Levers and Fulcrums Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 2

Question 21Question →

A technician is testing a mechanical linkage that uses a straight lever pinned at one end, which serves as the fulcrum. A control rod is attached to the lever at a distance of $4\text{ inches}$ from the fulcrum and creates a resistance of $150\text{ lbs}$ . An actuator applies an effort force at the free end of the lever, located $12\text{ inches}$ from the fulcrum. If all forces are applied perpendicular to the lever, what effort force (in pounds) is required to perfectly balance the control rod's resistance?

Show answer & explanation

Answer: 50

Answer

50

In a balanced lever system, the torque produced by the effort force must equal the torque produced by the resistance force. Torque is calculated by multiplying the force by its distance from the fulcrum. Therefore,

F_E \times 12 = 150 \times 4

, which simplifies to

12F_E = 600

, resulting in an effort force of

50\text{ lbs}

.

Step-by-Step Solution

1

Identify the known variables and their distances from the fulcrum.

Resistance Force (

F_R

) =

150\text{ lbs}

, Resistance Arm (

d_R

) =

4\text{ inches}

, Effort Arm (

d_E

) =

12\text{ inches}

.

Correctly measuring distances from the pivot point (fulcrum) is necessary to calculate torque.

2

Set up the torque balance equation for the lever.

F_E \times d_E = F_R \times d_R

For the lever system to be in equilibrium, the opposing torques must be equal.

3

Substitute the values and calculate the required effort force.

F_E \times 12 = 150 \times 4 \Rightarrow 12F_E = 600 \Rightarrow F_E = 50\text{ lbs}

.

Solving the algebraic equation provides the necessary balancing force.

Key Concept

Torque balance and lever mechanical advantage

Alternative Method

Calculate the mechanical advantage (MA) first. MA = Effort Arm

\div

Resistance Arm =

12 \div 4 = 3

. Since the lever provides a mechanical advantage of

3

, the required effort force is exactly one-third of the resistance force (

150 \div 3 = 50\text{ lbs}

).

Estimated Time:1m 0s

Question 22Question →

An automotive technician is inspecting a suspended brake pedal assembly in a vehicle. The pedal pivots on a stationary pin located at the very top. The master cylinder pushrod (which provides the load resistance) is connected to the pedal $3\text{ inches}$ directly below the pivot pin. The center of the driver's foot pad (where the effort force is applied) is located $15\text{ inches}$ directly below the pivot pin. If a driver applies exactly $60\text{ lbs}$ of horizontal force to the foot pad, what is the resulting horizontal force applied to the pushrod, and what class of lever does this assembly represent?

300\text{ lbs}

; Second-class lever

300\text{ lbs}

; Third-class lever

240\text{ lbs}

; First-class lever

12\text{ lbs}

; Third-class lever

Show answer & explanation

Answer: $300\text{ lbs}$ ; Second-class lever

Answer

The assembly represents a second-class lever, and the resulting force is

300\text{ lbs}

.

The brake pedal assembly features a fulcrum at the top pin, the load connection in the middle, and the effort application at the bottom. The Fulcrum-Load-Effort (F-L-E) arrangement defines a second-class lever. To find the output force, we use the law of levers (Torque = Force × Distance from fulcrum). The input torque is

60\text{ lbs} \times 15\text{ inches} = 900\text{ in-lbs}

. To maintain equilibrium, the output torque must also be

900\text{ in-lbs}

. Dividing this by the load distance of

3\text{ inches}

yields an output force of

300\text{ lbs}

.

Step-by-Step Solution

1

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

The pivot (fulcrum) is at one end, the pushrod (load) is in the middle, and the foot pad (effort) is at the other end. This F-L-E configuration is a second-class lever.

Correctly classifying the lever establishes the proper distances for the torque equation.

2

Set up the torque balance equation for the lever:

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

.

60\text{ lbs} \times 15\text{ inches} = F_{load} \times 3\text{ inches}

.

In equilibrium, the torque input from the effort must equal the torque output against the load.

3

Solve the algebraic equation for the unknown load force (

F_{load}

).

900 = 3 \times F_{load}

, which simplifies to

F_{load} = 300\text{ lbs}

.

Isolating the variable gives the exact force applied to the master cylinder pushrod.

