Levers and Fulcrums Alıştırma Soruları — Armed Services Vocational Aptitude Battery (ASVAB)

Soru 1Soru →

A construction worker uses a straight steel bar to lift a $150\text{-pound}$ concrete slab. A metal pipe is placed underneath the bar to serve as a pivot point (fulcrum). If the slab rests on the bar $3\text{ feet}$ from the fulcrum, and the worker applies downward force on the opposite end at a distance of $9\text{ feet}$ from the fulcrum, how much force is required to perfectly balance the slab?

37.5\text{ lbs}

50\text{ lbs}

150\text{ lbs}

450\text{ lbs}

Cevabı ve açıklamayı göster

Cevap: $50\text{ lbs}$

Cevap

The correct force required is

50\text{ lbs}

.

According to the principle of levers, a system is balanced when the input torque equals the output torque (

F_1 \times d_1 = F_2 \times d_2

). The concrete slab generates

450\text{ lb-ft}

of torque (

150\text{ lbs} \times 3\text{ ft}

). Dividing this torque by the effort arm distance of

9\text{ feet}

yields a required downward effort force of

50\text{ lbs}

.

Adım Adım Çözüm

1

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

Force_{effort} \times Distance_{effort} = Force_{load} \times Distance_{load}

For a lever to be perfectly balanced, the torque applied by the worker must equal the torque exerted by the load.

2

Calculate the torque of the load.

150\text{ lbs} \times 3\text{ ft} = 450\text{ lb-ft}

Torque is calculated by multiplying the weight of the object by its distance to the fulcrum.

3

Solve for the required effort force.

450\text{ lb-ft} / 9\text{ ft} = 50\text{ lbs}

Dividing the total load torque by the length of the effort arm determines the exact force the worker needs to apply.

Anahtar Kavram

Torque balance in first-class levers (Law of Levers)

Soru 2Soru →

Match each mechanical tool or scenario with its correct lever classification and mechanical advantage (MA) characteristic.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

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Cevap

Pry bar matches to Class 1 (MA > 1); Hand truck matches to Class 2 (MA > 1); Tweezers match to Class 3 (MA < 1); Symmetrical seesaw matches to Class 1 (MA = 1).

Each tool utilizes a specific lever class based on the relative arrangement of the Fulcrum, Load, and Effort. By identifying which component is positioned in the middle, we determine the correct class. The ratio of the effort arm to the load arm then dictates whether the mechanical advantage is greater than, equal to, or less than 1.

Adım Adım Çözüm

1

Analyze the pry bar scenario to locate the relative positions of the fulcrum, effort, and load.

The fulcrum (wooden block) is between the effort (hands) and the load (engine), but closer to the load.

This configuration defines a Class 1 lever that is optimized for force multiplication.

2

Analyze the hand truck to identify its lever class.

The wheels act as the fulcrum at one end, the effort is applied at the opposite end (handles), and the load (crates) rests in the middle.

Having the load in the middle defines a Class 2 lever, which always provides a mechanical advantage greater than 1.

3

Examine how tweezers apply force.

The fulcrum is the joined end, the effort is applied by fingers pinching the middle, and the load is grasped at the opposite open end.

Having the effort in the middle defines a Class 3 lever, which trades force for precision and distance (MA < 1).

4

Evaluate the symmetrical seesaw scenario.

The fulcrum is directly in the center with equal distances to both applied forces.

This is a Class 1 lever where neither distance nor force is multiplied, meaning MA equals 1.

Anahtar Kavram

Lever Classifications and Mechanical Advantage

Soru 3Soru →

A carpenter uses a claw hammer to extract a stubborn nail from a wooden board. The hammer functions as a first-class lever, pivoting on the curve of its head. The perpendicular distance from the pivot point to the nail is $6 \text{ cm}$ , and the carpenter applies an effort force on the handle at a perpendicular distance of $24 \text{ cm}$ from the pivot point. If the nail requires a resistance force of $480 \text{ N}$ to be pulled out, what minimum effort force must the carpenter apply?

96 \text{ N}

160 \text{ N}

120 \text{ N}

384 \text{ N}

Cevabı ve açıklamayı göster

Cevap: $120 \text{ N}$

Cevap

The minimum effort force required is

120 \text{ N}

.

In a first-class lever system in equilibrium, the effort torque must equal the resistance torque (Effort Force × Effort Arm = Resistance Force × Resistance Arm). Plugging in the given values yields: Effort Force × 24 cm = 480 N × 6 cm. Solving for the Effort Force gives 120 N.

Adım Adım Çözüm

1

Identify the type of lever and the location of the fulcrum.

The claw hammer is acting as a first-class lever. The pivot point (fulcrum) is located between the applied effort (the carpenter's hand) and the load (the nail).

Correctly identifying the lever setup is necessary to assign the proper lengths to the effort arm and resistance arm.

2

Determine the length of the effort arm and the resistance arm.

The effort arm is the distance from the pivot to the hand (

24 \text{ cm}

). The resistance arm is the distance from the pivot to the nail (

6 \text{ cm}

).

Torque is calculated using the distance from the pivot point to where the force is applied.

3

Calculate the mechanical advantage (MA) of the lever.

MA = \frac{\text{Effort Arm}}{\text{Resistance Arm}} = \frac{24}{6} = 4

.

