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

Question 1Question →

A traditional water well is operated by a hand crank that raises a bucket of water. The crank handle has a turning radius of $12 \text{ inches}$ , and it drives a wooden cylinder (the axle) that has a diameter of $8 \text{ inches}$ . If the total weight of the water and bucket is $45 \text{ pounds}$ , what is the ideal effort force required on the handle to lift the bucket at a constant speed?

15 \text{ pounds}

22.5 \text{ pounds}

30 \text{ pounds}

135 \text{ pounds}

Show answer & explanation

Answer: $15 \text{ pounds}$

Answer

The ideal effort force required is

15 \text{ pounds}

.

The system operates as a force multiplier because the effort is applied to the larger radius (the crank handle). First, find the axle's radius by dividing its

8 \text{ inch}

diameter by 2, yielding

4 \text{ inches}

. The mechanical advantage is the ratio of the effort radius to the load radius:

12 \text{ inches} / 4 \text{ inches} = 3

. This means the crank multiplies the applied force by 3. To lift a

45\text{-pound}

load, divide the load by the mechanical advantage (

45 / 3

), which results in a required effort force of

15 \text{ pounds}

.

Step-by-Step Solution

1

Determine the radius of the axle.

The radius of the axle (

r

) is

4 \text{ inches}

.

The diameter is given as

8 \text{ inches}

, and radius is half of the diameter (

8 / 2 = 4

).

2

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

The IMA is

3

.

The IMA for a wheel and axle where the effort is applied to the wheel is the radius of the wheel divided by the radius of the axle (

IMA = R / r = 12 / 4 = 3

).

3

Calculate the required ideal effort force.

The effort force is

15 \text{ pounds}

.

Effort force equals the load divided by the IMA (

45 / 3 = 15

). This means the crank multiplies the user's force by 3.

Key Concept

Ideal Mechanical Advantage in Wheel and Axle Systems

Question 2Question →

A manual boat winch is used to pull a small boat onto a trailer. The crank handle travels in a circle with a radius of $15 \text{ inches}$ , turning a central spool (the axle) with a radius of $3 \text{ inches}$ .

What is the Ideal Mechanical Advantage (IMA) of this winch?

Show answer & explanation

Answer: 5

Answer

The Ideal Mechanical Advantage is 5.

The Ideal Mechanical Advantage of a wheel and axle is calculated by dividing the radius of the wheel by the radius of the axle. Here, the wheel's radius is

15 \text{ inches}

and the axle's radius is

3 \text{ inches}

, which gives

15 \div 3 = 5

.

Step-by-Step Solution

1

Identify the given dimensions of the wheel and the axle.

R_{wheel} = 15 \text{ inches}

,

R_{axle} = 3 \text{ inches}

These are the required variables for the IMA equation.

2

Apply the Ideal Mechanical Advantage formula for a wheel and axle.

IMA = \frac{R_{wheel}}{R_{axle}}

IMA represents the ratio of the effort distance to the resistance distance.

3

Calculate the final value.

IMA = \frac{15}{3} = 5

Dividing the wheel radius by the axle radius yields the mechanical advantage.

Key Concept

Ideal Mechanical Advantage (IMA) of a Wheel and Axle

Question 3Question →

A manual winch is used to lift a $240 \text{ lb}$ engine block exactly $2 \text{ feet}$ vertically. The winch consists of a steering-wheel style control with a diameter of $3 \text{ feet}$ , which is rigidly attached to an axle drum with a radius of $3 \text{ inches}$ . Assuming an ideal system with no friction, what continuous effort force must be applied to the outer edge of the wheel, and what total distance must the effort force move along the wheel's rotation?

20 \text{ lbs}

of force;

24 \text{ feet}

of distance

480 \text{ lbs}

of force;

1 \text{ foot}

of distance

40 \text{ lbs}

of force;

12 \text{ feet}

of distance

1440 \text{ lbs}

of force;

1/3 \text{ foot}

of distance

Show answer & explanation

Answer: $40 \text{ lbs}$ of force; $12 \text{ feet}$ of distance

Answer

The correct answer is

40 \text{ lbs}

of force and

12 \text{ feet}

of distance.

The system's Ideal Mechanical Advantage is determined by the ratio of the effort radius (18 inches, half of the 3-foot diameter) to the load radius (3 inches), which yields an IMA of 6. This means the required effort force is divided by 6 (

240 \text{ lbs} / 6 = 40 \text{ lbs}

), while the required effort distance is multiplied by 6 (

2 \text{ feet} \times 6 = 12 \text{ feet}

) to conserve work.