Key Concept

Torque Balance and Lever Classification

Question 23Question →

A $60\text{-pound}$ crate is placed on one end of a first-class lever, exactly $4\text{ feet}$ from the fulcrum. What downward effort force must be applied at a distance of $8\text{ feet}$ on the opposite side of the fulcrum to perfectly balance the lever?

Show answer & explanation

Answer: 30

Answer

30\text{ lbs}

A lever is perfectly balanced when the turning force (torque) on one side of the fulcrum equals the torque on the other side. Torque is calculated by multiplying the force by its distance from the fulcrum. The crate creates a torque of

240\text{ ft-lbs}

(

60\text{ lbs} \times 4\text{ ft}

). Because the effort is being applied twice as far from the fulcrum (

8\text{ ft}

), it only needs to be half the force to generate the same

240\text{ ft-lbs}

of torque. Therefore,

240 \div 8 = 30\text{ lbs}

.

Step-by-Step Solution

1

Set up the torque balance equation.

F_1 \times d_1 = F_2 \times d_2

For a lever to be balanced, the torque (force times distance) on the left side must equal the torque on the right side.

2

Substitute the known values into the equation.

60 \times 4 = F_2 \times 8

The crate applies a

60\text{ lb}

force at a distance of

4\text{ ft}

, and the effort force is applied at a distance of

8\text{ ft}

.

3

Solve for the unknown effort force (

F_2

).

240 = 8 \times F_2 \rightarrow F_2 = 30

Dividing both sides of the equation by

8

isolates

F_2

, revealing the required balancing force.

Key Concept

Torque equilibrium and mechanical advantage in a first-class lever

Estimated Time:45s

Question 24Question →

A mechanic is using a rigid steel bar as a lever to lift a heavy engine block. To require the LEAST amount of downward effort force, where should the fulcrum be placed?

As close to the engine block as possible

Exactly halfway between the mechanic and the engine block

As close to the mechanic's hands as possible

The position of the fulcrum does not affect the required effort force

Show answer & explanation

Answer: As close to the engine block as possible

Answer

The fulcrum should be placed as close to the engine block as possible.

To minimize the effort force needed to lift a load using a lever, the mechanical advantage must be maximized. This is achieved by making the effort arm (the distance from where the mechanic pushes to the fulcrum) as long as possible, and the load arm (the distance from the engine block to the fulcrum) as short as possible. Placing the pivot point as close to the engine block as possible accomplishes this.

Step-by-Step Solution

1

Identify the relationship between the fulcrum position and mechanical advantage.

Moving the fulcrum closer to the load increases the length of the effort arm relative to the length of the load arm.

Mechanical advantage in a lever depends entirely on the ratio of the effort arm to the load arm.

2

Determine which position provides the greatest mechanical advantage to minimize force.

Placing the fulcrum right next to the engine block (the load) maximizes the effort arm and minimizes the load arm.

A larger effort arm requires less physical effort force from the user to lift the same heavy load.

Key Concept

Mechanical advantage and fulcrum placement in first-class levers

Question 25Question →

A landscaper is using a wheelbarrow to move a $120\text{-pound}$ rock. The rock's center of gravity is positioned exactly $2\text{ feet}$ behind the wheel's axle, which acts as the fulcrum. The handles where the landscaper applies upward force are located $6\text{ feet}$ behind the wheel's axle. What upward lifting force must be applied at the handles to hold the wheelbarrow level?

20\text{ pounds}

40\text{ pounds}

60\text{ pounds}

360\text{ pounds}

Show answer & explanation

Answer: $40\text{ pounds}$

Answer

The correct lifting force is

40\text{ pounds}

.

The correct answer is derived by balancing the torques around the fulcrum. The rock exerts a downward torque of

120\text{ pounds} \times 2\text{ feet} = 240\text{ foot-pounds}

. To keep the wheelbarrow level, the landscaper must exert an equal and opposite torque. By dividing the required torque (

240\text{ foot-pounds}

) by the distance of the handles from the fulcrum (

6\text{ feet}

), we find that

40\text{ pounds}

of upward force is required.

Step-by-Step Solution

1

Identify the type of simple machine and the appropriate formula.

A wheelbarrow is a second-class lever. The principle of torque balance applies:

\text{Effort Force} \times \text{Effort Arm} = \text{Load Force} \times \text{Load Arm}

.

To find the unknown force, we must set the rotational forces (torques) equal to each other.

2

Determine the values for the load, load arm, and effort arm from the problem.

