Simple Machines Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 8

Question 141Question →

An aviation mechanic is using a manual sheet metal bender. The operating lever pivots on a fixed hinge at its far end. The metal sheet being bent provides resistance against the lever at a point $3\text{ inches}$ from the hinge. If the mechanic applies a perpendicular effort force of $50\text{ pounds}$ to the handle at a distance of $24\text{ inches}$ from the hinge, which of the following correctly identifies the class of this lever and the resulting force applied to the metal sheet?

Second-class lever;

400\text{ pounds}

Third-class lever;

400\text{ pounds}

First-class lever;

350\text{ pounds}

Second-class lever;

6.25\text{ pounds}

Show answer & explanation

Answer: Second-class lever; $400\text{ pounds}$

Answer

The mechanism is a second-class lever, and the resulting force applied to the metal sheet is

400\text{ pounds}

.

In this scenario, the load (metal sheet) is positioned between the fulcrum (hinge) and the effort (mechanic's hand), which defines a second-class lever. The mechanical advantage is found by dividing the effort arm distance (

24\text{ inches}

) by the load arm distance (

3\text{ inches}

), yielding an MA of

8

. Multiplying the applied effort of

50\text{ pounds}

by the MA of

8

gives a total applied force of

400\text{ pounds}

.

Step-by-Step Solution

1

Determine the class of the lever based on the component positions.

The hinge (fulcrum) is at one end, the effort is at the other end, and the metal sheet (load) is between them. This is the definition of a second-class lever.

Correctly identifying the lever class ensures the proper understanding of how forces and distances are distributed.

2

Calculate the Mechanical Advantage (MA) of the lever.

MA = \frac{\text{Effort Arm}}{\text{Load Arm}} = \frac{24\text{ inches}}{3\text{ inches}} = 8

Mechanical advantage dictates how much the input effort force will be multiplied by the machine.

3

Calculate the resulting force applied to the load (metal sheet).

\text{Force} = \text{Effort Force} \times MA = 50\text{ pounds} \times 8 = 400\text{ pounds}

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

Key Concept

Lever Classification and Mechanical Advantage

Question 142Question →

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 143Question →

An automotive technician operates a hydraulic floor jack by pressing down on its handle. The handle pivots on a fixed pin at one end. The technician applies a steady downward force of $30\text{ lbs}$ at the opposite end of the handle. The total length of the handle from the pivot pin to the point of effort is $24\text{ inches}$ . The handle pushes down on a pump piston that is located $1.5\text{ inches}$ from the pivot pin. What is the total force, in pounds, applied to the pump piston? (Assume the handle is horizontal and ignore friction.)

Show answer & explanation

Answer: 480

Answer

480

The jack handle operates as a lever with the fulcrum at one end. Because the effort is applied much further from the fulcrum than the load, the lever multiplies the applied force. The mechanical advantage is the ratio of the effort arm to the load arm (

24 / 1.5 = 16

). Multiplying the

30\text{ lbs}

of effort by a mechanical advantage of 16 yields a

480\text{ lbs}

force on the piston.

Step-by-Step Solution

1

Identify the lever components.

Fulcrum = pivot pin; Effort Force (E) = 30 lbs; Effort Arm (d_E) = 24 inches; Load Arm (d_L) = 1.5 inches.

To apply the torque balance equation, all distances must be measured from the fulcrum.

2

Set up the torque balance equation.

E \times d_E = L \times d_L

In a static lever system, the torque generated by the effort equals the torque generated by the load.

3

Substitute the given values.

30 \times 24 = L \times 1.5

To prepare the equation for solving the unknown load force (L).

4

Solve for the load force.

720 = 1.5L \implies L = 480\text{ lbs}

Dividing the effort torque by the load arm distance yields the resulting force applied to the piston.

Key Concept

Torque Balance and Mechanical Advantage

Question 144Question →

An angler is reeling in a catch and holds the fishing rod perfectly horizontal. The butt of the rod is braced firmly against the angler's waist, which acts as the pivot point. The angler's top hand applies an upward lifting force $2 \text{ feet}$ away from the waist. If the fish creates a constant downward pull of $15 \text{ lbs}$ at the rod's tip, which is exactly $8 \text{ feet}$ away from the pivot point, how much upward effort force must the angler's top hand exert to maintain the rod's horizontal position? (Assume the weight of the rod itself is negligible.)