The mechanical advantage indicates how much the lever multiplies the applied force.

4

Calculate the required effort force to overcome the resistance.

\text{Effort Force} = \frac{\text{Resistance Force}}{MA} = \frac{480}{4} = 120 \text{ N}

.

The effort force multiplied by the mechanical advantage must equal the resistance force to balance the torques.

Anahtar Kavram

Torque Balance and Mechanical Advantage in First-Class Levers

Soru 4Soru →

A mechanic uses a 60-inch straight pry bar to lift a 300 lb machinery crate. Initially, they place a pivot block 12 inches from the crate, pressing down on the opposite end to lift the load. To lift the crate higher on the next attempt, they move the pivot block so it is 20 inches from the crate, again pressing down on the opposite end. By how much does the required effort force change, and what is the new mechanical advantage of the lever?

The effort force increases by 75 lbs; the new mechanical advantage is 2.

The effort force increases by 40 lbs; the new mechanical advantage is 3.

The effort force increases by 150 lbs; the new mechanical advantage is 0.5.

The effort force increases by 50 lbs; the new mechanical advantage is 2.4.

Cevabı ve açıklamayı göster

Cevap: The effort force increases by 75 lbs; the new mechanical advantage is 2.

Cevap

The effort force increases by 75 lbs; the new mechanical advantage is 2.

The correct answer accurately calculates that moving the pivot increases the load arm to 20 inches and shrinks the effort arm to 40 inches. This cuts the mechanical advantage in half (from 4 to 2), doubling the required effort force from 75 lbs to 150 lbs, which is an increase of 75 lbs.

Adım Adım Çözüm

1

Identify the lever class and the formula for torque balance.

This is a first-class lever. The formula is: Effort Force × Effort Arm = Load Force × Load Arm. The Mechanical Advantage (MA) is Effort Arm / Load Arm.

The fulcrum (pivot block) is located between the applied effort and the load.

2

Calculate the initial effort force and mechanical advantage (State 1).

Load Arm = 12 inches. Effort Arm = 60 - 12 = 48 inches. MA = 48 / 12 = 4. Initial Effort Force = 300 / 4 = 75 lbs.

The total length of the bar is 60 inches, so the effort arm is the remainder of the bar after subtracting the load arm.

3

Calculate the new effort force and mechanical advantage (State 2).

New Load Arm = 20 inches. New Effort Arm = 60 - 20 = 40 inches. New MA = 40 / 20 = 2. New Effort Force = 300 / 2 = 150 lbs.

Moving the pivot 8 inches away from the load increases the load arm and correspondingly decreases the effort arm.

4

Determine the change in the required effort force.

150 lbs (New) - 75 lbs (Initial) = 75 lbs increase.

The question asks for the difference between the two required forces.

Anahtar Kavram

Torque balance and changing fulcrum positions on a first-class lever.

Tahmini Süre:2m 0s

Soru 5Soru →

A uniform $10$ -foot steel pry bar weighing $40$ pounds is used to lift a rock. A fulcrum is placed exactly $2$ feet from the load end of the bar. A $480$ -pound downward force from the rock acts exactly at the load end. How much downward effort force must be applied at the opposite end of the pry bar to keep it perfectly balanced and horizontal?

84 lbs

105 lbs

120 lbs

135 lbs

Cevabı ve açıklamayı göster

Cevap: 105 lbs

Cevap

105 lbs

The system is balanced when clockwise and counter-clockwise torques are equal. The rock creates

960 \text{ lb-ft}

of counter-clockwise torque (

480 \times 2

). The bar's weight creates

120 \text{ lb-ft}

of clockwise torque (

40 \times 3

, since its center is

3 \text{ ft}

from the fulcrum). The remaining clockwise torque must be provided by the effort force applied at

8 \text{ ft}

from the fulcrum. Therefore,

960 = 120 + 8E

, which solves to

E = 105 \text{ lbs}

.

Adım Adım Çözüm

1

Calculate the counter-clockwise torque created by the rock.

480 \text{ lbs} \times 2 \text{ ft} = 960 \text{ lb-ft}

Torque is force multiplied by the distance to the fulcrum. The rock acts

2

feet from the fulcrum.

2

Determine the location of the pry bar's center of gravity relative to the fulcrum.

The center is

5 \text{ ft}

from either end. The fulcrum is at

2 \text{ ft}

. The distance from the fulcrum to the center of gravity is

5 - 2 = 3 \text{ ft}

on the effort side.

A uniform bar's weight acts at its exact center point.

3

Calculate the clockwise torque created by the pry bar's own weight.

40 \text{ lbs} \times 3 \text{ ft} = 120 \text{ lb-ft}

The bar's weight assists the effort force because its center of gravity is on the same side of the fulcrum as the applied effort.

4

Set up the torque balance equation to find the required effort force.

960 \text{ lb-ft} = 120 \text{ lb-ft} + (E \times 8 \text{ ft})

For the lever to be balanced, total counter-clockwise torque must equal total clockwise torque. The effort is applied

10 - 2 = 8 \text{ ft}

from the fulcrum.

5

Solve for the effort force (E).

8E = 960 - 120 \Rightarrow 8E = 840 \Rightarrow E = 105 \text{ lbs}

Isolating E gives the exact downward force required at the end of the bar.