\text{Load Force} = 120\text{ pounds}

,

\text{Load Arm} = 2\text{ feet}

, and

\text{Effort Arm} = 6\text{ feet}

.

Correctly identifying distances relative to the fulcrum (the wheel's axle) is critical for calculating torque.

3

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

\text{Effort Force} \times 6 = 120 \times 2

. Simplifying gives

6 \times \text{Effort Force} = 240

. Dividing both sides by

6

gives

\text{Effort Force} = 40\text{ pounds}

.

This mathematically isolates the unknown variable to find the exact force required to balance the load.

Key Concept

Torque balance in a second-class lever dictates that the mechanical advantage is the ratio of the effort arm to the load arm.

Question 26Question →

Match each everyday tool to its correct lever classification based on the relative positions of the fulcrum, effort, and load.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Scissors match with First-class lever, a nutcracker matches with Second-class lever, and a swinging baseball bat matches with Third-class lever.

Each tool can be classified by identifying the component in the middle: Scissors have the fulcrum in the middle (Class 1), nutcrackers have the load in the middle (Class 2), and baseball bats have the effort applied in the middle (Class 3).

Step-by-Step Solution

1

Analyze a pair of scissors

The fulcrum is in the middle, making it a first-class lever.

First-class levers always have the fulcrum positioned between the effort and the load.

2

Analyze a nutcracker

The load is in the middle, making it a second-class lever.

Second-class levers always have the load positioned between the fulcrum and the effort.

3

Analyze a swinging baseball bat

The effort is in the middle, making it a third-class lever.

Third-class levers always have the effort positioned between the fulcrum and the load.

Key Concept

Lever classes are defined by the component that is positioned in the middle: Fulcrum (1st class), Load (2nd class), or Effort (3rd class).

Estimated Time:45s

Question 27Question →

A person uses a straight-handled snow shovel to lift a $20\text{-pound}$ load of snow. The person's top hand acts as a stationary pivot (fulcrum) at the very end of the handle. Their lower hand applies an upward lifting effort $10\text{ inches}$ down the shaft from the top hand. If the center of the snow load is located $40\text{ inches}$ further down the shaft from the lower hand, what is the upward lifting force required by the lower hand to hold the shovel horizontally?

4\text{ lbs}

20\text{ lbs}

80\text{ lbs}

100\text{ lbs}

Show answer & explanation

Answer: $100\text{ lbs}$

Answer

100\text{ lbs}

In a third-class lever, the effort is located between the fulcrum and the load. The distance from the fulcrum to the effort is

10\text{ inches}

, and the distance from the fulcrum to the load is

10 + 40 = 50\text{ inches}

. Setting the torques equal gives

F_E \times 10 = 20 \times 50

. Solving for

F_E

results in

100\text{ lbs}

.

Step-by-Step Solution

1

Identify the lever class and the locations of the fulcrum, effort, and load.

The shovel acts as a third-class lever. Fulcrum is at the top hand, effort is applied at the lower hand, and load is at the snow.

This establishes the pivot point for calculating torque (moments).

2

Determine the effort arm (

d_E

) and the load arm (

d_L

).

d_E = 10\text{ inches}

. The load arm

d_L

is the total distance from the fulcrum to the load:

10\text{ inches} + 40\text{ inches} = 50\text{ inches}

.

Torque is always calculated as force multiplied by the distance from the fulcrum.

3

Apply the principle of moments (Torque equilibrium) to find the effort force (

F_E

).

F_E \times d_E = F_L \times d_L \Rightarrow F_E \times 10 = 20 \times 50 \Rightarrow F_E \times 10 = 1000 \Rightarrow F_E = 100\text{ lbs}

.

For the shovel to be held horizontally in equilibrium, the upward torque from the effort must equal the downward torque from the load.

Key Concept

Third-Class Levers and Torque Balance

Estimated Time:1m 0s

Question 28Question →

Match each lever configuration with its corresponding mechanical advantage and force direction properties.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Second-class levers multiply force without reversing direction. Third-class levers multiply distance without reversing direction. First-class levers with the fulcrum near the load multiply force and reverse direction. First-class levers with a centered fulcrum only reverse direction.

The correct pairings rely on evaluating two factors: whether the fulcrum sits between the effort and load (which determines if force direction is reversed), and the relative lengths of the effort and load arms (which determines if the lever multiplies force or distance).