120 \text{ lbs}

60 \text{ lbs}

45 \text{ lbs}

3.75 \text{ lbs}

Show answer & explanation

Answer: $60 \text{ lbs}$

Answer

60 \text{ lbs}

In a third-class lever, the effort is located between the fulcrum and the load. To maintain equilibrium, the effort torque must equal the load torque. Since the effort arm (

2 \text{ feet}

) is one-quarter the length of the load arm (

8 \text{ feet}

), the mechanical advantage is

0.25

. Therefore, the required effort force must be four times the load force (

15 \text{ lbs} \times 4 = 60 \text{ lbs}

).

Step-by-Step Solution

1

Identify the lever class and pivot point.

The waist is the fulcrum, the top hand is the effort, and the fish is the load. This is a third-class lever.

Understanding the arrangement is required to correctly measure distances for the torque equation.

2

Calculate the torque exerted by the load (the fish).

\text{Load Torque} = 15 \text{ lbs} \times 8 \text{ ft} = 120 \text{ lb-ft}

.

Torque is the product of force and distance from the fulcrum.

3

Set the effort torque equal to the load torque to find the required force.

\text{Effort} \times 2 \text{ ft} = 120 \text{ lb-ft}

, which means

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

.

For the rod to remain horizontally stable, the opposing torques must be perfectly balanced.

Key Concept

Third-Class Levers and Torque Equilibrium

Alternative Method

Calculate the Mechanical Advantage (MA) first:

\text{MA} = \frac{\text{Effort Arm}}{\text{Load Arm}} = \frac{2}{8} = 0.25

. Then find the effort using

\text{Effort} = \frac{\text{Load}}{\text{MA}} = \frac{15}{0.25} = 60 \text{ lbs}

.

Estimated Time:1m 0s

Question 145Question →

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 146Question →

Match each operational adjustment made to a mechanical lever system with its correct effect on the system's mechanical advantage.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Grasping the pry bar further out increases the effort arm. Shifting the wheelbarrow load closer to the wheel decreases the load arm. Pinching tweezers closer to the hinge decreases the effort arm. Doubling all proportions maintains the original ratio.

Mechanical advantage for any lever is calculated by dividing the length of the effort arm by the length of the load arm (

MA = \frac{d_e}{d_l}

). Any adjustment that increases the effort arm or decreases the load arm will result in a higher mechanical advantage, making the work feel easier. Conversely, reducing the effort arm or increasing the load arm lowers the mechanical advantage. Proportional changes to both arms maintain the exact same ratio.

Step-by-Step Solution

1

Recall the formula for Mechanical Advantage (MA) of a lever.

MA = \frac{\text{Effort Arm}}{\text{Load Arm}}

.

This fundamental relationship determines how any physical adjustment to the lever's dimensions will affect the required force.

2

Analyze the pry bar adjustment.

Grasping the tool further from the fulcrum increases the numerator (Effort Arm), which increases the overall MA.

A higher MA means less force is required to move the same load.

3

Analyze the wheelbarrow adjustment.

The wheel is the fulcrum. Moving the load closer to the wheel decreases the denominator (Load Arm), which increases the overall MA.

A smaller denominator results in a larger ratio, requiring less lifting force at the handles.

4

Analyze the tweezers adjustment.

Tweezers are a third-class lever with the fulcrum at the joined end. Pinching closer to the joined end decreases the Effort Arm.

Decreasing the numerator lowers the MA, meaning more pinch force is required, though the tips will close faster and over a larger distance.

5

Analyze the proportional scaling adjustment.

If both the effort arm and load arm are multiplied by 2, the new MA is

\frac{2 \times \text{Effort Arm}}{2 \times \text{Load Arm}}

.

The factor of 2 cancels out, leaving the mechanical advantage exactly the same as the original lever.

Key Concept

Mechanical Advantage Ratio and Lever Adjustments

Question 147Question →

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.

Question 148Question →

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 149Question →

A dump truck bed pivots on a rear hinge, which acts as the fulcrum. A hydraulic cylinder provides the upward effort force and is attached to the bed $2\text{ ft}$ from the rear hinge. The center of gravity of the loaded bed is located $6\text{ ft}$ from the rear hinge. If the total weight of the loaded bed is $12,000\text{ lbs}$ , what is the minimum purely vertical force the hydraulic cylinder must apply to begin lifting the horizontally resting bed?

