Levers and Fulcrums Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 3

Match each of the following common mechanical tools to the correct description of its lever configuration.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Pliers are matched to a first-class lever (fulcrum in the middle). The nutcracker is matched to a second-class lever (load in the middle). The tongs are matched to a third-class lever (effort in the middle).

Each tool physically represents a specific lever class based on the arrangement of the fulcrum (pivot), the effort (input force), and the load (output resistance). By tracing the pivot point, where the user applies force, and where the work is done, each tool accurately maps directly to a specific lever configuration.

Step-by-Step Solution

1

Analyze the mechanical structure of a pair of pliers.

Identify that the pivot point is in the center, between the handles and the jaws.

Locating the fulcrum first helps classify the lever.

2

Analyze the structure of a nutcracker.

Identify that the hinge is at one end and the nut sits in the middle.

This establishes that the load is between the fulcrum and the applied effort.

3

Analyze the structure of a pair of tongs.

Identify that they pivot at the top hinge, but the user's hand squeezes the middle.

This establishes that the effort is applied between the fulcrum and the load.

Key Concept

Classification of Simple Levers based on Fulcrum, Load, and Effort arrangement

Question 42Question →

A shipyard rigger needs to lift a $200\text{-pound}$ machinery casing. The rigger places one end of a rigid $60\text{-inch}$ steel pry bar under the casing and positions a fulcrum beneath the bar, creating a first-class lever. If the rigger applies a downward effort force of $50\text{ pounds}$ at the opposite end of the bar, how far from the casing must the fulcrum be placed to perfectly balance the load?

15\text{ inches}

12\text{ inches}

20\text{ inches}

48\text{ inches}

Show answer & explanation

Answer: $12\text{ inches}$

Answer

12\text{ inches}

To balance a first-class lever, the torque generated by the load must exactly equal the torque generated by the effort force. Setting the fulcrum

12\text{ inches}

from the

200\text{-pound}

load leaves

48\text{ inches}

of the bar for the effort arm (

60 - 12 = 48

). The torque on the load side is

200\text{ lbs} \times 12\text{ inches} = 2400\text{ inch-pounds}

, which perfectly matches the effort torque of

50\text{ lbs} \times 48\text{ inches} = 2400\text{ inch-pounds}

.

Step-by-Step Solution

1

Set up the torque balance equation for a first-class lever.

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

For the lever to balance, the rotational force (torque) applied by the load must equal the torque applied by the effort.

2

Define the lever arms in terms of the unknown distance from the load to the fulcrum (

x

).

d_{\text{load}} = x

d_{\text{effort}} = 60 - x

The fulcrum splits the total

60\text{-inch}

bar into two segments. If one side is

x

, the remaining length is

60 - x

.

3

Substitute the known values into the equation and solve for

x

.

200 \cdot x = 50 \cdot (60 - x)

200x = 3000 - 50x

250x = 3000

x = 12\text{ inches}

Using basic algebra isolates

x

, revealing the exact fulcrum placement required for balance.

Key Concept

Torque balance and lever arms in a first-class lever

Question 43Question →

A painter uses a flathead screwdriver to pry open a tightly sealed paint can. The rim of the can acts as the fulcrum. The tip of the screwdriver is inserted $2\text{ inches}$ under the lid from the fulcrum, while the handle extends $8\text{ inches}$ in the opposite direction from the fulcrum. If opening the lid requires an upward force of $40\text{ pounds}$ , how much downward effort force must the painter apply to the handle?

8 pounds

40 pounds

10 pounds

160 pounds

Show answer & explanation

Answer: 10 pounds

Answer

10 pounds

The screwdriver is functioning as a first-class lever. To find the required effort, the torque of the load must be balanced by the torque of the effort. The load torque is

40\text{ pounds} \times 2\text{ inches} = 80\text{ inch-pounds}

. Dividing this torque by the effort arm of

8\text{ inches}

yields an effort force of

10\text{ pounds}

.

Step-by-Step Solution

1

Identify the lever class and the lengths of the load arm and effort arm.

The screwdriver acts as a first-class lever. The load arm (distance from the fulcrum to the tip) is 2 inches. The effort arm (distance from the fulcrum to the handle) is 8 inches.

Accurate arm lengths from the fulcrum are required to calculate the torque balance.

2

Apply the principle of moments (torque balance equation).