Step-by-Step Solution

1

Convert all measurements to the same unit (inches).

Wheel diameter =

3 \text{ feet} = 36 \text{ inches}

.

Units must be identical to calculate a valid, dimensionless mechanical advantage ratio.

2

Identify the effort radius and the load radius.

Wheel radius (

R

) =

18 \text{ inches}

. Axle drum radius (

r

) =

3 \text{ inches}

.

The lever arms in a wheel and axle system are the distances from the central axis (the radii), not the diameters.

3

Calculate the Ideal Mechanical Advantage (IMA).

\text{IMA} = \frac{R}{r} = \frac{18}{3} = 6

.

This establishes that the system multiplies the user's force by a factor of 6.

4

Calculate the required effort force.

\text{Effort Force} = \frac{\text{Load}}{\text{IMA}} = \frac{240 \text{ lbs}}{6} = 40 \text{ lbs}

.

Because the machine provides a mechanical advantage of 6, the required input force is one-sixth of the load.

5

Calculate the required effort distance.

\text{Effort Distance} = \text{Load Distance} \times \text{IMA} = 2 \text{ feet} \times 6 = 12 \text{ feet}

.

According to the principle of work, whatever factor force is reduced by, distance must be multiplied by that same factor.

Key Concept

Ideal Mechanical Advantage (IMA) and Work in a Wheel and Axle

Estimated Time:1m 30s

Question 4Question →

A bucket is being raised from a well using a windlass mechanism (a type of wheel and axle). The cylindrical drum (axle) around which the rope winds has a diameter of $8 \text{ inches}$ . The crank handle used to turn the drum has a length (radius) from the center of rotation of $20 \text{ inches}$ . If the bucket is raised vertically by exactly $15 \text{ feet}$ , what is the total distance, in feet, that the crank handle must travel along its circular path? (Assume the rope does not overlap itself on the drum).

Show answer & explanation

Answer: 75

Answer

75

In any simple machine, assuming 100% efficiency, the work put in equals the work put out. The Ideal Mechanical Advantage (IMA) for a wheel and axle is the ratio of the effort radius to the load radius. Here, the load radius (the drum) is

4 \text{ inches}

(half of the

8 \text{ inch}

diameter) and the effort radius (the crank) is

20 \text{ inches}

. Thus, the IMA is

20 / 4 = 5

. This geometric relationship dictates that the effort force must travel

5

times further than the load to conserve work. If the load is lifted

15 \text{ feet}

, the effort distance is

15 \times 5 = 75 \text{ feet}

.

Step-by-Step Solution

1

Calculate the drum's radius.

4 \text{ inches}

The Ideal Mechanical Advantage compares radii, but the drum's dimension was provided as a diameter.

2

Calculate the Ideal Mechanical Advantage (IMA).

IMA =

5

Dividing the effort radius (

20 \text{ inches}

) by the load radius (

4 \text{ inches}

) gives the mechanical advantage of the system.

3

Calculate the total distance traveled by the crank handle.

75 \text{ feet}

By the principle of conservation of energy, the effort distance must be equal to the load distance (

15 \text{ feet}

) multiplied by the IMA (

5

).

Key Concept

Ideal Mechanical Advantage (IMA) and Force-Distance Tradeoff

Estimated Time:1m 30s

Question 5Question →

An electric motor directly turns the rear solid axle of an industrial transport cart. The axle has a diameter of $4 \text{ inches}$ , and the attached drive wheels have a diameter of $24 \text{ inches}$ . If the motor applies a rotational force of $1200 \text{ lbs}$ at the outer surface of the axle, what is the resulting forward force exerted by the wheels against the ground? (Assume zero friction losses).

200 \text{ lbs}

7200 \text{ lbs}

1200 \text{ lbs}

100 \text{ lbs}

Show answer & explanation

Answer: $200 \text{ lbs}$

Answer

200 \text{ lbs}

In a vehicle drive system, the effort is applied to the smaller axle and the output is at the outer edge of the larger wheel. The mechanical advantage is the ratio of the input radius (

2 \text{ inches}

) to the output radius (

12 \text{ inches}

), which is

1/6

. Therefore, the

1200 \text{ lbs}

of input force is reduced to

1/6

of its value, resulting in

200 \text{ lbs}

of forward force.

