Mechanical Comprehension Practice Test 10

Correct answer: C — Friction

Brake pads create friction against the rotating rim, converting kinetic energy into thermal energy to slow down or stop motion.

Correct answer: B — A gear

A toothed wheel designed to engage with a chain or other teeth functions identically to a gear in transmitting rotational motion.

Correct answer: B — It changes the direction of the applied force.

A fixed pulley allows a person to pull downward using gravity and body weight to lift a load upward, changing the direction of the applied force without reducing the required effort.

Correct answer: B — Lever B

Lever B has a longer handle (longer effort arm relative to the fulcrum), providing greater mechanical advantage (Torque = Force × Distance). Thus, less input force is required on Lever B.

Correct answer: A — 60 rpm

The idler gear (Gear B) does not affect the speed ratio between Gear A and Gear C. $$\text{Speed of Gear C} = \text{Speed of Gear A} \times (\text{Teeth on Gear A} / \text{Teeth on Gear C}) = 120\text{ rpm} \times (\frac{15}{30}) = 60$$ rpm.

Correct answer: C — Metal (Steel/Cast Iron)

Metals like steel or cast iron have the high density required to store rotational kinetic energy effectively and the tensile strength to resist centrifugal forces.

Correct answer: B — 30 rpm

Rotational speed is inversely proportional to circumference/diameter: $$\text{Speed}_X \times \text{Circumference}_X = \text{Speed}_Y \times \text{Circumference}_Y$$. $$\text{Speed}_X = (120\text{ rpm} \times 1) / 4 = 30$$ rpm.

Correct answer: A — 2 inches

For parallel springs, equivalent stiffness adds up: $$k_{\text{eq}} = 10 + 10 + 10 = 30$$ lb/in. Total compression $$x = \text{Force} / k_{\text{eq}} = 60\text{ lb} / 30\text{ lb/in} = 2$$ inches.

Correct answer: B — 12.25 inches

In series, total compression is the sum of individual compressions ($$x_{\text{total}} = F/k_1 + F/k_2 + F/k_3$$). Compression $$= (\frac{21}{3}) + (\frac{21}{6}) + (\frac{21}{12}) = 7 + 3.5 + 1.75 = 12.25$$ inches.

Correct answer: A — 200 rpm

Stage 1: $$\text{Speed}_B = 1800 \times (\frac{4}{12}) = 600$$ rpm. Shaft 2: $$\text{Speed}_C = \text{Speed}_B = 600$$ rpm. Stage 2: $$\text{Speed}_D = 600 \times (\frac{5}{15}) = 200$$ rpm.

Correct answer: A — 10 inches

Parallel pair equivalent stiffness: $$k_p = 8 + 8 = 16$$ lb/in.

Series combination with 4 lb/in spring: $$x_{\text{total}} = F/k_p + F/k_3 = (\frac{32}{16}) + (\frac{32}{4}) = 2 + 8 = 10$$ inches.

Correct answer: B — 62.5 N·m

Torque is directly proportional to radius in belt drives:

$$\text{Torque}_2 / \text{Torque}_1 = \text{Radius}_2 / \text{Radius}_1$$.

$$\text{Torque}_2 = 25\text{ N}\cdot\text{m} \times (25\text{ cm} / 10\text{ cm}) = 62.5\text{ N}\cdot\text{m}$$.

Correct answer: B — 299.45 N

Buoyant Force $$F_b = \text{density} \times g \times \text{volume} = 1{,}025\text{ kg/m}^3 \times 9.81\text{ m/s}^2 \times 0.01\text{ m}^3 = 100.55\text{ N}$$. Net Downward Force = Air Weight - Buoyant Force $$= 400\text{ N} - 100.55\text{ N} = 299.45\text{ N}$$