Effort Force × Effort Arm = Load Force × Load Arm

For the lever to move the load, the torque applied by the effort must equal the torque of the load.

3

Substitute the known values into the equation and solve for the Effort Force.

Effort Force × 8 = 40 × 2. Effort Force × 8 = 80. Effort Force = 10 pounds.

Solving the equation provides the exact downward force needed on the handle.

Key Concept

In a first-class lever, the mechanical advantage is determined by the ratio of the effort arm to the load arm. A longer effort arm reduces the force required to lift a load.

Estimated Time:45s

Question 44Question →

Match each mechanical lever configuration to its corresponding functional outcome and mechanical advantage (MA).

Click a left item, then click its matching right item

Show answer & explanation

Answer

The configurations map as follows: Effort in middle is a distance multiplier (MA < 1); Load in middle is a force multiplier (MA > 1) same direction; Fulcrum in middle with long effort arm is a force multiplier (MA > 1) reversed direction; Fulcrum exact middle is a directional change only (MA = 1).

Each lever class uniquely dictates whether the force direction is changed, while the placement of the fulcrum relative to the effort and load dictates the mechanical advantage. First-class levers reverse direction, second-class multiply force in the same direction, and third-class multiply distance in the same direction.

Step-by-Step Solution

1

Identify the class of each lever configuration based on the middle component.

Fulcrum in middle = 1st Class. Load in middle = 2nd Class. Effort in middle = 3rd Class.

The central component dictates the fundamental behavior and force direction of the lever.

2

Determine the force direction for each class.

First-class levers reverse the direction of force. Second and third-class levers maintain the original direction of force.

Only when the fulcrum acts as a pivot between the input and output (1st class) does the motion arc in opposite directions.

3

Evaluate the mechanical advantage (MA) based on the relative lengths of the effort arm and load arm.

Effort arm > Load arm yields MA > 1 (force multiplier). Effort arm < Load arm yields MA < 1 (distance multiplier). Equal arms yield MA = 1.

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

4

Synthesize the direction and MA rules to match each configuration to its functional outcome.

Matches are made based on combining the lever class rules with the specific arm length constraints given.

This guarantees an accurate mapping of physical setup to mechanical performance.

Key Concept

Lever Classes and Mechanical Advantage

Question 45Question →

A construction worker uses a rigid pry bar to lift a concrete slab that weighs $400\text{ lbs}$ . The fulcrum is placed exactly $1\text{ foot}$ away from the point where the pry bar contacts the slab. If the worker applies a downward force exactly $5\text{ feet}$ away from the fulcrum, how many pounds of effort force must be applied to just balance the slab?

Show answer & explanation

Answer: 80

Answer

80 lbs

To balance the lever, the torque on both sides of the fulcrum must be equal. The load torque is generated by the slab:

400\text{ lbs} \times 1\text{ ft} = 400\text{ ft-lbs}

. To create an equal effort torque on the other side at a distance of

5\text{ feet}

, an effort force of

80\text{ lbs}

is required because

80\text{ lbs} \times 5\text{ ft} = 400\text{ ft-lbs}

.

Step-by-Step Solution

1

Identify the known variables given in the problem statement.

Load Force (

F_L

) =

400\text{ lbs}

, Load Distance (

d_L

) =

1\text{ ft}

, Effort Distance (

d_E

) =

5\text{ ft}

.

These values are required to calculate the unknown effort force using the law of the lever.

2

Apply the law of the lever (torque balance equation).

F_E \times d_E = F_L \times d_L

A lever is in static equilibrium when the input torque exactly equals the output torque.

3

Substitute the known values and solve for the effort force (

F_E

).

F_E \times 5 = 400 \times 1 \rightarrow 5F_E = 400 \rightarrow F_E = 80

Isolating

F_E

provides the exact pounds of force needed to balance the concrete slab.

Key Concept

Torque Balance and the Law of the Lever

Estimated Time:45s

Question 46Question →

A carpenter uses a claw hammer to pull a stubborn nail out of a wooden board. The hammer's head rests on the surface of the board while the carpenter pulls back on the handle. In this lever system, which component serves as the fulcrum?