Step-by-Step Solution

1

Identify the input and output components of the machine.

Input is applied to the axle. Output is delivered at the edge of the wheel.

Determining where force is applied dictates whether the system acts as a force multiplier or a speed/distance multiplier.

2

Determine the radius of both the input and output components.

Axle radius =

2 \text{ inches}

(

4 \div 2

). Wheel radius =

12 \text{ inches}

(

24 \div 2

).

Mechanical advantage must be calculated using consistent dimensions, typically the radius of rotation.

3

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

\text{IMA} = \frac{\text{Input Radius}}{\text{Output Radius}} = \frac{2}{12} = \frac{1}{6}

.

When an axle drives a wheel, the output force is distributed over a larger distance, resulting in a mechanical advantage of less than 1.

4

Calculate the output force.

\text{Output Force} = \text{Input Force} \times \text{IMA} = 1200 \text{ lbs} \times \frac{1}{6} = 200 \text{ lbs}

.

The output force is the product of the applied effort and the mechanical advantage.

Key Concept

Mechanical Advantage of Axle-Driven Systems

Alternative Method

You can also use diameters directly without converting to radii: IMA = (Input Diameter) / (Output Diameter) = 4 / 24 = 1/6. Then multiply by the 1200 lbs force to get 200 lbs.

Estimated Time:1m 30s

Question 6Question →

Match each wheel and axle mechanism to its corresponding force-distance trade-off profile and Ideal Mechanical Advantage (IMA).

Click a left item, then click its matching right item

Show answer & explanation

Answer

Screwdriver maps to IMA 6 (force multiplier). Bicycle maps to IMA 1/9 (distance multiplier). Steering wheel maps to IMA 14 (force multiplier). Ceiling fan maps to IMA 1/100 (speed multiplier).

Each mechanism is correctly paired by identifying where the effort is applied (wheel or axle) and calculating the ratio of the effort diameter to the load diameter. This ratio directly dictates whether the machine sacrifices distance for force (IMA > 1) or sacrifices force for distance/speed (IMA < 1).

Step-by-Step Solution

1

Identify the effort and load components for each mechanism.

Determine which part acts as the 'wheel' (larger radius) and which acts as the 'axle' (smaller radius), and definitively locate where the input effort is applied.

The formula for Ideal Mechanical Advantage requires knowing whether the input is applied to the wheel or the axle.

2

Calculate the Ideal Mechanical Advantage (IMA) for each item.

Using the formula IMA = (Diameter of Effort) / (Diameter of Load). Screwdriver: 1.5/0.25 = 6. Bicycle: 3/27 = 1/9. Steering Wheel: 14/1 = 14. Ceiling Fan: 0.5/50 = 1/100.

This mathematical ratio represents the mechanical trade-off of the machine.

3

Classify each mechanism based on its IMA value.

IMA > 1 (Screwdriver, Steering Wheel) means the machine multiplies force. IMA < 1 (Bicycle, Ceiling Fan) means the machine multiplies distance or speed.

A machine cannot multiply both force and distance simultaneously due to the conservation of energy.

Key Concept

Ideal Mechanical Advantage (IMA) of a Wheel and Axle

Question 7Question →

A $180 \text{ lb}$ load is suspended by a cable wrapped around a solid cylindrical axle. The cable hangs freely from the right side of the axle. A steering handwheel with a radius of $12 \text{ inches}$ is rigidly attached to this axle. The axle has a diameter of $8 \text{ inches}$ . To lift the load, a worker applies a tangential force to the outer edge of the handwheel. Viewed from the front, what is the required direction of rotation and the ideal effort force needed to lift the load at a constant speed?

Counterclockwise;

60 \text{ lbs}

Clockwise;

60 \text{ lbs}

Counterclockwise;

120 \text{ lbs}

Counterclockwise;

540 \text{ lbs}

Show answer & explanation

Answer: Counterclockwise; $60 \text{ lbs}$

Answer

Counterclockwise;

60 \text{ lbs}

The correct answer accurately determines that raising a load from a cable hanging off the right side of a cylinder requires counterclockwise rotation. Furthermore, it accurately calculates the mechanical advantage by dividing the handwheel radius (

12 \text{ inches}

) by the axle radius (

4 \text{ inches}

), yielding an IMA of

3

. Dividing the

180 \text{ lb}

load by this IMA gives a required effort of

60 \text{ lbs}

.

Step-by-Step Solution

1

Determine the required direction of rotation.