Anahtar Kavram

To balance a lever, the sum of all clockwise torques must equal the sum of all counter-clockwise torques. For heavy levers, the object's own weight acting at its center of gravity must be included in the calculation.

Soru 6Soru →

Match each mechanical scenario to its correct lever classification and mechanical advantage (MA) property.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Pry bar matches Class 1 (MA > 1). Hand truck matches Class 2 (MA > 1). Forceps match Class 3 (MA < 1). Trebuchet matches Class 1 (MA < 1).

Each mechanism must be evaluated by locating the fulcrum, effort, and load. A pry bar and trebuchet both have the fulcrum in the middle (Class 1), but differ in their arm ratios, resulting in MA > 1 and MA < 1 respectively. A hand truck has the load in the middle (Class 2, MA > 1). Forceps have the effort in the middle (Class 3, MA < 1).

Adım Adım Çözüm

1

Analyze the pry bar scenario.

Identified as a Class 1 lever with MA > 1.

The fulcrum is located between the effort (operator) and the load (crate). Because the fulcrum is placed closer to the load, the effort arm is longer than the load arm, creating a mechanical advantage greater than 1.

2

Analyze the hand truck scenario.

Identified as a Class 2 lever with MA > 1.

The wheels serve as the fulcrum at one end. The refrigerator's weight (load) acts in the middle, and the operator's hands apply effort at the opposite end. This configuration is a Class 2 lever, which inherently multiplies force (MA > 1).

3

Analyze the forceps scenario.

Identified as a Class 3 lever with MA < 1.

The joined end is the fulcrum. The surgeon applies effort to the middle of the tool, and the tissue is the load at the far end. Effort in the middle defines a Class 3 lever, which always acts to multiply distance and speed at the expense of force (MA < 1).

4

Analyze the trebuchet scenario.

Identified as a Class 1 lever with MA < 1.

The pivot is between the dropping counterweight (effort) and the projectile (load), making it a Class 1 lever. Because the effort arm is much shorter than the load arm, the mechanical advantage is less than 1, which serves to drastically increase the speed of the projectile.

Anahtar Kavram

Identifying the relative positions of the fulcrum, effort, and load to classify levers (Class 1, 2, or 3) and analyzing arm lengths to determine if the mechanism multiplies force (MA > 1) or speed (MA < 1).

Tahmini Süre:2m 0s

Soru 7Soru →

Match each real-world application of a lever to its correct mechanical configuration and primary mechanical advantage characteristic.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Wheelbarrow matches the configuration where resistance is in the middle (Class 2). Baseball bat matches the configuration where applied force is in the middle (Class 3). Crowbar matches the pivot-in-middle configuration with a long effort arm (Class 1, high torque). Balance scale matches the pivot-in-middle configuration with equal arms (Class 1, neutral torque).

Each tool's primary function dictates its lever configuration. Class 1 levers separate effort and resistance with a pivot (crowbar maximizes force; balance scale equalizes force). Class 2 levers place resistance in the middle to multiply force (wheelbarrow). Class 3 levers place effort in the middle to multiply speed and distance (baseball bat).

Adım Adım Çözüm

1

Analyze the mechanics of the wheelbarrow.

Identified that the wheel is the pivot, the load is in the center, and the handles receive the effort.

Determining the relative positions of the pivot, load, and effort confirms it is a Class 2 lever, which always reduces the effort force required.

2

Analyze the mechanics of the baseball bat.

Identified that the wrists act as the pivot, the arms apply force near the base, and the load is at the far end.

This confirms it is a Class 3 lever, which trades force for increased speed and distance at the resistance end.

3

Analyze the mechanics of the crowbar.

Identified that the pivot rests between the long handle (effort) and the short prying tip (resistance).

This confirms it is a Class 1 lever designed with a much longer effort arm to maximize the mechanical advantage and torque.

4

Analyze the mechanics of the double-pan balance scale.

Identified that the pivot is perfectly centered between the two pans.

This confirms it is a Class 1 lever with equal arm lengths, yielding a mechanical advantage of exactly 1.

Anahtar Kavram

Lever Classes, Torque, and Mechanical Advantage

Soru 8Soru →

A rescue crew uses a uniform $14$ -foot steel beam as a first-class lever to lift a $900\text{ lb}$ concrete slab. The beam itself weighs $240\text{ lbs}$ . They place a pivot block exactly $2\text{ feet}$ from the end where the concrete slab rests. How much downward effort force must the crew apply at the opposite end of the beam to perfectly balance the slab?

50\text{ lbs}

150\text{ lbs}

190\text{ lbs}

250\text{ lbs}

Cevabı ve açıklamayı göster

Cevap: $50\text{ lbs}$

Cevap

The correct answer is

50\text{ lbs}

, because the uniform beam's own weight assists the effort force in balancing the heavy slab.

The correct answer properly applies the principle of rotational equilibrium. The slab creates

1800\text{ lb-ft}

of counter-clockwise torque. The beam's center of gravity is

5\text{ feet}

to the right of the fulcrum, creating

1200\text{ lb-ft}

of clockwise torque. This means the effort force only needs to provide the remaining

600\text{ lb-ft}

of torque. At a distance of

12\text{ feet}

from the fulcrum, a

50\text{ lb}

force provides exactly

600\text{ lb-ft}

(

50 \times 12

).