The point where the hammer head contacts the wooden board

The hand applying force to the hammer handle

The nail being gripped and pulled by the claw

The point exactly halfway down the length of the handle

Show answer & explanation

Answer: The point where the hammer head contacts the wooden board

Answer

The correct answer is the point where the hammer head contacts the wooden board.

The fulcrum is defined as the fixed pivot point around which a lever moves. When using a claw hammer to pull a nail, the point where the curved head of the hammer rests against the board acts as this stationary pivot point. It allows the lever to multiply the force applied at the handle to overcome the resistance of the nail.

Step-by-Step Solution

1

Define the role of a fulcrum in any lever system.

A fulcrum is the stationary point or hinge around which a lever pivots or rotates.

Understanding the mechanical definition is required before analyzing the specific tool.

2

Analyze the physical action of the claw hammer pulling the nail.

As the carpenter pulls the handle (effort) to lift the nail (load), the entire hammer rotates around the point where the curved head presses into the wood.

Mapping the definitions of effort, load, and fulcrum to the real-world scenario identifies the correct component.

Key Concept

Lever Components: Identifying the Fulcrum, Effort, and Load

Estimated Time:30s

Question 47Question →

Pair the following everyday tools with their correct mechanical lever classification.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Scissors match to a First-Class Lever, the paper cutter matches to a Second-Class Lever, and the broom matches to a Third-Class Lever.

Lever classes are defined by which component is located in the middle: the Fulcrum (Class 1), the Load (Class 2), or the Effort (Class 3). The mnemonic 'FLE 123' is commonly used to remember this rule.

Step-by-Step Solution

1

Analyze the mechanical arrangement of the scissors.

The fulcrum is between the effort and the load, making it a Class 1 lever.

To classify the lever based on its central component.

2

Analyze the mechanical arrangement of the paper cutter.

The load is positioned between the fulcrum and the effort, making it a Class 2 lever.

To classify the lever based on its central component.

3

Analyze the mechanical arrangement of the broom.

The effort is applied between the fulcrum and the load, making it a Class 3 lever.

To classify the lever based on its central component.

Key Concept

Lever Classification by Central Component

Estimated Time:45s

Question 48Question →

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 49Question →

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 50Question →

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 51Question →

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 52Question →

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 53Question →

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 54Question →

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 55Question →

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 56Question →

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 57Question →

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

Question 58Question →

A specialized clamping arm on a manufacturing machine operates as a straight lever, hinged at one end to a fixed support. A pneumatic actuator is attached to the arm $0.4 \text{ m}$ away from the hinge, applying a downward effort force. The arm presses down onto a workpiece located $2.0 \text{ m}$ from the hinge. The workpiece exerts an upward resistance of $400 \text{ N}$ against the arm.

Assuming the arm is level and its weight is negligible, what downward force must the actuator apply to hold the arm steady, and what class of lever does this mechanism represent?

2000 \text{ N}

; Third-class lever

2000 \text{ N}

; Second-class lever

1600 \text{ N}

; Third-class lever

80 \text{ N}

; Third-class lever

Show answer & explanation

Answer: $2000 \text{ N}$ ; Third-class lever

Answer

The actuator must apply

2000 \text{ N}

of force, and the mechanism represents a third-class lever.

The correct answer accurately classifies the setup as a third-class lever because the effort (actuator) is located between the fulcrum (hinge) and the load (workpiece). To maintain equilibrium, the moments around the hinge must balance. The load creates a moment of

400 \text{ N} \times 2.0 \text{ m} = 800 \text{ N}\cdot\text{m}

. The actuator must create an equal moment:

F_e \times 0.4 \text{ m} = 800 \text{ N}\cdot\text{m}

. Solving for

F_e

gives

2000 \text{ N}

.

Step-by-Step Solution

1

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

Third-class lever.

The hinge is the fulcrum at one end (0.0 m), the actuator is the effort (0.4 m), and the workpiece is the load (2.0 m). Because the effort is applied between the fulcrum and the load, it is a third-class lever.

2

Set up the torque balance equation (Principle of Moments).

F_e \times 0.4 \text{ m} = 400 \text{ N} \times 2.0 \text{ m}

For the lever to be held steady, the torque generated by the effort force must perfectly balance the torque generated by the load resistance around the fulcrum.

3

Solve the equation for the unknown effort force (

F_e

).

F_e = 2000 \text{ N}

Evaluating

(400 \times 2.0) / 0.4

yields

800 / 0.4 = 2000 \text{ N}

.

