Wheel and Axle Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 2

Question 21Question →

A worker uses a hand-operated winch to vertically lift a bundle of materials weighing $300 \text{ pounds}$ . The winch is driven by a crank handle that has a length of $15 \text{ inches}$ from its center of rotation. The cable lifting the load wraps around a drum with a diameter of $5 \text{ inches}$ .

Assuming a $100\%$ efficient system, what is the theoretical effort force required on the crank handle to lift the bundle?

Show answer & explanation

Answer: 50

Answer

50

The system operates as a wheel and axle where the crank handle is the wheel and the drum is the axle. By finding the ratio of their radii (

15 \text{ inches}

to

2.5 \text{ inches}

), we determine that the mechanical advantage is

6

. This means the effort force required is one-sixth of the total load, yielding

50 \text{ pounds}

.

Step-by-Step Solution

1

Determine the effective radius of both the wheel and the axle.

Wheel radius =

15 \text{ inches}

. Axle radius =

5 \text{ inches} \div 2 = 2.5 \text{ inches}

.

The formula for mechanical advantage requires consistent measurements, typically the radius of each component.

2

Calculate the Ideal Mechanical Advantage (IMA) of the system.

IMA = \frac{15}{2.5} = 6

.

The IMA indicates how many times the simple machine multiplies the input effort force.

3

Calculate the theoretical effort force required.

F_{effort} = \frac{300}{6} = 50 \text{ pounds}

.

In a frictionless (

100\%

efficient) system, the effort force is equal to the load force divided by the mechanical advantage.

Key Concept

Ideal Mechanical Advantage (IMA) of a Wheel and Axle

Question 22Question →

A driver operates a heavy-duty steering wheel connected directly to a steering column (axle). The steering wheel has a radius of $16 \text{ inches}$ , and the steering column has a diameter of $4 \text{ inches}$ . If the driver applies an effort force of $20 \text{ pounds}$ to the outer rim of the steering wheel, what is the resulting output force exerted by the steering column? (Assume a perfectly efficient system.)

2.5 \text{ pounds}

5 \text{ pounds}

80 \text{ pounds}

160 \text{ pounds}

Show answer & explanation

Answer: $160 \text{ pounds}$

Answer

160 \text{ pounds}

The mechanical advantage of a wheel and axle when effort is applied to the wheel is the radius of the wheel divided by the radius of the axle. The wheel's radius is

16 \text{ inches}

. The axle's diameter is

4 \text{ inches}

, meaning its radius is

2 \text{ inches}

. The mechanical advantage is

16 / 2 = 8

. Multiplying the

20 \text{ pound}

input force by the mechanical advantage of

8

gives a total output force of

160 \text{ pounds}

.

Step-by-Step Solution

1

Determine the radius of the steering column (axle).

The diameter is

4 \text{ inches}

, so the radius is

2 \text{ inches}

.

Mechanical advantage must be calculated by comparing equivalent dimensions, such as the radii of both components.

2

Calculate the Ideal Mechanical Advantage (IMA) of the wheel and axle.

\text{IMA} = \frac{\text{Radius of Wheel}}{\text{Radius of Axle}} = \frac{16}{2} = 8

.

When effort is applied to the larger wheel, the system acts as a force multiplier based on the ratio of the radii.

3

Calculate the output force exerted by the steering column.

\text{Output Force} = \text{Effort Force} \times \text{IMA} = 20 \times 8 = 160 \text{ pounds}

.

The system multiplies the driver's input force by the mechanical advantage to turn the steering mechanism.

Key Concept

Ideal Mechanical Advantage of a Wheel and Axle

Estimated Time:1m 0s

Question 23Question →

Match each real-world mechanism to its correct functional 'wheel' and 'axle' components based on how it operates.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Manual Screwdriver matches 'Wheel = Outer wide grip; Axle = Narrow metal shaft'. Bicycle Rear Wheel Assembly matches 'Wheel = Outer tire and rim; Axle = Central toothed gear'. Hand-cranked Water Well matches 'Wheel = Circular path traced by the handle; Axle = Central drum'. Ceiling Fan matches 'Wheel = Extended rotating blades; Axle = Central motorized spindle'.

Each mechanism is correctly categorized by identifying the component with the larger rotational diameter as the wheel and the smaller central component as the axle, accurately reflecting how they transmit rotational force and speed.

Step-by-Step Solution

1

Define the physical characteristics of a wheel and axle system.

The 'wheel' is always the component or path with the larger radius, and the 'axle' is the rigidly attached component with the smaller radius.

Establishing the definitions allows us to look past the physical shape of the object and focus on its mechanical function.

2

Analyze tools designed to multiply force (Screwdriver and Windlass).

In these tools, effort is applied to the larger radius (the wide grip or the crank's path) to turn the smaller radius (the shaft or the drum) with greater force.

Identifying where the effort is applied helps determine which part acts as the wheel.

3

Analyze mechanisms designed to multiply speed or distance (Bicycle and Ceiling Fan).

In these systems, effort is applied to the smaller central radius (the sprocket or the motor spindle), which forces the larger radius (the tire or the blades) to spin faster and cover more distance.

This confirms the axle is driving the wheel in these specific applications.

Key Concept

In a wheel and axle simple machine, the 'wheel' is the component with the larger diameter, and the 'axle' is the connected component with the smaller diameter. A wheel does not have to be a solid disk; it can be a crank path or fan blades.