Counterclockwise rotation.

Since the cable hangs freely from the right side of the axle, pulling the right side upwards (winding the cable) requires turning the axle to the left, which is a counterclockwise direction.

2

Determine the working radii of the wheel and the axle.

Wheel radius (

R

) =

12 \text{ inches}

; Axle radius (

r

) =

4 \text{ inches}

.

The handwheel's radius is given as

12 \text{ inches}

. The axle's diameter is

8 \text{ inches}

, so its radius is half of that (

8 / 2 = 4 \text{ inches}

).

3

Calculate the Ideal Mechanical Advantage (IMA).

\text{IMA} = 3

For a wheel and axle system where effort is applied to the wheel,

\text{IMA} = R / r = 12 / 4 = 3

.

4

Calculate the required ideal effort force.

\text{Effort} = 60 \text{ lbs}

\text{Effort} = \text{Load} / \text{IMA} = 180 / 3 = 60 \text{ lbs}

.

Key Concept

Wheel and Axle Mechanical Advantage and Rotational Kinematics

Question 8Question →

A carpenter uses a standard screwdriver to drive a screw into a piece of wood. The screwdriver handle, acting as the wheel, has a radius of $2 \text{ cm}$ . The metal shaft, acting as the axle, has a radius of $4 \text{ mm}$ . What is the ideal mechanical advantage (IMA) of the screwdriver?

5

0.5

2

0.2

Show answer & explanation

Answer: 5

Answer

The ideal mechanical advantage is 5.

The correct answer is 5. To find the ideal mechanical advantage, both radii must first be in the same unit. Converting the

2 \text{ cm}

handle radius yields

20 \text{ mm}

. Dividing the wheel radius (

20 \text{ mm}

) by the axle radius (

4 \text{ mm}

) gives a mechanical advantage of 5, meaning the screwdriver multiplies the user's turning force by 5 times.

Step-by-Step Solution

1

Convert the given measurements to the same unit (millimeters).

The handle (wheel) radius is

2 \text{ cm}

, which equals

20 \text{ mm}

.

Units must match before calculating any ratios to ensure accuracy.

2

Identify the values for the wheel and the axle.

Wheel radius =

20 \text{ mm}

; Axle radius =

4 \text{ mm}

.

In a screwdriver, the wider handle functions as the wheel, while the narrower shaft functions as the axle.

3

Calculate the Ideal Mechanical Advantage (IMA) by dividing the wheel radius by the axle radius.

IMA = 20 / 4 = 5

.

The formula for the IMA of a wheel and axle system driven by the wheel is the radius of the wheel divided by the radius of the axle.

Key Concept

Ideal Mechanical Advantage (IMA) of a Wheel and Axle

Question 9Question →

A worker operates a well crank (a wheel and axle mechanism) to lift a heavy bucket of water. The crank handle travels in a circle with a radius of $12 \text{ inches}$ . The lifting rope is wound around an axle that has a diameter of $6 \text{ inches}$ . What is the Ideal Mechanical Advantage (IMA) of this system?

4

2

1

0.5

Show answer & explanation

Answer: $4$

Answer

The correct Ideal Mechanical Advantage (IMA) is

4

.

The correct answer is calculated by finding the ratio of the wheel's radius to the axle's radius. The wheel radius is given as

12 \text{ inches}

. The axle diameter is

6 \text{ inches}

, meaning its radius is

3 \text{ inches}

. Dividing

12

by

3

yields an IMA of

4

.

Step-by-Step Solution

1

Identify the radius of the wheel (input distance).

The radius of the crank handle circle is

12 \text{ inches}

.

The IMA formula requires the radius of the wheel.

2

Calculate the radius of the axle (output distance).

The diameter is

6 \text{ inches}

, so the radius is

6 \div 2 = 3 \text{ inches}

.

The IMA formula requires the radius of the axle, not the diameter.

3

Calculate the Ideal Mechanical Advantage (IMA).

IMA = \frac{\text{Radius of Wheel}}{\text{Radius of Axle}} = \frac{12}{3} = 4

This ratio shows the mechanical advantage gained by applying force to the larger wheel to turn the smaller axle.

Key Concept

Ideal Mechanical Advantage of a Wheel and Axle

Estimated Time:45s

Question 10Question →

A mechanic is inspecting a steering mechanism. If the steering wheel has a radius of $14 \text{ inches}$ and the attached steering column (axle) has a radius of $2 \text{ inches}$ , what is the ideal mechanical advantage of this wheel and axle system?