Adım Adım Çözüm

1

Determine the moment arms (distances from the fulcrum) for all forces involved.

Load arm =

2\text{ ft}

. Effort arm =

14 - 2 = 12\text{ ft}

. The beam's center of gravity (CG) is at

7\text{ ft}

, making its distance to the fulcrum

7 - 2 = 5\text{ ft}

.

Torque is calculated as force multiplied by the perpendicular distance to the pivot point.

2

Calculate the counter-clockwise torque exerted by the concrete slab.

Torque =

900\text{ lbs} \times 2\text{ ft} = 1800\text{ lb-ft}

.

This is the primary torque that must be overcome to balance the lever.

3

Calculate the clockwise torque exerted by the beam's own weight.

Torque =

240\text{ lbs} \times 5\text{ ft} = 1200\text{ lb-ft}

.

Because the beam's CG is to the right of the fulcrum, its weight assists the downward effort force.

4

Set up the torque equilibrium equation and solve for the required effort force (

F

).

1800 = 1200 + (F \times 12)

. Solving for

F

:

12F = 600

, which gives

F = 50\text{ lbs}

.

For the lever to be perfectly balanced, the total counter-clockwise torque must equal the total clockwise torque.

Anahtar Kavram

Torque equilibrium in levers accounting for uniform mass distribution.

Soru 9Soru →

A hydraulic piston is attached to a hinged ramp $2$ feet away from the hinge (the fulcrum) and pushes perpendicularly to close it. A $200$ -pound heavy crate is positioned on the ramp $8$ feet away from the hinge. Assuming the weight of the ramp itself is negligible, how much effort force must the hydraulic piston apply to balance the crate, and which class of lever does this mechanism represent?

800 pounds; Third-class lever

50 pounds; Second-class lever

800 pounds; Second-class lever

50 pounds; Third-class lever

Cevabı ve açıklamayı göster

Cevap: 800 pounds; Third-class lever

Cevap

800 pounds; Third-class lever

The correct answer accurately identifies the system as a third-class lever because the effort (the piston) is located between the fulcrum (the hinge) and the load (the crate). Using the principle of moments, the torque from the load is

200 \text{ pounds} \times 8 \text{ feet} = 1600 \text{ lb-ft}

. To balance this, the piston must apply an equal opposing torque at a distance of

2

feet:

1600 \div 2 = 800 \text{ pounds}

.

Adım Adım Çözüm

1

Identify the lever class based on the arrangement of components.

The arrangement is Fulcrum (hinge) - Effort (piston at 2 ft) - Load (crate at 8 ft).

Because the effort is applied between the fulcrum and the load, this is by definition a third-class lever.

2

Set up the torque balance equation to find the effort force.

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

For the lever to be balanced, the clockwise torque must equal the counterclockwise torque.

3

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

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

, which simplifies to

2\text{E} = 1600

. Solving for E gives

800

pounds.

A third-class lever always has a mechanical advantage of less than 1, requiring an effort force greater than the load.

Anahtar Kavram

Torque Balance and Third-Class Levers

Soru 10Soru →

A worker is using a wheelbarrow, which acts as a second-class lever, to transport materials. The wheel's axle serves as the fulcrum. The empty wheelbarrow has a downward weight of $200 \text{ N}$ , with its center of gravity located $0.5 \text{ m}$ horizontally from the axle. A load of bricks weighing $800 \text{ N}$ is placed in the tray so that its center of gravity is $0.4 \text{ m}$ horizontally from the axle. The worker lifts upward at the ends of the handles, which are located $1.2 \text{ m}$ horizontally from the axle.

Assuming the wheelbarrow is in a horizontal position, what is the minimum upward force required to lift the rear supports off the ground?

Cevabı ve açıklamayı göster

Cevap: 350

Cevap

The minimum upward force required to lift the wheelbarrow is 350 N.

The correct answer is derived by applying the principle of moments (torque). Because there are two separate masses acting on the lever, their torques must be calculated independently and summed. The wheelbarrow contributes

100 \text{ N}\cdot\text{m}

of torque and the bricks contribute

320 \text{ N}\cdot\text{m}

. Balancing the total

420 \text{ N}\cdot\text{m}

torque over a

1.2 \text{ m}

lever arm requires exactly

350 \text{ N}

of effort.

Adım Adım Çözüm

1

Calculate the torque from the empty wheelbarrow.

\tau_{cart} = 100 \text{ N}\cdot\text{m}

The intrinsic weight of the lever system applies a downward force that must be accounted for.

2

Calculate the torque from the added load.

\tau_{load} = 320 \text{ N}\cdot\text{m}

The load applies an additional independent downward force at its specific distance from the fulcrum.

3

Sum the downward torques.

\tau_{total} = 420 \text{ N}\cdot\text{m}

In equilibrium, the total clockwise torque must be balanced by the total counterclockwise torque.

4

Divide the total torque by the effort distance to find the effort force.

F_{effort} = 350 \text{ N}

Torque equals force times distance. Rearranging

420 = F \times 1.2

isolates the required lifting force.