Question 24Question →

A mechanic uses a manual screwdriver to remove a tightly secured machine screw. The handle of the screwdriver has a diameter of $1.5 \text{ inches}$ , and the metal shaft has a diameter of $0.25 \text{ inches}$ . If the mechanic applies a rotational effort force of $12 \text{ pounds}$ to the outer edge of the handle, how much output force is exerted by the shaft onto the screw? (Assume $100\%$ efficiency.)

72 \text{ pounds}

36 \text{ pounds}

18 \text{ pounds}

2 \text{ pounds}

Show answer & explanation

Answer: $72 \text{ pounds}$

Answer

72 \text{ pounds}

In a wheel and axle system where effort is applied to the wheel (the handle), the force is multiplied by the ratio of the wheel's diameter to the axle's diameter. The ratio of

1.5 \text{ inches}

to

0.25 \text{ inches}

provides a mechanical advantage of

6

. Multiplying the

12 \text{ pounds}

of effort by

6

yields an output force of

72 \text{ pounds}

.

Step-by-Step Solution

1

Identify the components of the wheel and axle system.

The handle acts as the wheel (Diameter =

1.5 \text{ in}

), and the shaft acts as the axle (Diameter =

0.25 \text{ in}

).

Establishing which component is the wheel and which is the axle determines how force is multiplied.

2

Calculate the Ideal Mechanical Advantage (IMA).

IMA = \frac{1.5}{0.25} = 6

The IMA for a wheel and axle is found by dividing the diameter of the wheel by the diameter of the axle.

3

Calculate the total output force applied to the screw.

\text{Output Force} = \text{Input Force} \times IMA = 12 \text{ lbs} \times 6 = 72 \text{ lbs}

.

The system multiplies the input effort force by the mechanical advantage to produce the final output force.

Key Concept

Ideal Mechanical Advantage of a Wheel and Axle

Question 25Question →

A manual meat grinder is operated by turning a crank handle. The crank arm has a length of $8.0 \text{ inches}$ (acting as the radius of the wheel). The internal grinding shaft, which acts as the axle, has a diameter of $1.6 \text{ inches}$ . If a user applies a steady perpendicular force of $15 \text{ pounds}$ to the crank handle, what is the resulting theoretical output force (in pounds) exerted at the outer surface of the grinding shaft? Assume $100\%$ efficiency.

Show answer & explanation

Answer: 150

Answer

150

The machine multiplies the input force based on the ratio of the crank arm radius to the grinding shaft radius. Converting the

1.6 \text{ inch}

axle diameter to a

0.8 \text{ inch}

radius gives an Ideal Mechanical Advantage (IMA) of

10

(

8.0 \div 0.8

). Multiplying the

15 \text{ pound}

input force by this IMA results in a

150 \text{ pound}

theoretical output force.

Step-by-Step Solution

1

Determine the radius of the axle.

Axle radius =

0.8 \text{ inches}

The Ideal Mechanical Advantage formula requires comparing equivalent radial measurements. Since the wheel dimension is given as a radius, converting the axle's diameter to a radius ensures units are consistent.

2

Calculate the Ideal Mechanical Advantage (IMA).

IMA = 10

Dividing the wheel radius (

8.0 \text{ inches}

) by the axle radius (

0.8 \text{ inches}

) determines the mechanical advantage multiplier of the machine.

3

Calculate the theoretical output force.

Output Force =

150 \text{ pounds}

Multiplying the input force (

15 \text{ pounds}

) by the mechanical advantage (

10

) yields the final output force exerted at the axle surface.

Key Concept

The Ideal Mechanical Advantage (IMA) of a wheel and axle is the ratio of the wheel's radius to the axle's radius. The resulting output force is equal to the input force multiplied by this mechanical advantage.

Question 26Question →

Match each real-world wheel and axle application to its corresponding Ideal Mechanical Advantage (IMA). Remember to identify whether the input force is applied to the wheel or the axle.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Hand winch: IMA = 5; Bicycle rear wheel: IMA = 1/7; Steering mechanism: IMA = 6; Ceiling fan: IMA = 1/20.

Each real-world application must be evaluated by identifying the radius where the input force is applied and dividing it by the radius where the output force is exerted.

Step-by-Step Solution

1

Identify the input and output components for each mechanism.

Determine if the input force is applied to the larger radius (wheel) or the smaller radius (axle).

IMA depends on where the input force is applied.

2

Recall the formula for Ideal Mechanical Advantage (IMA) of a wheel and axle.

\text{IMA} = \frac{\text{Radius of Input}}{\text{Radius of Output}}

.

This formula provides the theoretical mechanical trade-off of the system.

3

Calculate IMA for mechanisms acting as force multipliers (input on the larger wheel).

Winch:

\frac{15}{3} = 5

. Steering mechanism:

\frac{12}{2} = 6

.

These devices decrease the required input force but require it to be applied over a greater distance, resulting in an IMA > 1.

4

Calculate IMA for mechanisms acting as speed or distance multipliers (input on the smaller axle).

Bicycle:

\frac{2}{14} = \frac{1}{7}

. Ceiling fan:

\frac{1}{20}

.

These devices increase output speed and distance, but require a larger input force, resulting in an IMA < 1.

Key Concept

Ideal Mechanical Advantage (IMA) of a Wheel and Axle

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