Show answer & explanation

Answer: 7

Answer

7

The ideal mechanical advantage (IMA) is the ratio of the distance the effort moves to the distance the load moves. In a wheel and axle, this is proportional to their radii. Dividing the wheel's radius (

14 \text{ inches}

) by the axle's radius (

2 \text{ inches}

) gives an IMA of

7

.

Step-by-Step Solution

1

Identify the relevant formula

\text{IMA} = \frac{R}{r}

The ideal mechanical advantage of a wheel and axle is the ratio of the effort radius (wheel) to the load radius (axle).

2

Substitute the known values

\text{IMA} = \frac{14}{2}

The wheel radius

R

is

14 \text{ inches}

and the axle radius

r

is

2 \text{ inches}

.

3

Perform the division

7

Dividing

14

by

2

yields an ideal mechanical advantage of

7

.

Key Concept

For a wheel and axle system where the input force is applied to the wheel, the ideal mechanical advantage is calculated by dividing the radius of the wheel by the radius of the axle.

Question 11Question →

A pipeline technician operates a circular handwheel to control the flow of water through a large valve. The handwheel has an overall diameter of $16 \text{ inches}$ and is attached to a central valve stem (the axle) that has a radius of $2 \text{ inches}$ . Assuming the effort force is applied to the outside edge of the handwheel, what is the ideal mechanical advantage (IMA) of this mechanism?

0.25

2

4

8

Show answer & explanation

Answer: $4$

Answer

The ideal mechanical advantage is

4

.

The ideal mechanical advantage (IMA) of a wheel and axle system is found by dividing the radius of the input force (the wheel) by the radius of the output force (the axle). Since the handwheel's diameter is

16 \text{ inches}

, its radius is

8 \text{ inches}

. Dividing the wheel radius (

8 \text{ inches}

) by the axle radius (

2 \text{ inches}

) gives an IMA of

4

.

Step-by-Step Solution

1

Convert the handwheel diameter to a radius.

The handwheel radius is

16 \div 2 = 8 \text{ inches}

.

To calculate mechanical advantage accurately, both the wheel and the axle must use the same type of measurement (either both radii or both diameters).

2

Divide the wheel radius by the axle radius.

\text{IMA} = \frac{8 \text{ inches}}{2 \text{ inches}} = 4

.

The formula for the Ideal Mechanical Advantage of a wheel and axle (when force is applied to the wheel) is the radius of the wheel divided by the radius of the axle.

Key Concept

The ideal mechanical advantage of a wheel and axle is the ratio of the radius of the wheel to the radius of the axle.

Question 12Question →

Match each wheel and axle configuration to its resulting Ideal Mechanical Advantage (IMA).

Click a left item, then click its matching right item

Show answer & explanation

Answer

Match each configuration by correctly dividing the effort radius by the load radius.

The correct matches are found by calculating the ratio of the input (effort) distance to the output (load) distance. For a wheel and axle, this is equivalent to

\frac{\text{Effort Radius}}{\text{Load Radius}}

. When the effort is on the wheel, the IMA is greater than 1. When the effort is on the axle, the IMA is less than 1.

Step-by-Step Solution

1

Identify the effort radius and the load radius for each given configuration.

You determine whether the force is applied to the wheel (larger radius) or the axle (smaller radius).

The location of the applied force dictates which radius is the numerator in the IMA formula.

2

Apply the Ideal Mechanical Advantage formula:

\text{IMA} = \frac{\text{Effort Radius}}{\text{Load Radius}}

.

Calculations yield

5

,

4

,

0.5

, and

0.2

for the respective configurations.

This formula provides the theoretical multiplication of force for a wheel and axle mechanism.

Key Concept

Ideal Mechanical Advantage of a Wheel and Axle

Estimated Time:45s

Question 13Question →

A technician uses a T-handle socket wrench to tighten a nut. The wide horizontal T-handle functions as the wheel, and the narrow vertical shaft functions as the axle. If the technician turns the T-handle clockwise, what is the direction of the shaft's rotation, and how does the output force at the shaft compare to the effort force applied to the handle?

The shaft rotates clockwise, and the output force is greater than the effort force.

The shaft rotates counterclockwise, and the output force is greater than the effort force.

The shaft rotates clockwise, and the output force is less than the effort force.

The shaft rotates counterclockwise, and the mechanism acts as a first-class lever.