Key Concept

Lever classifications and torque balance in simple machines

Estimated Time:1m 0s

Question 59Question →

A technician uses a pair of long industrial tweezers to securely hold a sensitive electronic component. The tweezers pivot at the joined top end and have a total length of $40 \text{ cm}$ from the pivot to the gripping tips. The technician's fingers apply pressure at a point $8 \text{ cm}$ from the pivot. If the component requires a holding force of $50 \text{ N}$ at the tips, what effort force (in Newtons) must the technician apply?

Show answer & explanation

Answer: 250

Answer

The technician must apply an effort force of 250 N.

In a third-class lever like tweezers, the effort is applied between the fulcrum and the load. Since the effort arm (

8 \text{ cm}

) is shorter than the load arm (

40 \text{ cm}

), the mechanical advantage is less than 1. This means a much larger effort force is required to generate a smaller load force at the tips. Using the torque balance equation

F_e \times d_e = F_l \times d_l

, we get

F_e \times 8 = 50 \times 40

. Solving this gives

250 \text{ N}

.

Step-by-Step Solution

1

Identify the lever configuration and its parameters.

The tweezers are a third-class lever (fulcrum at one end, effort in the middle). Effort arm

d_e = 8 \text{ cm}

, load arm

d_l = 40 \text{ cm}

, load force

F_l = 50 \text{ N}

.

Correctly identifying the lever class ensures the correct distances are used relative to the fulcrum.

2

Set up the torque equilibrium equation.

F_e \times d_e = F_l \times d_l

For the lever to remain stable and provide the required grip, the effort torque must balance the load torque.

3

Calculate the required effort force.

F_e \times 8 = 50 \times 40 \Rightarrow 8 F_e = 2000 \Rightarrow F_e = 250 \text{ N}

To find the unknown variable by substituting the given values.

Key Concept

Torque Balance in Third-Class Levers

Question 60Question →

A maintenance crew uses a horizontal boom mounted on a utility truck to lift a $600 \text{ N}$ generator. The boom pivots at a hinge on the truck bed. A vertical hydraulic cylinder supports the boom, attaching exactly $1.0 \text{ m}$ from the hinge. The generator hangs from the far end of the boom, $4.0 \text{ m}$ from the hinge. Assuming the weight of the boom itself is negligible, how much upward force must the hydraulic cylinder apply to hold the generator stationary, and what class of lever does this system represent?

2400 \text{ N}

; third-class lever

150 \text{ N}

; second-class lever

1800 \text{ N}

; third-class lever

2400 \text{ N}

; first-class lever

Show answer & explanation

Answer: $2400 \text{ N}$ ; third-class lever

Answer

2400 \text{ N}

; third-class lever

The system represents a third-class lever because the effort force (the hydraulic cylinder) is applied between the fulcrum (the hinge) and the load (the generator). To maintain static equilibrium, the counterclockwise torque provided by the cylinder must perfectly equal the clockwise torque from the generator. The generator creates a torque of

600 \text{ N} \times 4.0 \text{ m} = 2400 \text{ N}\cdot\text{m}

. Therefore, the cylinder must apply

2400 \text{ N}

of force at a distance of

1.0 \text{ m}

to balance the system (

2400 \text{ N} \times 1.0 \text{ m} = 2400 \text{ N}\cdot\text{m}

).

Step-by-Step Solution

1

Identify the lever class based on the arrangement of the fulcrum, effort, and load.

The effort force (hydraulic cylinder) is located between the fulcrum (hinge) and the load (generator). This arrangement defines a third-class lever.

Determining the lever arrangement is essential to properly visualize the system and answers the classification part of the question.

2

Set up the torque balance equation (principle of moments) around the fulcrum.

To remain stationary, upward torque must equal downward torque:

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

A lever in static equilibrium must have balanced rotational forces acting around its pivot point.

3

Substitute the known distances and forces into the equation to solve for the effort force.

F_{\text{effort}} \times 1.0 \text{ m} = 600 \text{ N} \times 4.0 \text{ m}

, which simplifies to

F_{\text{effort}} = 2400 \text{ N}

.

Applying the specific measurements yields the precise upward force required by the cylinder.

Key Concept

Third-Class Levers and Torque Balance

Estimated Time:1m 15s

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