Anahtar Kavram

Torque Equilibrium and Composite Loads in Second-Class Levers

Soru 11Soru →

A specialized industrial press uses a compound lever system composed of two uniform steel bars to compress a heavy coil spring.

- Lever A (Top handle): is $20\text{ inches}$ long, weighs $10\text{ lbs}$ , and pivots on a fulcrum at its left end. A vertical rigid connecting link is attached to Lever A at a point $5\text{ inches}$ from its fulcrum.
- Lever B (Bottom pressing bar): is $30\text{ inches}$ long, weighs $30\text{ lbs}$ , and pivots on a fulcrum at its left end. The connecting link from Lever A attaches to the far right end of Lever B ( $30\text{ inches}$ from its fulcrum). A heavy coil spring is placed under Lever B, located $6\text{ inches}$ from Lever B's fulcrum.

If a worker applies a $40\text{ lb}$ downward force at the far right end of Lever A ( $20\text{ inches}$ from its fulcrum), what is the total downward compressive force (in lbs) exerted on the spring?

*Assume both levers are horizontal, the connecting link is perfectly vertical, and the system is in static equilibrium.*

Cevabı ve açıklamayı göster

Cevap: 975

Cevap

975 lbs

In static equilibrium, the sum of clockwise and counterclockwise torques about any pivot point must be exactly zero. By calculating the torques step-by-step through the compound system—incorporating both the applied forces and the center of gravity of each uniform bar—we find the final compressive force transmitted to the spring is 975 lbs.

Adım Adım Çözüm

1

Calculate total clockwise torque on Lever A

900 lb-in

The worker's force (40 lbs at 20 inches) and the lever's own uniform weight (10 lbs at its 10-inch center of gravity) both create clockwise rotation around the fulcrum.

2

Determine the force transmitted through the connecting link

180 lbs

The link must balance the 900 lb-in torque at a distance of 5 inches (900 / 5 = 180 lbs). This 180 lbs force is transmitted downward to Lever B.

3

Calculate total clockwise torque on Lever B

5850 lb-in

The downward force from the link (180 lbs at 30 inches) and Lever B's uniform weight (30 lbs at its 15-inch center of gravity) create clockwise rotation around Lever B's fulcrum.

4

Calculate the final compressive force applied to the spring

975 lbs

The spring must balance Lever B's total 5850 lb-in torque at a distance of 6 inches (5850 / 6 = 975 lbs).

Anahtar Kavram

Torque Equilibrium in Compound and Uniform Levers

Soru 12Soru →

A worker uses a uniform $60$ -inch steel pry bar to lift a $150\text{ lb}$ concrete block. The worker places a pivot exactly $20$ inches from the load end of the bar, configuring it as a first-class lever. If the pry bar itself weighs $20\text{ lbs}$ , how much downward vertical force must the worker apply at the opposite end of the bar to perfectly balance the concrete block?

75\text{ lbs}

300\text{ lbs}

70\text{ lbs}

80\text{ lbs}

Cevabı ve açıklamayı göster

Cevap: $70\text{ lbs}$

Cevap

The correct force required is

70\text{ lbs}

.

To achieve equilibrium, the torques around the pivot must balance. The concrete block creates a counter-clockwise torque of

3000\text{ in-lbs}

(

150\text{ lbs} \times 20\text{ inches}

). The pry bar's weight acts at its center (

30

inches from the end), which places it

10

inches from the pivot on the effort side. This provides an assisting clockwise torque of

200\text{ in-lbs}

(

20\text{ lbs} \times 10\text{ inches}

). The effort force is applied at the very end,

40

inches from the pivot. The equilibrium equation is

3000 = 200 + (F \times 40)

, resolving to

2800 = F \times 40

, which means the required force is

70\text{ lbs}

.

Adım Adım Çözüm

1

Calculate the counter-clockwise torque generated by the concrete block.

150\text{ lbs} \times 20\text{ inches} = 3000\text{ in-lbs}

.

The load is placed

20

inches from the pivot, creating the primary rotational force to overcome.

2

Determine the location of the pry bar's center of gravity and its resulting torque.

The center of the

60

-inch bar is at

30

inches. Since the pivot is at

20

inches, the center of gravity is

10

inches from the pivot on the effort side. Torque =

20\text{ lbs} \times 10\text{ inches} = 200\text{ in-lbs}

(clockwise).

A uniform bar's weight acts at its exact center. Because this center is on the same side of the pivot as the worker's effort, it provides an assisting clockwise torque.

3

Set up the torque equilibrium equation to find the required effort force.

Counter-clockwise Torque = Clockwise Torque

3000 = 200 + (F \times 40)

For the lever to balance, the sum of opposing torques must be equal. The effort arm is the total length minus the load arm (

60 - 20 = 40

inches).

4

Solve the equation for the effort force (

F

).

2800 = F \times 40

, which simplifies to

F = 70\text{ lbs}

.

Isolating

F

gives the exact supplemental force the worker needs to apply to achieve balance.

Anahtar Kavram

Torque Equilibrium Involving Lever Mass

Soru 13Soru →

Match each mechanical device or system to its correct lever classification and corresponding mechanical advantage (MA) characteristics.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

The correct matches pair each mechanical system with its corresponding lever classification based on the sequence of Effort, Fulcrum, and Load.