4,000 \text{ lbs}

12,000 \text{ lbs}

24,000 \text{ lbs}

36,000 \text{ lbs}

Show answer & explanation

Answer: $36,000 \text{ lbs}$

Answer

The minimum purely vertical force the hydraulic cylinder must apply is

36,000 \text{ lbs}

.

In this 3rd class lever, the effort arm (

2\text{ ft}

) is significantly shorter than the load arm (

6\text{ ft}

). To balance the torque created by the

12,000\text{ lbs}

load (

12,000 \times 6 = 72,000\text{ ft-lbs}

), the hydraulic cylinder must apply a proportionally larger force of

36,000\text{ lbs}

at its

2\text{ ft}

distance (

36,000 \times 2 = 72,000

).

Step-by-Step Solution

1

Identify the lever class and the pivot point.

The rear hinge is the fulcrum. The hydraulic cylinder provides the effort between the fulcrum and the load, making this a 3rd class lever.

To correctly apply the principles of torque, the locations of the effort and load relative to the fulcrum must be established.

2

Calculate the torque exerted by the loaded bed.

12,000 \text{ lbs} \times 6 \text{ ft} = 72,000 \text{ ft-lbs}

.

Torque is equal to force multiplied by the distance from the fulcrum.

3

Set up the torque balance equation for the effort force (

F_e

).

F_e \times 2 \text{ ft} = 72,000 \text{ ft-lbs}

.

To begin lifting the bed, the effort torque must be equal to the load torque.

4

Solve the equation for the effort force.

F_e = 36,000 \text{ lbs}

.

Dividing both sides by

2 \text{ ft}

isolates the required force.

Key Concept

Mechanical advantage and torque balance in a 3rd class lever

Estimated Time:1m 0s

Question 150Question →

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 151Question →

An automotive brake pedal operates as a second-class lever. The pivot point (fulcrum) is located at the top of the pedal assembly. The pushrod, which connects to the brake master cylinder, is attached $4 \text{ inches}$ below the pivot. The driver applies force to the pedal pad, which is located $14 \text{ inches}$ below the pivot. If the driver applies a perpendicular force of $50 \text{ lbs}$ to the pedal pad, what is the force, in pounds, transmitted to the pushrod? Assume the system is in equilibrium.

Show answer & explanation

Answer: 175

Answer

175

The brake pedal acts as a second-class lever in equilibrium. According to the principle of moments, the effort torque (

50 \text{ lbs} \times 14 \text{ inches} = 700 \text{ in-lbs}

) must equal the load torque. Dividing the total torque by the load arm distance (

4 \text{ inches}

) gives an output force of

175 \text{ lbs}

.

Step-by-Step Solution

1

Identify the lever components and their distances from the fulcrum.

Effort arm =

14 \text{ inches}

, Load arm =

4 \text{ inches}

.

In torque calculations, all distances must be measured directly from the pivot point (fulcrum).

2

Apply the torque balance equation for a lever in equilibrium.

F_{effort} \times d_{effort} = F_{load} \times d_{load}

To find the unknown force, the applied torque must equal the output torque.

3

Substitute the known values into the equation and solve for the load force.

50 \times 14 = F_{load} \times 4 \rightarrow 700 = F_{load} \times 4 \rightarrow F_{load} = 175 \text{ lbs}

.

Isolating the unknown variable gives the final transmitted force.

Key Concept

Torque Equilibrium and Mechanical Advantage in Second-Class Levers

Estimated Time:1m 0s

Question 152Question →

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 153Question →

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 154Question →

A manual water pump features a straight, rigid handle that pivots on a fixed hinge at one end. A vertical piston rod, which provides the resistance, is attached to the handle $0.20 \text{ meters}$ from the hinge. A user pushes down on the opposite end of the handle, located $1.00 \text{ meter}$ from the hinge. If the piston rod requires $400 \text{ N}$ of force to operate, what is the minimum effort force the user must apply, and what class of lever does this mechanism represent?