Show answer & explanation

Answer: The shaft rotates clockwise, and the output force is greater than the effort force.

Answer

The shaft rotates clockwise, and the output force is greater than the effort force.

A wheel and axle mechanism consists of two parts that rotate together on the same axis. Therefore, if the wheel (T-handle) is turned clockwise, the axle (shaft) also turns clockwise. Because the input effort is applied to the larger radius of the handle, the mechanism multiplies the force, resulting in a greater output force at the smaller radius of the axle.

Step-by-Step Solution

1

Determine the direction of rotation for the axle (shaft).

The shaft rotates clockwise, the same as the handle.

In a wheel and axle mechanism, both components are rigidly attached to the same axis and rotate as a single solid unit.

2

Evaluate the force relationship between the input and output.

The output force at the shaft is greater than the effort force applied to the handle.

Applying effort to the component with the larger diameter (the wheel/handle) to drive the component with the smaller diameter (the axle/shaft) creates a mechanical advantage greater than 1, which multiplies the force.

Key Concept

Wheel and Axle Rotational Direction and Mechanical Advantage

Question 14Question →

Wheel and axle mechanisms can be designed to multiply either force or speed. Match each real-world application of a wheel and axle to its correct mechanical description.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Doorknob matches the internal latch description; Automobile drive wheels match the engine and tire description; Manual hand winch matches the handle and cable drum description; Ceiling fan matches the motor and blades description.

Each mechanism uses the wheel and axle principle based on its specific need. The doorknob and hand winch are force multipliers because effort is applied to the larger wheel. The drive wheels and ceiling fan are speed multipliers because effort is applied to the smaller axle.

Step-by-Step Solution

1

Identify the mechanisms that act as force multipliers (input on the wheel).

The doorknob and manual hand winch both receive input force on the larger radius to produce a larger output force on the smaller radius.

Determining the location of the input force helps classify the mechanical advantage.

2

Identify the mechanisms that act as speed multipliers (input on the axle).

The automobile drive wheels and ceiling fan both receive input force on the smaller central radius to produce greater speed at the larger outer radius.

Speed multipliers take advantage of the larger circumference of the wheel covering more distance in the same amount of time.

3

Pair each specific device with its corresponding functional description.

The doorknob operates an internal latch; the drive wheels rotate the outer tire; the winch winds a cable around a drum; the ceiling fan spins blades.

Matching the physical components completes the logical pairing for each wheel and axle application.

Key Concept

Force Multipliers vs. Speed Multipliers in Wheel and Axle Systems

Question 15Question →

In a wheel and axle system, the mechanical advantage profile changes depending on whether the input (effort) force is applied to the larger radius (the wheel) or the smaller radius (the axle). Match each specific component of the following real-world mechanisms to its exact mechanical role.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Screwdriver handle matches to 'Wheel receiving input effort'; Screwdriver shaft matches to 'Axle delivering output load'; Motorcycle sprocket matches to 'Axle receiving input effort'; Motorcycle tire matches to 'Wheel delivering output speed/distance'.

Mechanical advantage in a wheel and axle depends entirely on where the input force is applied. In a screwdriver, applying force to the larger radius (the handle/wheel) multiplies the torque delivered by the smaller radius (the shaft/axle) to drive a screw. Conversely, on a motorcycle, the engine applies high force to the small radius (the sprocket/axle), which turns the larger radius (the tire/wheel). This configuration requires more effort but results in the tire's outer edge traveling a much greater distance at high speed.

Step-by-Step Solution

1

Analyze the screwdriver mechanism to determine component roles.

The handle is larger (the wheel) and is turned by the user (receives effort). The narrow shaft (the axle) turns the screw (delivers load).

Identifying which component receives the input force and which applies the output force establishes the system as a force multiplier.

2

Analyze the motorcycle rear wheel mechanism to determine component roles.

The chain drives the small central sprocket (the axle receives effort). This turns the large outer tire (the wheel delivers speed/distance).

Systems driven by the axle are designed as speed multipliers, sacrificing force for greater distance covered per rotation.

Key Concept

A wheel and axle can either act as a force multiplier (when effort is applied to the wheel) or a speed and distance multiplier (when effort is applied to the axle).

Estimated Time:1m 30s

Question 16Question →

A maintenance worker operates a hand-cranked rotary fluid pump. The pump's crank handle rotates in a circle with a radius of $18 \text{ inches}$ . It turns a central drive shaft with a diameter of $6 \text{ inches}$ . If the worker applies a steady perpendicular force of $60 \text{ pounds}$ to the handle, what is the output force exerted at the outer edge of the drive shaft? (Assume an ideal simple machine with no friction losses.)