Each system perfectly aligns with the fundamental definitions of the three lever classes based on the relative positions of the fulcrum, effort, and load. Class 1 has the fulcrum in the middle, Class 2 has the load in the middle, and Class 3 has the effort in the middle.

Adım Adım Çözüm

1

Analyze the excavator arm's force arrangement.

The hydraulic cylinder (effort) acts between the hinge (fulcrum) and bucket (load).

Having the effort in the middle defines a third-class lever, which always has a mechanical advantage less than 1 to multiply speed and distance.

2

Analyze the nutcracker's force arrangement.

The walnut (load) is located between the hinge (fulcrum) and hands (effort).

Having the load in the middle defines a second-class lever, which always has a mechanical advantage greater than 1 to multiply force.

3

Analyze the pry bar's force arrangement.

The wood block (fulcrum) is between the hands (effort) and slab (load), close to the slab.

Having the fulcrum in the middle makes it a first-class lever. Placing it close to the load creates a longer effort arm, yielding a mechanical advantage greater than 1.

4

Analyze the trebuchet's force arrangement.

The pivot (fulcrum) is between the counterweight (effort) and projectile (load), but closer to the counterweight.

Having the fulcrum in the middle makes it a first-class lever. The longer load arm means the mechanical advantage is less than 1, multiplying the speed of the projectile.

Anahtar Kavram

Identifying Lever Classes and Mechanical Advantage

Tahmini Süre:2m 0s

Soru 14Soru →

A steam boiler uses a horizontal safety valve lever hinged at one end (the fulcrum). The uniform steel lever is $50\text{ inches}$ long and weighs $20\text{ lbs}$ . A pressure valve is attached to the lever $10\text{ inches}$ from the fulcrum and exerts an upward force of $150\text{ lbs}$ . To keep the valve closed, what is the minimum downward-acting weight that must be hung at the very end of the lever ( $50\text{ inches}$ from the fulcrum)?

20\text{ lbs}

30\text{ lbs}

90\text{ lbs}

750\text{ lbs}

Cevabı ve açıklamayı göster

Cevap: $20\text{ lbs}$

Cevap

The minimum weight that must be hung at the end is

20\text{ lbs}

.

The system is in equilibrium when the total upward torque equals the total downward torque. The upward torque from the valve (

1500\text{ in-lbs}

) must be balanced by the downward torque of the lever's center of gravity (

500\text{ in-lbs}

) plus the torque of the applied weight at the end (

20\text{ lbs} \times 50\text{ inches} = 1000\text{ in-lbs}

).

Adım Adım Çözüm

1

Identify the opposing torques acting around the fulcrum.

Upward torque is generated by the steam valve. Downward torque is generated by the lever's own weight plus the applied weight at the end.

For the lever to remain balanced (closed), the total clockwise torque must equal the total counter-clockwise torque.

2

Calculate the upward torque from the steam valve.

150\text{ lbs} \times 10\text{ inches} = 1500\text{ in-lbs}

(upward).

Torque equals force multiplied by the perpendicular distance from the fulcrum.

3

Calculate the downward torque contributed by the uniform lever's own weight.

The center of gravity is at

25\text{ inches}

(half of

50\text{ inches}

). Torque =

20\text{ lbs} \times 25\text{ inches} = 500\text{ in-lbs}

(downward).

A uniform object's weight acts through its geometric center.

4

Determine the remaining downward torque required and solve for the unknown weight.

1500\text{ in-lbs} - 500\text{ in-lbs} = 1000\text{ in-lbs}

of downward torque needed.

W \times 50\text{ inches} = 1000\text{ in-lbs}

, so

W = 20\text{ lbs}

.

The weight at the end must provide exactly the remaining torque to maintain equilibrium.

Anahtar Kavram

Torque Equilibrium and Lever Mechanics

Alternatif Yöntem

Instead of calculating full torques, recognize that an idealized weightless lever would require

30\text{ lbs}

of force at the end (

150 \div 5

). The lever's own weight (

20\text{ lbs}

) acts at its midpoint, which effectively provides

10\text{ lbs}

of downward force translated to the very end (

20 \div 2

). Thus, the required external weight is

30 - 10 = 20\text{ lbs}

.

Tahmini Süre:1m 30s

Soru 15Soru →

A sheet metal fabricator uses a custom compound lever tool to punch holes in aluminum panels. The tool consists of two connected levers:

1. The Handle (Lever A): A second-class lever with its fulcrum at the very top. The handle is $25\text{ inches}$ long overall. A vertical connecting rod is attached to this handle $5\text{ inches}$ away from the fulcrum.
2. The Punch Arm (Lever B): A first-class lever positioned horizontally. The connecting rod pulls down on one end, which is $8\text{ inches}$ away from its fulcrum. The punching die is located on the opposite side of the fulcrum, $2\text{ inches}$ away.

If the fabricator applies a perpendicular force of $24\text{ lbs}$ to the extreme end of the handle, what is the total downward force, in pounds, exerted by the punch die onto the aluminum panel?

Cevabı ve açıklamayı göster

Cevap: 480

Cevap

The correct answer is

480\text{ lbs}

.

