Simple Machines Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 7

Question 121Question →

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 122Question →

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 123Question →

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 124Question →

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 125Question →

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 126Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

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

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

Step-by-Step Solution

1

Analyze the screwdriver mechanism to determine component roles.

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

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

2

Analyze the motorcycle rear wheel mechanism to determine component roles.

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

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

Key Concept

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

Estimated Time:1m 30s

Question 127Question →

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 128Question →

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 129Question →

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 130Question →

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 131Question →

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)

Question 132Question →

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 133Question →

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 134Question →

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

180 \text{ pounds}

10 \text{ pounds}

360 \text{ pounds}

20 \text{ pounds}

Show answer & explanation

Answer: $360 \text{ pounds}$

Answer

The output force is

360 \text{ pounds}

.

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

3 \text{ inches}

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

60 \text{ pounds}

by the mechanical advantage of 6 yields an output force of

360 \text{ pounds}

.

Step-by-Step Solution

1

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

Wheel radius (

R

) is given as

18 \text{ inches}

. Axle diameter is

6 \text{ inches}

, so axle radius (

r

) is

3 \text{ inches}

.

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

2

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

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

.

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

3

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

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

.

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

Key Concept

Mechanical Advantage of a Wheel and Axle

Question 135Question →

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 136Question →

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 137Question →

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

Show answer & explanation

Answer: 125

Answer

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

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

Step-by-Step Solution

1

Identify the given dimensions and force.

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

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

2

Calculate the Ideal Mechanical Advantage (IMA).

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

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

3

Calculate the output force.

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

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

Key Concept

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

Question 138Question →

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 139Question →

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 140Question →

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

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