100 \text{ N}

; First-class lever

80 \text{ N}

; Third-class lever

80 \text{ N}

; Second-class lever

2000 \text{ N}

; Second-class lever

Show answer & explanation

Answer: $80 \text{ N}$ ; Second-class lever

Answer

80 \text{ N}

; Second-class lever

In this mechanism, the fulcrum (hinge) is at one end, the effort is applied at the opposite end, and the resistance (load) is located between them. This arrangement defines a second-class lever. The effort arm is the total distance from the fulcrum to the effort (

1.00 \text{ m}

), while the load arm is the distance from the fulcrum to the resistance (

0.20 \text{ m}

). The mechanical advantage (MA) is found by dividing the effort arm by the load arm (

1.00 / 0.20 = 5

). To find the required effort force, the load is divided by the MA (

400 \text{ N} / 5 = 80 \text{ N}

).

Step-by-Step Solution

1

Identify the fulcrum, effort, and load positions to determine the lever class.

The fulcrum is at one end (hinge), the effort is at the opposite end, and the load is located between them.

This specific physical arrangement defines a second-class lever.

2

Determine the lengths of the effort arm and load arm.

Effort arm =

1.00 \text{ m}

; Load arm =

0.20 \text{ m}

.

Lever arms are strictly measured from the fulcrum to the respective points where the force and resistance are applied.

3

Calculate the Mechanical Advantage (MA).

\text{MA} = 1.00 \text{ m} / 0.20 \text{ m} = 5

.

The MA of a lever is the ratio of the effort arm to the load arm.

4

Calculate the required effort force.

400 \text{ N} / 5 = 80 \text{ N}

.

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

Key Concept

Second-Class Levers and Mechanical Advantage

Alternative Method

Instead of using the Mechanical Advantage (MA) ratio, the static equilibrium torque equation can be used directly: (Effort Force) × (Distance from Fulcrum to Effort) = (Load Force) × (Distance from Fulcrum to Load). This gives: Effort × 1.00 = 400 × 0.20, solving directly to Effort = 80 N.

Estimated Time:1m 0s

Question 155Question →

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 156Question →

An industrial sheet metal shear operates as a second-class lever. The pivot point (fulcrum) is located at the far end of the cutting arm. A mechanic applies a downward effort force of $150 \text{ N}$ at the handle, which is located $80 \text{ cm}$ from the fulcrum. The metal sheet is positioned $16 \text{ cm}$ from the fulcrum. Assuming the lever is rigid and frictionless, what is the downward cutting force, in newtons, applied to the metal sheet?

Show answer & explanation

Answer: 750

Answer

750

A second-class lever always places the resistance between the effort and the fulcrum, providing a mechanical advantage greater than 1. The torque applied by the effort (

150 \text{ N} \times 80 \text{ cm}

) must equal the opposing torque exerted on the resistance (

F_r \times 16 \text{ cm}

). Solving this equilibrium equation yields a cutting force of

750 \text{ N}

.

Step-by-Step Solution

1

Identify the lever parameters and equation

F_e = 150 \text{ N}

,

d_e = 80 \text{ cm}

,

d_r = 16 \text{ cm}

Establishing the values needed for the principle of moments (torque balance).

2

Apply the law of the lever

150 \times 80 = F_r \times 16

In equilibrium, the input torque equals the output torque.

3

Calculate the unknown resistance force

F_r = \frac{12000}{16} = 750 \text{ N}

Isolating

F_r

gives the mechanical force applied to the metal.

Key Concept

Mechanical Advantage and Torque Balance in Second-Class Levers

Estimated Time:1m 0s

Question 157Question →

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

Question 158Question →

A construction worker is hoisting a bucket of tools to a scaffold using the pulley arrangement shown in the schematic. If the bucket must be raised a vertical distance of $15 \text{ feet}$ , what total length of rope must the worker pull downward?

7.5 \text{ feet}

15 \text{ feet}

30 \text{ feet}

45 \text{ feet}

Show answer & explanation

Answer: $30 \text{ feet}$

Answer

30 \text{ feet}

To find the total length of rope that must be pulled, you multiply the vertical lifting distance of the load by the Ideal Mechanical Advantage (IMA) of the system. In the given schematic, there are 2 rope strands that pull upward on the movable pulley, so the IMA is 2. Therefore, the rope must be pulled

15 \text{ feet} \times 2 = 30 \text{ feet}

.