In a compound lever system, the mechanical advantage of each lever multiplies sequentially to produce the final force. The handle (Lever A) provides a mechanical advantage of

25/5 = 5

. Therefore, the

24\text{ lbs}

input is multiplied by

5

, becoming

120\text{ lbs}

of force on the connecting rod. The punch arm (Lever B) provides a mechanical advantage of

8/2 = 4

. The

120\text{ lbs}

input to Lever B is then multiplied by

4

, resulting in a total output force of

480\text{ lbs}

at the punching die.

Adım Adım Çözüm

1

Calculate the input torque generated by the fabricator on Lever A.

24\text{ lbs} \times 25\text{ inches} = 600\text{ in-lbs}

Torque is determined by multiplying the applied force by the perpendicular distance from the fulcrum.

2

Determine the force exerted on the connecting rod by Lever A.

600\text{ in-lbs} / 5\text{ inches} = 120\text{ lbs}

In equilibrium, the input torque equals the output torque. The rod is attached

5\text{ inches}

from the fulcrum.

3

Calculate the input torque on Lever B caused by the connecting rod.

120\text{ lbs} \times 8\text{ inches} = 960\text{ in-lbs}

The connecting rod applies

120\text{ lbs}

of force at a distance of

8\text{ inches}

from Lever B's fulcrum.

4

Calculate the final output force at the punching die.

960\text{ in-lbs} / 2\text{ inches} = 480\text{ lbs}

The punching die is

2\text{ inches}

from the fulcrum, balancing the

960\text{ in-lbs}

of torque on the opposite side.

Anahtar Kavram

Compound Lever Systems and Torque Balance

Tahmini Süre:1m 30s

Soru 16Soru →

A fisherman holds an $8\text{-foot}$ fishing rod. He rests the base of the rod against his waist, which acts as the pivot point (fulcrum). His upper hand pulls upward on the rod at a distance of $2\text{ feet}$ from his waist. If a fish pulls downward on the tip of the rod with a force of $20\text{ lbs}$ , what is the required effort force applied by his upper hand to keep the rod balanced, and which class of lever does this represent? (Assume the weight of the rod is negligible).

80\text{ lbs}

; Third-class lever

5\text{ lbs}

; Third-class lever

60\text{ lbs}

; First-class lever

80\text{ lbs}

; Second-class lever

Cevabı ve açıklamayı göster

Cevap: $80\text{ lbs}$ ; Third-class lever

Cevap

The required effort force is

80\text{ lbs}

, and the fishing rod acts as a Third-class lever.

In this scenario, the effort (the upper hand) is located between the fulcrum (the waist) and the load (the fish). This specific arrangement defines a third-class lever. To find the required effort force, set up a torque balance equation:

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

. Substituting the given values yields

\text{Effort Force} \times 2\text{ ft} = 20\text{ lbs} \times 8\text{ ft}

. Solving for the effort force gives

160 / 2 = 80\text{ lbs}

.

Adım Adım Çözüm

1

Identify the locations of the fulcrum, effort, and load on the rod.

The fulcrum is at the base (waist), the effort is applied

2\text{ feet}

from the base, and the load is at the tip,

8\text{ feet}

from the base.

Establishing the sequence and distances of these points is necessary to determine the lever class and torque equations.

2

Determine the class of the lever.

It is a Third-class lever.

Because the effort force (upper hand) is located between the fulcrum (waist) and the load (fish).

3

Set up the torque balance equation.

\text{Effort Force} \times 2\text{ ft} = 20\text{ lbs} \times 8\text{ ft}

.

For the lever to be in equilibrium, the torque from the effort must equal the torque from the load.

4

Solve the equation for the effort force.

\text{Effort Force} \times 2 = 160

, so

\text{Effort Force} = 80\text{ lbs}

.

To find the specific pulling force the fisherman must exert.

Anahtar Kavram

Torque balance and Third-class levers

Soru 17Soru →

A carpenter uses a claw hammer to pull a stubborn nail out of a board. The point where the hammer head rests on the board acts as the fulcrum. The distance from the fulcrum to the nail is $1.5\text{ inches}$ , and the carpenter grips the handle $15\text{ inches}$ away from the fulcrum. If the carpenter applies a $25\text{-lb}$ pull at the handle, what is the maximum nail resistance that can be overcome, and what class of lever is being used?

275\text{ lbs}

; second-class lever

250\text{ lbs}

; second-class lever

250\text{ lbs}

; first-class lever

2.5\text{ lbs}

; third-class lever

Cevabı ve açıklamayı göster

Cevap: $250\text{ lbs}$ ; first-class lever

Cevap

The maximum resistance is

250\text{ lbs}

, and the system acts as a first-class lever.

The correct answer accurately identifies the claw hammer as a first-class lever because the fulcrum is located between the applied effort and the load. Applying the principle of moments (torque balance), the effort force multiplied by the effort arm (

25 \times 15 = 375

) must equal the load force multiplied by the load arm. Dividing

375

by

1.5

yields a maximum resistance of

250\text{ lbs}

.

Adım Adım Çözüm

1

Determine the relative positions of the fulcrum, effort, and load to identify the lever class.

The fulcrum (where the hammer rests on the board) is located between the effort (the hand pulling the handle) and the load (the nail). This arrangement defines a first-class lever.

Knowing the correct lever class confirms how the forces interact around the pivot point.