Step-by-Step Solution

1

Determine if the lever configuration reverses the direction of the applied force.

First-class levers reverse the direction of force. Second and third-class levers do not reverse the direction.

The fulcrum's position relative to the effort and load determines directional changes.

2

Compare the length of the effort arm to the load arm for each configuration.

Second-class levers and offset first-class levers have longer effort arms. Third-class levers have shorter effort arms. Centered first-class levers have equal arms.

The ratio of the effort arm to the load arm dictates the mechanical advantage (MA).

3

Match the mechanical advantage outcome to the lever configuration.

Longer effort arm = MA > 1 (multiplies force). Shorter effort arm = MA < 1 (multiplies distance/speed). Equal arms = MA of 1.

This establishes the direct relationship between physical lever structure and mechanical properties.

Key Concept

Mechanical Advantage of Lever Configurations

Question 29Question →

A worker uses an industrial paper cutter to slice through thick cardboard. The cutter operates as a second-class lever with the pivot (fulcrum) at one end of the blade. The cardboard is positioned $8\text{ inches}$ from the pivot and creates a cutting resistance of $120\text{ lbs}$ . If the worker pushes down on the handle at a distance of $40\text{ inches}$ from the pivot, what is the minimum downward effort force required to cut the cardboard?

24\text{ lbs}

30\text{ lbs}

20\text{ lbs}

600\text{ lbs}

Show answer & explanation

Answer: $24\text{ lbs}$

Answer

The minimum downward effort force required is

24\text{ lbs}

.

In a second-class lever, the effort torque must equal the load torque. By multiplying the cutting resistance (

120\text{ lbs}

) by its distance from the fulcrum (

8\text{ inches}

), we find a load torque of

960\text{ inch-lbs}

. Dividing this torque by the total effort distance from the fulcrum (

40\text{ inches}

) gives a minimum effort force of

24\text{ lbs}

.

Step-by-Step Solution

1

Identify the lever class and the given variables.

This is a second-class lever. Load (

F_L

) =

120\text{ lbs}

, Load arm (

d_L

) =

8\text{ inches}

, Effort arm (

d_E

) =

40\text{ inches}

.

Correctly identifying the lever class ensures that the distances are properly measured from the fulcrum.

2

Apply the torque balance (principle of moments) equation.

F_E \times d_E = F_L \times d_L

For the lever to balance and cut the cardboard, the input effort torque must equal the output load torque.

3

Substitute the known values and solve for the effort force (

F_E

).

F_E \times 40 = 120 \times 8 \rightarrow 40 F_E = 960 \rightarrow F_E = 24\text{ lbs}

.

Calculating this equation yields the minimum effort force required.

Key Concept

Torque Balance in Second-Class Levers

Estimated Time:1m 0s

Question 30Question →

A warehouse worker uses a first-class lever to lift a machinery part weighing $200\text{ lbs}$ . The fulcrum is placed exactly $2\text{ feet}$ from the machinery part. If the worker applies a downward effort force at a distance of $8\text{ feet}$ from the fulcrum on the opposite side, how much effort force is required to balance the load? (Assume the lever itself is weightless.)

Show answer & explanation

Answer: 50

Answer

50

A first-class lever is balanced when the torque on one side of the fulcrum equals the torque on the other side. Torque is calculated by multiplying the force by its distance from the fulcrum. The machinery part creates a torque of

200\text{ lbs} \times 2\text{ ft} = 400\text{ ft-lbs}

. To balance this, the effort force applied at

8\text{ ft}

must also create

400\text{ ft-lbs}

of torque. Therefore, the required effort force is

\frac{400}{8} = 50\text{ lbs}

.

Step-by-Step Solution

1

Set up the torque balance equation for a lever in equilibrium.

\text{Load} \times \text{Load Arm} = \text{Effort Force} \times \text{Effort Arm}

For a lever to balance, the rotational forces (torques) on both sides of the fulcrum must be equal.

2

Substitute the given values into the equation.

200 \times 2 = \text{Effort Force} \times 8

The load is

200\text{ lbs}

located

2\text{ ft}

from the fulcrum, and the effort is applied

8\text{ ft}

from the fulcrum.

3

Calculate the unknown effort force.

400 = \text{Effort Force} \times 8 \implies \text{Effort Force} = \frac{400}{8} = 50

Dividing the total load torque by the effort arm distance gives the required effort force.