180 \text{ pounds}

10 \text{ pounds}

360 \text{ pounds}

20 \text{ pounds}

Show answer & explanation

Answer: $360 \text{ pounds}$

Answer

The output force is

360 \text{ pounds}

.

The correct answer is derived by first converting the axle's diameter to a radius (

3 \text{ inches}

). The system's mechanical advantage is the ratio of the wheel's radius to the axle's radius (18 / 3 = 6). Multiplying the input force of

60 \text{ pounds}

by the mechanical advantage of 6 yields an output force of

360 \text{ pounds}

.

Step-by-Step Solution

1

Determine the radius of both the wheel (crank path) and the axle (drive shaft).

Wheel radius (

R

) is given as

18 \text{ inches}

. Axle diameter is

6 \text{ inches}

, so axle radius (

r

) is

3 \text{ inches}

.

Mechanical advantage formulas require the radii of the rotating components, not the diameter.

2

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

\text{IMA} = \frac{R}{r} = \frac{18}{3} = 6

.

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

3

Calculate the output force by multiplying the input force by the IMA.

\text{Output Force} = 60 \text{ pounds} \times 6 = 360 \text{ pounds}

.

In an ideal simple machine, the input force multiplied by the mechanical advantage equals the output load force.

Key Concept

Mechanical Advantage of a Wheel and Axle

Question 17Question →

A machinist operates a manual drill press. The feed lever (acting as the wheel) has an effective length of $8 \text{ inches}$ from the center of rotation. It turns a pinion gear (acting as the axle) with a radius of $1.6 \text{ inches}$ . If the machinist applies a tangential pull of $25 \text{ lbs}$ to the end of the feed lever, what downward force is applied by the pinion gear to the drill quill? (Assume $100\%$ efficiency).

Show answer & explanation

Answer: 125

Answer

The downward force applied by the pinion gear is 125 lbs.

In a wheel and axle system where effort is applied to the larger wheel, the mechanism multiplies the input force. The Ideal Mechanical Advantage (IMA) is found by dividing the effort radius (lever length of 8 inches) by the load radius (pinion gear radius of 1.6 inches). With an IMA of 5, the input force of 25 lbs is multiplied by 5 to yield 125 lbs of output force.

Step-by-Step Solution

1

Identify the given dimensions and force.

Wheel radius (lever) = 8 inches, Axle radius (pinion) = 1.6 inches, Input Force = 25 lbs.

Identifying known variables is the first step in solving a mechanics problem.

2

Calculate the Ideal Mechanical Advantage (IMA).

IMA = Wheel Radius / Axle Radius = 8 / 1.6 = 5.

The IMA determines how much the wheel and axle mechanism multiplies the input force.

3

Calculate the output force.

Output Force = Input Force × IMA = 25 × 5 = 125 lbs.

In a 100% efficient system, the output force is the product of the input force and the mechanical advantage.

Key Concept

Ideal Mechanical Advantage of a wheel and axle, calculated as the ratio of their radii.

Question 18Question →

Sailors are using a manual capstan to pull an anchor chain. The capstan drum, around which the chain wraps, has a diameter of $1.6 \text{ feet}$ . The sailors push on the capstan bars at a horizontal distance of $4.0 \text{ feet}$ from the central axis of rotation. If the anchor chain exerts a steady pull of $1,500 \text{ pounds}$ , what is the total effort force the sailors must apply to the bars to keep the anchor moving at a constant speed? (Assume 100% efficiency and ignore friction)

600 \text{ pounds}

7,500 \text{ pounds}

300 \text{ pounds}

3,750 \text{ pounds}

Show answer & explanation

Answer: $300 \text{ pounds}$

Answer

300 \text{ pounds}

A capstan operates as a wheel and axle mechanism. The effort is applied at a distance of

4.0 \text{ feet}

(the wheel's radius), and the load is pulled by a drum with a radius of

0.8 \text{ feet}

(half of the

1.6\text{-foot}

diameter). The Ideal Mechanical Advantage is the ratio of these radii:

4.0 / 0.8 = 5

. To find the required effort force, divide the

1,500\text{-pound}

load by the mechanical advantage of

5

, which equals

300 \text{ pounds}

.

Step-by-Step Solution

1

Determine the radii of both the wheel (effort arm) and the axle (drum).