Step-by-Step Solution

1

Determine the Ideal Mechanical Advantage (IMA) by counting the supporting strands.

IMA = 2

There are exactly 2 segments of rope pulling upward directly on the movable block holding the bucket. The downward-pulling effort strand does not support the load.

2

Relate the IMA to the effort distance and load distance.

Effort Distance = Load Distance × IMA

In simple machines, force is multiplied at the expense of distance. You must pull the rope further than the distance the load actually rises.

3

Calculate the required pull distance.

Effort Distance =

15 \text{ feet} \times 2 = 30 \text{ feet}

Applying the values to the formula gives the final answer.

Key Concept

In a pulley system, the distance the effort rope must be pulled is equal to the distance the load is lifted multiplied by the Ideal Mechanical Advantage (IMA).

Estimated Time:1m 0s

Question 159Question →

A combat engineer uses a $120 \text{ cm}$ rigid steel pry bar to lift a $600 \text{ N}$ concrete block. The pivot point (fulcrum) is placed exactly $30 \text{ cm}$ from the block. If the engineer applies a downward force at the far opposite end of the pry bar, what is the effort force required to balance the block?

150 \text{ N}

1800 \text{ N}

200 \text{ N}

2400 \text{ N}

Show answer & explanation

Answer: $200 \text{ N}$

Answer

200 \text{ N}

The pry bar acts as a first-class lever. The load arm (

d_L

) is the distance from the fulcrum to the concrete block, which is given as

30 \text{ cm}

. The effort arm (

d_E

) is the distance from the fulcrum to the opposite end of the bar, calculated as the total length minus the load arm (

120 \text{ cm} - 30 \text{ cm} = 90 \text{ cm}

). According to the law of moments, the system is balanced when the effort torque equals the load torque (

F_E \times d_E = F_L \times d_L

). Substituting the known values yields

F_E \times 90 = 600 \times 30

. Solving for

F_E

gives

200 \text{ N}

.

Step-by-Step Solution

1

Identify the lever class and applicable formula.

This is a first-class lever. The principle of moments states

F_E \times d_E = F_L \times d_L

.

To set up the torque equilibrium equation required to find the unknown force.

2

Determine the length of the load arm (

d_L

).

d_L = 30 \text{ cm}

The load arm is defined as the distance from the fulcrum to the load.

3

Calculate the length of the effort arm (

d_E

).

d_E = 120 \text{ cm} - 30 \text{ cm} = 90 \text{ cm}

The effort arm is the distance from the fulcrum to the point where the effort force is applied. Since the force is at the far opposite end, we subtract the load arm from the total length.

4

Calculate the required effort force (

F_E

).

F_E \times 90 = 600 \times 30 \Rightarrow 90 F_E = 18000 \Rightarrow F_E = 200 \text{ N}

Solving the equilibrium equation for the unknown variable.

Key Concept

First-class lever mechanical advantage and torque balance

Question 160Question →

Match each lever arm configuration or adjustment with its corresponding effect on mechanical advantage (MA) and force requirements.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The configurations match their mathematical outcomes based directly on the Mechanical Advantage formula: MA = Effort Arm / Load Arm.

Mechanical advantage is governed by the ratio of the effort arm to the load arm. An MA greater than 1 multiplies the force, an MA less than 1 multiplies distance at the cost of force, an MA of 1 only changes direction, and dynamically moving the fulcrum away from the load directly decreases the effort arm, thereby reducing the mechanical advantage.

Step-by-Step Solution

1

Analyze the configuration where the effort arm is twice the load arm.

MA = 2 / 1 = 2.

Mechanical advantage is calculated by dividing the length of the effort arm by the length of the load arm.

2

Analyze the configuration where the effort arm is half the load arm.

MA = 0.5 / 1 = 0.5.

A smaller effort arm results in an MA less than 1, requiring more force but trading it for increased distance.

3

Analyze the configuration with equal arm lengths.

MA = 1 / 1 = 1.

Equal arms mean the force magnitude doesn't change, only the direction of the application.

4

Analyze the effect of shifting the fulcrum away from the load.

The effort arm decreases while the load arm increases, reducing the overall MA.

Moving the pivot point closer to the effort force always lowers the mechanical advantage ratio.

Key Concept

Mechanical Advantage (MA) of Levers

Simple Machines

Answer

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