Key Concept

The principle of moments (torque balance) dictates that a lever is in equilibrium when the clockwise torque equals the counterclockwise torque.

Question 31Question →

A carpenter uses a claw hammer to extract a stubborn nail from a wooden board. The hammer acts as a first-class lever. The distance from the nail to the pivot point (fulcrum) on the hammer's head is $2\text{ inches}$ . The carpenter applies an effort force perpendicular to the handle at a distance of $16\text{ inches}$ from the pivot point. If a resistance force of $280\text{ pounds}$ is required to extract the nail, how many pounds of effort force must the carpenter apply?

Show answer & explanation

Answer: 35

Answer

35 pounds

In a first-class lever, the mechanical advantage is determined by the ratio of the effort arm to the resistance arm. The hammer provides a mechanical advantage of

16 / 2 = 8

. Therefore, the effort force required is the load divided by 8, which is

280 / 8 = 35\text{ pounds}

. Using the fundamental torque balance equation (

F_1 \times d_1 = F_2 \times d_2

),

280 \times 2 = 560

, and

560 / 16 = 35

.

Step-by-Step Solution

1

State the torque balance equation for a lever.

F_E \times d_E = F_R \times d_R

For the lever to overcome the resistance, the torque applied by the effort must equal the torque exerted by the load.

2

Substitute the given values into the equation.

F_E \times 16 = 280 \times 2

The effort distance (

d_E

) is

16\text{ inches}

, the resistance force (

F_R

) is

280\text{ pounds}

, and the resistance distance (

d_R

) is

2\text{ inches}

.

3

Calculate the total torque of the resistance.

280 \times 2 = 560\text{ inch-pounds}

This represents the total rotational force required to loosen the nail.

4

Solve for the effort force (

F_E

).

F_E = 560 / 16 = 35\text{ pounds}

Dividing the total required torque by the longer effort arm gives the smaller effort force needed.

Key Concept

Torque Balance and Mechanical Advantage

Question 32Question →

A demolition worker uses a straight steel bar to pry up a concrete slab. The bar rests on a small block acting as a fulcrum. The distance from the fulcrum to the slab is $1\text{ foot}$ , and the distance from the fulcrum to the worker's hands is $4\text{ feet}$ . If the worker pushes down with a force of $50\text{ pounds}$ , what is the maximum weight of the slab that can be lifted?

12.5\text{ pounds}

50\text{ pounds}

250\text{ pounds}

200\text{ pounds}

Show answer & explanation

Answer: $200\text{ pounds}$

Answer

The maximum weight of the slab that can be lifted is

200\text{ pounds}

.

In a balanced lever system, the torque produced by the worker must equal the torque produced by the load. Torque is calculated as Force multiplied by Distance to the fulcrum. The worker generates an effort torque of

50\text{ pounds} \times 4\text{ feet} = 200\text{ foot-pounds}

. To maintain balance, the load torque must also be

200\text{ foot-pounds}

. Since the load is

1\text{ foot}

away from the fulcrum, the maximum weight of the slab is

200\text{ foot-pounds} / 1\text{ foot} = 200\text{ pounds}

.

Step-by-Step Solution

1

Determine the mechanical advantage of the lever system.

Mechanical Advantage =

4\text{ feet} / 1\text{ foot} = 4

.

The mechanical advantage of a first-class lever is found by dividing the length of the effort arm by the length of the load arm.

2

Calculate the maximum load weight.

50\text{ pounds} \times 4 = 200\text{ pounds}

.

To find the load that can be lifted, multiply the applied effort force by the mechanical advantage.

Key Concept

Mechanical Advantage and Torque Balance in First-Class Levers

Question 33Question →

A farmer uses a long, rigid pole to move a $180\text{-pound}$ boulder. The pole is positioned over a smaller rock that serves as a fulcrum. The fulcrum is placed exactly $2\text{ feet}$ away from the boulder. The farmer pushes down on the opposite end of the pole, which is $6\text{ feet}$ away from the fulcrum. What is the minimum downward force the farmer must apply to lift the boulder?

30\text{ pounds}

45\text{ pounds}

60\text{ pounds}

540\text{ pounds}

Show answer & explanation

Answer: $60\text{ pounds}$

Answer

The farmer must apply

60\text{ pounds}

of downward force.