Wheel radius (

R

) =

4.0 \text{ feet}

. Axle radius (

r

) =

1.6 \text{ feet} / 2 = 0.8 \text{ feet}

.

The Ideal Mechanical Advantage (IMA) of a wheel and axle depends on the ratio of their radii. The problem provides the diameter of the drum, which must be halved.

2

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

IMA = \frac{R}{r} = \frac{4.0}{0.8} = 5

.

Because the effort is applied to the larger wheel (the bars), the system multiplies the input force by a factor of 5.

3

Calculate the required effort force to overcome the load.

\text{Effort Force} = \frac{\text{Load}}{IMA} = \frac{1500}{5} = 300 \text{ pounds}

.

The effort force is the total load divided by the mechanical advantage.

Key Concept

Ideal Mechanical Advantage of a Wheel and Axle

Question 19Question →

Match each wheel and axle concept to its correct definition or mechanical characteristic.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches are: Wheel and Axle System pairs with its definition of two rotating circular objects; Ideal Mechanical Advantage (IMA) pairs with the ratio of input radius to output radius; Force Multiplier Setup pairs with effort applied to the larger wheel; and Distance/Speed Multiplier Setup pairs with effort applied to the smaller axle.

Each concept describes a specific operational mode or structural property of the wheel and axle. The fundamental physical relationship dictates that applying effort to the larger diameter (wheel) multiplies force at the expense of distance, while applying effort to the smaller diameter (axle) multiplies speed and distance at the expense of force. The mathematical ratio of these radii determines the exact Ideal Mechanical Advantage.

Step-by-Step Solution

1

Identify the basic definition of the wheel and axle system.

It is matched as a simple machine with two connected circular objects of different radii rotating together.

Establishes the structural baseline for the mechanism.

2

Determine how Ideal Mechanical Advantage (IMA) is calculated for rotational machines.

IMA is matched to the ratio of the input component's radius to the output component's radius.

IMA represents the theoretical multiplication factor of the machine based on lever arms.

3

Analyze the conditions for a force multiplier setup.

It is matched to the scenario where effort is on the larger wheel, resulting in greater output force at the smaller axle (MA > 1).

A larger input radius creates a longer effective lever arm for the effort.

4

Analyze the conditions for a distance/speed multiplier setup.

It is matched to the scenario where effort is on the smaller axle, resulting in greater output speed at the larger wheel (MA < 1).

One rotation of the small input axle forces the large outer wheel to travel a much larger circumference in the same amount of time.

Key Concept

Mechanical principles and force-distance trade-offs of the wheel and axle simple machine.

Question 20Question →

A technician is operating a large industrial pipeline valve. The valve's central stem, which acts as the axle, has a diameter of $3 \text{ inches}$ . The handwheel attached to the stem has a radius of $12 \text{ inches}$ . If the technician applies a tangential effort force of $60 \text{ pounds}$ to the outer edge of the handwheel, what is the resulting output force applied to the valve stem?

7.5 \text{ pounds}

480 \text{ pounds}

240 \text{ pounds}

960 \text{ pounds}

Show answer & explanation

Answer: $480 \text{ pounds}$

Answer

The resulting output force is

480 \text{ pounds}

.

The correct answer accurately calculates the mechanical advantage by first finding the axle's radius (

1.5 \text{ inches}

), then dividing the wheel's radius (

12 \text{ inches}

) by the axle's radius to get an Ideal Mechanical Advantage (IMA) of

8

. Multiplying this IMA by the

60 \text{ pounds}

of input force yields an output force of

480 \text{ pounds}

.

Step-by-Step Solution

1

Determine the radius of the axle.

Axle radius =

3 \text{ inches} \div 2 = 1.5 \text{ inches}

.

Mechanical advantage must be calculated using the same dimension type for both components (either both radii or both diameters). The stem provided the wheel's radius but the axle's diameter.

2

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

IMA =

\frac{12 \text{ inches}}{1.5 \text{ inches}} = 8

.

IMA is the ratio of the radius of the wheel (input) to the radius of the axle (output).

3

Calculate the output force applied to the valve stem.

Output Force =

8 \times 60 \text{ pounds} = 480 \text{ pounds}

.

The output force is the product of the input effort force and the mechanical advantage.

Key Concept

The mechanical advantage of a wheel and axle is determined by the ratio of their radii (or diameters), which multiplies the input force to overcome a larger load.

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