2

Set up the torque balance equation using the principle of moments.

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

For the lever to overcome the nail's resistance, the torque applied by the carpenter must at least equal the opposing torque of the nail.

3

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

25 \times 15 = \text{Load Force} \times 1.5 \rightarrow 375 = \text{Load Force} \times 1.5 \rightarrow \text{Load Force} = 375 / 1.5 = 250\text{ lbs}

.

This calculation yields the maximum resistance force the lever can lift under the applied effort.

Anahtar Kavram

Torque balance and lever classification

Soru 18Soru →

Match each lever classification or concept to its correct structural arrangement or definition.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

First-Class Lever matches with fulcrum in the middle; Second-Class Lever matches with load in the middle; Third-Class Lever matches with effort in the middle; Mechanical Advantage matches with the force multiplier ratio.

Lever classes are universally categorized by which component (Fulcrum, Load, or Effort) is located in the middle of the structural arrangement. Matching these central components correctly defines the machine.

Adım Adım Çözüm

1

Identify the arrangement for a first-class lever.

Matches with the fulcrum positioned between the effort and the load.

This is the fundamental definition of a first-class lever system, like a standard seesaw.

2

Identify the arrangement for a second-class lever.

Matches with the load positioned between the fulcrum and the effort.

This represents systems designed to multiply force by keeping the load closer to the pivot point, like a nutcracker.

3

Identify the arrangement for a third-class lever.

Matches with the applied effort positioned between the fulcrum and the load.

This represents systems designed to increase range of motion, like a broom or tweezers.

4

Define mechanical advantage in the context of simple machines.

Matches with the factor by which a machine multiplies the input force.

Mechanical advantage is the primary metric for evaluating how much a lever amplifies the applied force.

Anahtar Kavram

Lever Classification and Mechanical Advantage

Soru 19Soru →

A carpenter uses a claw hammer to pull a stubborn nail out of a wooden plank. The hammer head rests on the plank, creating a pivot point (fulcrum) that is exactly $2\text{ inches}$ away from the nail. The carpenter applies a perpendicular pulling force of $40\text{ lbs}$ on the hammer's handle, at a distance of $12\text{ inches}$ away from the pivot point. Assuming the hammer acts as a simple first-class lever, how much upward force is exerted on the nail?

80\text{ lbs}

200\text{ lbs}

240\text{ lbs}

480\text{ lbs}

Cevabı ve açıklamayı göster

Cevap: $240\text{ lbs}$

Cevap

The upward force exerted on the nail is

240\text{ lbs}

.

The claw hammer acts as a first-class lever where torque is balanced around the fulcrum. The formula for a balanced lever is

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

. Plugging in the given values yields

40 \times 12 = \text{Load} \times 2

, which simplifies to

480 = 2 \times \text{Load}

. Dividing both sides by 2 gives a load force of

240\text{ lbs}

.

Adım Adım Çözüm

1

Identify the lever arms for the hammer setup.

The effort arm (distance from the pulling hand to the fulcrum) is

12\text{ inches}

. The load arm (distance from the nail to the fulcrum) is

2\text{ inches}

.

Correctly identifying the distances from the fulcrum is necessary for calculating mechanical advantage and balancing torque.

2

Calculate the mechanical advantage (MA) of the lever.

\text{MA} = \text{Effort Arm} / \text{Load Arm} = 12 / 2 = 6

.

The mechanical advantage indicates how many times the lever multiplies the input force.

3

Calculate the total upward force exerted on the nail.

\text{Force on nail} = \text{Effort force} \times \text{MA} = 40 \times 6 = 240\text{ lbs}

.

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

Anahtar Kavram

Mechanical Advantage and Torque Balance in First-Class Levers

Soru 20Soru →

Identify the lever classification and mechanical advantage principles for the following common tools. Match each real-world application to its correct mechanical description.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

The pry bar is a Class 1 lever, the paper cutter is a Class 2 lever, and the tweezers are a Class 3 lever.

By analyzing the geometry of each tool, we find that the pry bar places the fulcrum between the applied force and the resistance (Class 1). The paper cutter places the resistance between the fulcrum and the applied force (Class 2). Finally, the tweezers place the applied force between the fulcrum and the resistance (Class 3).

Adım Adım Çözüm

1

Identify the three main components of a lever for each tool.

Pry bar: Fulcrum is the block. Paper cutter: Load is the paper. Tweezers: Effort is applied at the middle.

The relative physical position of the fulcrum (pivot), effort (input force), and load (output force or resistance) determines the lever class.

2

Determine the lever class based on the component located in the middle.

Pry bar: Fulcrum is in the middle (Class 1). Paper cutter: Load is in the middle (Class 2). Tweezers: Effort is in the middle (Class 3).

Applying standard mechanical classification definitions (using the FLE mnemonic: Fulcrum in middle = Class 1, Load in middle = Class 2, Effort in middle = Class 3).

3

Evaluate the mechanical advantage (MA) characteristics for each class.

Class 1 MA can vary. Class 2 always has MA > 1 (multiplies force). Class 3 always has MA < 1 (multiplies distance).

MA is defined as the ratio of the effort arm length to the load arm length.

Anahtar Kavram

Lever Classification and Mechanical Advantage

Levers and Fulcrums

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Alternatif Yöntem

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