The system acts as a first-class lever. To lift the boulder, the torque (force times distance) on the farmer's side of the fulcrum must match the torque on the boulder's side. The boulder creates a torque of

360\text{ ft-lbs}

(

180\text{ lbs} \times 2\text{ ft}

). To balance this over a

6\text{-foot}

distance, the required force is

360 / 6

, which equals

60\text{ pounds}

.

Step-by-Step Solution

1

Set up the equation for a lever in equilibrium based on the principle of moments.

\text{Effort Force} \times \text{Effort Distance} = \text{Load} \times \text{Load Distance}

For a lever to lift a load, the torque generated by the effort must equal the torque exerted by the load.

2

Substitute the provided values into the equilibrium equation.

\text{Effort Force} \times 6\text{ ft} = 180\text{ lbs} \times 2\text{ ft}

The load is

180\text{ lbs}

at a distance of

2\text{ ft}

, and the effort is applied at a distance of

6\text{ ft}

.

3

Solve the equation for the Effort Force.

\text{Effort Force} = \frac{360\text{ ft-lbs}}{6\text{ ft}} = 60\text{ lbs}

Dividing the total torque of the load by the effort distance gives the required downward force.

Key Concept

Torque and Mechanical Advantage in 1st Class Levers

Estimated Time:45s

Question 34Question →

A hinged lifting ramp is supported by a vertical hydraulic cylinder. The hinge acts as the fulcrum at one end of the ramp. The hydraulic cylinder applies an upward force exactly $3\text{ feet}$ from the hinge. A heavy crate weighing $600\text{ lbs}$ rests on the ramp exactly $12\text{ feet}$ from the hinge. Assuming the ramp is horizontal and ignoring its weight, how much upward force must the hydraulic cylinder exert to keep the ramp in equilibrium?

2400\text{ lbs}

1800\text{ lbs}

800\text{ lbs}

150\text{ lbs}

Show answer & explanation

Answer: $2400\text{ lbs}$

Answer

2400\text{ lbs}

In this third-class lever setup, the fulcrum is at one end (the hinge), the effort is applied in the middle, and the load is at the other end. To maintain equilibrium, the upward torque must equal the downward torque. The load torque is

600\text{ lbs} \times 12\text{ feet} = 7200\text{ ft-lbs}

. To find the required effort force, divide the total torque by the effort distance from the fulcrum (

3\text{ feet}

). Therefore,

7200 / 3 = 2400\text{ lbs}

.

Step-by-Step Solution

1

Identify the lever configuration and locate the fulcrum.

The hinge is the fulcrum. The effort is applied at

3\text{ feet}

from the fulcrum, and the load is at

12\text{ feet}

from the fulcrum. This makes it a third-class lever.

Correctly identifying the distances from the pivot point is necessary for torque calculations.

2

Calculate the torque exerted by the load.

600\text{ lbs} \times 12\text{ feet} = 7200\text{ ft-lbs}

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

3

Set up the equilibrium equation and solve for the required effort force.

F_{effort} \times 3\text{ feet} = 7200\text{ ft-lbs}

. Solving for

F_{effort}

yields

7200 / 3 = 2400\text{ lbs}

.

For the ramp to be in equilibrium, the upward torque provided by the hydraulic cylinder must equal the downward torque of the load.

Key Concept

Torque balance in third-class levers

Question 35Question →

Match each common tool or device to its correct lever classification based on the relative positions of the fulcrum, effort, and load.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Seesaw matches with First-class lever. Handheld bottle opener matches with Second-class lever. Tweezers match with Third-class lever.

The classification of a lever depends entirely on which component is located in the middle. By breaking down the physical action of each tool, we determine that a seesaw has its pivot in the center (first-class), a bottle opener lifts a load situated between the pivot and the handle (second-class), and tweezers are squeezed in the middle to grasp a load at the ends (third-class).

Step-by-Step Solution

1

Analyze the mechanics of a seesaw.

The fulcrum is in the center.

When the fulcrum is located between the effort and the load, it is a first-class lever.

2

Analyze the mechanics of a handheld bottle opener.

The load (bottle cap) is in the middle.

When the load is located between the fulcrum and the effort, it is a second-class lever.

3

Analyze the mechanics of tweezers.

The effort (fingers squeezing) is in the middle.

When the effort is applied between the fulcrum and the load, it is a third-class lever.

