the primary effect of crossing the belt (forming a figure-8 layout) between two parallel pulleys?

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Correct answer: B — It reverses the direction of rotation of the driven pulley relative to the driver pulley

An open belt causes both pulleys to rotate in the same direction. Crossing the belt into a figure-8 layout causes the driven pulley to rotate in the opposite direction relative to the driver pulley.

Question 9 hard

An industrial worker can pull with a maximum effort force equivalent to 70 kilograms. If he switches from lifting directly to using a 6-pulley block and tackle system (assuming 100% efficiency), what maximum load mass can he lift using the same effort?

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Correct answer: C — 420 kilograms

In an ideal block and tackle system with 6 supporting rope segments (MA = 6), the lifting capacity is multiplied by 6: Maximum Load = Effort × MA = $$70\text{ kg} \times 6 = 420$$ kilograms.

Question 10 hard

A mountain bike has a front chainring with 44 teeth and a rear cassette sprocket with 11 teeth. If the rider pedals at a cadence of 75 revolutions per minute (rpm), what is the rotational speed of the rear wheel?

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Correct answer: A — 300 rpm

Gear ratio = Number of front teeth / Number of rear teeth = $$\frac{44}{11} = 4$$.

The rear wheel rotates 4 times faster than the pedals. Rear wheel rotation speed (RPM) = Pedal stroke × Gear ratio $$= 75\text{ rpm} \times 4 = 300$$ rpm.

Question 11 hard

A rigger uses a compound pulley system where a mechanical advantage of 4 is combined with a $$2:1$$ lever arm ratio. What total load can be lifted if an input effort of 150 newtons is applied?

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Correct answer: C — 1,200 newtons

In combined/integrated mechanical systems:

The total mechanical advantage ($$\text{MA}_{\text{total}}$$) is the product of the individual advantages: $$\text{MA}_{\text{total}} = \text{MA}_{\text{pulley}} \times \text{MA}_{\text{lever}} = 4 \times 2 = 8$$.

Total load $$= \text{Effort} \times \text{MA}_{\text{total}} = 150\text{ N} \times 8 = 1{,}200$$ Newtons.

Question 12 hard

Gear A (60 teeth) turns clockwise at 120 rpm and drives Gear B (20 teeth), which is keyed to the same shaft as Gear C (40 teeth). Gear C then meshes with Gear D (10 teeth). What is the speed and rotational direction of Gear D?

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Correct answer: A — 1,440 rpm, clockwise

1) Gear A (60t, CW, 120 rpm) drives Gear B (20t): Gear B turns CCW at $$120 \times (\frac{60}{20}) = 360$$ rpm.

2) Gear C (40t) is on the same shaft as B, so Gear C turns CCW at 360 rpm.

3) Gear C drives Gear D (10t): Gear D turns CW at $$360 \times (\frac{40}{10}) = 1{,}440$$ rpm.

Question 13 hard

A winch system with an ideal mechanical advantage of 5 is used to lift a 2,000-newton load to a height of 4 meters. Due to friction, the winch operates at 80% efficiency. How much total input work must be exerted by the operator?

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Correct answer: B — 10,000 joules

Useful work output = Load weight × Height = $$2{,}000\text{ N} \times 4\text{ m} = 8{,}000$$ joules.

Efficiency = Work output / Work input => $$0.80 = \frac{8{,}000\text{ J}}{\text{Work input}}$$.

Work input $$= \frac{8{,}000\text{ J}}{0.80} = 10{,}000$$ joules.