Key Concept

Lever Classifications (First, Second, and Third Class)

Question 36Question →

A mechanic presses a brake pedal with a force of $25\text{ pounds}$ . The pedal pivots on a hinge at the top. The mechanic's foot applies the force $16\text{ inches}$ from the pivot point. The brake's pushrod, which transfers the force to the master cylinder, is attached to the pedal $4\text{ inches}$ from the pivot point. Calculate the force, in pounds, applied to the pushrod.

Show answer & explanation

Answer: 100

Answer

100 pounds

The brake pedal functions as a second-class lever. To find the force transmitted to the pushrod, use the principle of moments where Effort Force × Effort Distance = Load Force × Load Distance. By substituting the known values, we get

25\text{ lbs} \times 16\text{ inches} = F_{\text{load}} \times 4\text{ inches}

. Multiplying

25

by

16

yields

400\text{ inch-pounds}

of torque. Dividing this torque by the load distance of

4\text{ inches}

reveals that a force of

100\text{ pounds}

is applied to the pushrod.

Step-by-Step Solution

1

Identify the known variables from the problem.

Effort Force = 25 lbs, Effort Distance = 16 in, Load Distance = 4 in.

Organizing the known values is the first step to applying the lever equation.

2

Apply the mechanical advantage principle for levers.

Effort Force × Effort Distance = Load Force × Load Distance

In a state of equilibrium, the torque applied by the effort must equal the torque applied to the load.

3

Substitute the known values into the equation.

25 × 16 = Load Force × 4

This sets up the equation to isolate the unknown load force.

4

Calculate the effort torque.

400 = Load Force × 4

Multiplying the effort force by its distance from the pivot gives the total torque in inch-pounds.

5

Solve for the final load force.

Load Force = 100

Dividing the total torque by the load's distance from the pivot gives the force applied to the pushrod.

Key Concept

Torque and Mechanical Advantage in Second-Class Levers

Question 37Question →

Match each lever classification or configuration to its correct mechanical characteristic regarding force and distance.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Third-class levers multiply distance; second-class levers multiply force; general first-class levers can do either depending on fulcrum position; and a centered first-class lever only changes force direction.

Each lever configuration strictly dictates the geometric relationship between the effort arm and the load arm. This relationship determines the mechanical advantage and specifies whether the simple machine multiplies force or multiplies distance.

Step-by-Step Solution

1

Analyze the structural properties of third-class levers.

The effort is applied between the fulcrum and the load.

Because the effort arm is shorter, more force is required to move the load, but the load moves a greater distance, yielding a mechanical advantage less than 1.

2

Analyze the structural properties of second-class levers.

The load is located between the fulcrum and the effort force.

Because the effort arm is longer, less force is required to move a heavier load, multiplying the input force with a mechanical advantage greater than 1.

3

Analyze the structural properties of general first-class levers.

The fulcrum is located anywhere between the effort and the load.

The mechanical advantage is entirely dependent on the specific placement of the fulcrum relative to the effort and load arms.

4

Analyze the specific case of a centered first-class lever.

The effort arm length is exactly equal to the load arm length.

This configuration provides a mechanical advantage of 1, meaning force magnitude is unchanged, but the physical direction of the input force is reversed.

Key Concept

Mechanical Advantage of Lever Classes

Estimated Time:1m 0s

Question 38Question →

A camper operates a manual water pump. The pump handle acts as a first-class lever pivoting on a central pin. The distance from the handle's grip to the pivot pin is $32\text{ inches}$ , and the distance from the pivot pin to the pump rod connection is $4\text{ inches}$ . If the camper pushes down on the grip with $12\text{ pounds}$ of force, what is the upward force exerted on the pump rod?

96\text{ pounds}

108\text{ pounds}

12\text{ pounds}

1.5\text{ pounds}

Show answer & explanation

Answer: $96\text{ pounds}$

Answer

The correct force exerted on the pump rod is

96\text{ pounds}

.

For a first-class lever, the mechanical advantage is calculated by dividing the effort arm by the load arm. Here,

32\text{ inches} / 4\text{ inches} = 8

. By multiplying the applied effort force of

12\text{ pounds}

by the mechanical advantage of 8, we find the resulting upward force is

96\text{ pounds}

.

Step-by-Step Solution

1

Calculate the mechanical advantage (MA) of the first-class lever.

\text{MA} = \frac{32\text{ inches}}{4\text{ inches}} = 8

The mechanical advantage is the ratio of the effort arm length to the load arm length.

2

Calculate the output force at the pump rod.

12\text{ pounds} \times 8 = 96\text{ pounds}

The output force (load) is the effort force multiplied by the mechanical advantage.

Key Concept

Mechanical Advantage in First-Class Levers

Question 39Question →

A landscaping worker uses a shovel to lift $30\text{ lbs}$ of soil. The worker's rear hand pushes down on the grip to act as the fulcrum, while the front hand lifts the shaft at a point $2\text{ feet}$ away from the rear hand. The soil is centered on the shovel's blade, which is $5\text{ feet}$ away from the rear hand. Ignoring the weight of the shovel itself, what is the minimum upward effort force the front hand must apply to lift the soil?

75\text{ lbs}

50\text{ lbs}

45\text{ lbs}

12\text{ lbs}

Show answer & explanation

Answer: $75\text{ lbs}$

Answer

The minimum upward effort force required is

75\text{ lbs}

.

A shovel functions as a third-class lever, where the effort (front hand) is between the fulcrum (rear hand) and the load (soil). To find the required effort force, set the effort torque equal to the load torque:

\text{Effort} \times \text{Effort Arm} = \text{Load} \times \text{Load Arm}

. Substituting the given values yields

E \times 2 = 30 \times 5

. This simplifies to

2E = 150

, giving an effort force of

75\text{ lbs}

. Third-class levers always require an effort force greater than the load because they multiply distance at the expense of force.

Step-by-Step Solution

1

Identify the lever class and the pivot point (fulcrum).

The shovel acts as a third-class lever where the rear hand is the fulcrum.

Establishing the fulcrum is necessary to measure the lengths of the effort arm and load arm accurately.

2

Determine the lengths of the effort arm and the load arm relative to the fulcrum.

Effort Arm =

2\text{ feet}

(rear hand to front hand). Load Arm =

5\text{ feet}

(rear hand to blade).

Torque is calculated by multiplying force by its perpendicular distance from the fulcrum.

3

Apply the principle of moments (torque balance) to solve for the unknown effort force.

\text{Effort} \times 2 = 30 \times 5 \Rightarrow 2E = 150 \Rightarrow E = 75\text{ lbs}

.

For the lever to be balanced and lift the load, the effort torque must equal the load torque.

Key Concept

Torque balance in third-class levers

Question 40Question →

A maintenance worker uses a $50\text{-inch}$ rigid steel bar to lift one edge of a heavy generator. The fulcrum is positioned at the far end of the bar, and the worker applies an upward force at the opposite end. If the weight of the generator applies a downward force of $320\text{ pounds}$ at a point exactly $10\text{ inches}$ from the fulcrum, how much upward effort force (in pounds) must the worker apply to maintain static equilibrium?

Show answer & explanation

Answer: 64

Answer

64 pounds

The problem describes a second-class lever setup where the fulcrum is at one end, the effort is at the opposite end, and the resistance (load) is located between them. According to the principle of moments (law of the lever), the system is in static equilibrium when the effort torque equals the resistance torque (

F_e \times d_e = F_r \times d_r

). The effort arm length is the total length of the bar (

50\text{ inches}

), and the resistance arm length is the distance from the pivot to the load (

10\text{ inches}

). Substituting the values gives

F_e \times 50 = 320 \times 10

, which simplifies to

50F_e = 3200

. Dividing both sides by 50 yields an effort force of

64\text{ pounds}

.

Step-by-Step Solution

1

Identify the lever class and the lengths of the moment arms.

The setup is a second-class lever. The effort arm (

d_e

) is

50\text{ inches}

and the resistance arm (

d_r

) is

10\text{ inches}

.

Correctly identifying the distances from the fulcrum to the forces is required to calculate torques.

2

Set up the torque balance equation.

F_e \times d_e = F_r \times d_r

For the lever to be in static equilibrium, the effort torque must equal the resistance torque.

3

Substitute the known values and solve for the effort force (

F_e

).

F_e \times 50 = 320 \times 10 \rightarrow 50F_e = 3200 \rightarrow F_e = 64

Isolating the unknown variable determines the exact force required.

Key Concept

The Law of the Lever (Torque Balance)

Levers and Fulcrums

Answer

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