Simple Machines Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 4

Question 61Question →

A logistics team can use either a $12\text{-foot}$ ramp or a $24\text{-foot}$ ramp to slide heavy crates onto a loading platform that is $4\text{ feet}$ above the ground. Assuming friction is negligible, which of the following statements is true regarding the use of the $24\text{-foot}$ ramp compared to the $12\text{-foot}$ ramp?

The required effort force is halved, but the crate must be pushed twice as far, resulting in the same amount of total work.

The required effort force is halved, and the total work done to move the crate is also reduced by half.

The greater surface area of the

24\text{-foot}

ramp decreases the friction applied to the crate, which reduces the total work required.

Since the crate's mass is supported over a longer distance, its overall weight decreases, making it easier to lift.

Show answer & explanation

Answer: The required effort force is halved, but the crate must be pushed twice as far, resulting in the same amount of total work.

Answer

The longer ramp reduces the effort force by half because it is twice as long, but because the pushing distance is doubled, the total work remains exactly the same.

An inclined plane trades distance for reduced force. Doubling the ramp's length doubles the mechanical advantage, thus halving the effort force required. However, because the force must be applied over twice the distance to reach the same vertical height, the total work (Force × Distance) remains exactly the same, obeying the law of conservation of energy.

Step-by-Step Solution

1

Calculate the ideal mechanical advantage (IMA) of both ramps.

The

12\text{-foot}

ramp has an IMA of

3

(

12/4

), and the

24\text{-foot}

ramp has an IMA of

6

(

24/4

).

IMA is found by dividing the slope length by the vertical height. It determines the factor by which effort force is reduced.

2

Compare the effort forces required for each ramp.

Because the

24\text{-foot}

ramp has twice the IMA of the

12\text{-foot}

ramp (

6

vs

3

), it requires exactly half the effort force to push the same crate.

Effort force is inversely proportional to mechanical advantage.

3

Analyze the total work done in both scenarios.

Work equals force multiplied by distance (

W = F \times d

). Applying half the force over twice the distance yields identical total work.

Simple machines do not change the total amount of work required; they only trade increased distance for decreased force.

Key Concept

Conservation of work and ideal mechanical advantage (IMA) on an inclined plane.

Estimated Time:1m 0s

Question 62Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

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.

Step-by-Step Solution

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.

Key Concept

Identifying Lever Classes and Mechanical Advantage

Estimated Time:2m 0s

Question 63Question →

A manufacturer is redesigning a manual screw press used for stamping metal. The press currently uses a single-start threaded spindle turned by a handwheel. To reduce the physical effort required by the operator to apply the necessary stamping force, which of the following design modifications should be made?

Increase the thread pitch of the spindle

Decrease the diameter of the handwheel

Increase the number of threads per inch (TPI) on the spindle

Replace the right-handed threaded spindle with a left-handed one of the same pitch

Show answer & explanation

Answer: Increase the number of threads per inch (TPI) on the spindle

Answer

Increase the number of threads per inch (TPI) on the spindle

A screw is essentially an inclined plane wrapped around a central cylinder. Increasing the number of threads per inch (TPI) decreases the pitch (the distance between adjacent threads). A smaller pitch acts like a shallower inclined plane, which increases the mechanical advantage of the system. With a higher mechanical advantage, the operator needs to apply less effort force to achieve the exact same clamping or stamping pressure.

Step-by-Step Solution

1

Identify the factors that determine the mechanical advantage of a screw mechanism.

Mechanical advantage depends on the ratio of the effort distance (handwheel circumference) to the load distance (thread pitch).

Understanding this relationship is required to determine how design changes affect the required input effort.

2

Evaluate the effect of increasing the threads per inch (TPI).

A higher TPI means the threads are closer together, resulting in a smaller thread pitch.

TPI and thread pitch have a direct inverse relationship.

3

Relate the smaller pitch to mechanical advantage.

A smaller pitch means the screw advances less per revolution, acting like a shallower inclined plane. This increases the overall mechanical advantage.

A higher mechanical advantage dictates that less input effort force is required to achieve the same output stamping force.

Key Concept

Mechanical Advantage of Screws

Estimated Time:45s

Question 64Question →

Match each complex pulley system scenario to the correct required effort force to lift the load.

Click a left item, then click its matching right item

Show answer & explanation

Answer

System 1 requires 250 N, System 2 requires 200 N, System 3 requires 150 N, and System 4 requires 50 N.

Each system's effort force is determined by dividing the given load by the correct overall mechanical advantage, properly factoring in mechanical efficiency, geometric constraints, and compound combinations.

Step-by-Step Solution

1

Calculate the required effort for the block and tackle system.

Ideal MA = 5. Actual MA = 4. Effort = 250 N.

Actual MA is the product of ideal MA and mechanical efficiency. Effort is found by dividing the load by the actual MA.

2

Calculate the required effort for the compound series pulley system.

Total MA = 8. Effort = 200 N.

For multiple independent movable pulleys connected in series, their individual mechanical advantages (2) are multiplied together (

2 \times 2 \times 2

).

3

Calculate the required effort for the combined incline and movable pulley system.

Incline MA = 5/3, Pulley MA = 2. Total MA = 10/3. Effort = 150 N.

The total mechanical advantage of a compound simple machine system is the product of the component machines' MAs.

4

Calculate the required effort for the differential pulley hoist.

MA = 30. Effort = 50 N.

The MA formula for a differential pulley relies on the difference between the outer and inner radii:

2R / (R - r)

.

Key Concept

Calculating advanced mechanical advantage across varied complex pulley configurations

Estimated Time:4m 0s

Question 65Question →

A portable hand-cranked emergency generator utilizes a compound gear train to step up the rotational speed of its internal rotor. The hand crank is directly attached to the driving gear (Gear A), which has $60$ teeth. Gear A meshes with Gear B, which has $15$ teeth. Gear B is fixed to the same solid shaft as Gear C, which has $45$ teeth. Finally, Gear C meshes with Gear D, which has $10$ teeth and is attached to the generator's rotor.

If a person turns the hand crank at a steady speed of $40\text{ RPM}$ , what is the resulting speed of the generator's rotor in $\text{RPM}$ ?

Show answer & explanation

Answer: 720

Answer

720

In a compound gear train, the overall mechanical advantage (or speed ratio) is found by evaluating each meshing pair separately. The first pair (A to B) increases the speed by a factor of

4

(since

60/15 = 4

). The input speed is

40\text{ RPM}

, so Gear B spins at

160\text{ RPM}

. Because Gear B and Gear C are fixed to the same shaft, Gear C also spins at

160\text{ RPM}

. The second meshing pair (C to D) increases the speed by a factor of

4.5

(since

45/10 = 4.5

). Multiplying Gear C's speed by this ratio gives

160 \times 4.5 = 720\text{ RPM}

.

Step-by-Step Solution

1

Determine the rotational speed of Gear B.

160\text{ RPM}

Gear A (

60

teeth) drives Gear B (

15

teeth). The speed multiplier is

60 / 15 = 4

. Therefore, Gear B spins at

40 \times 4 = 160\text{ RPM}

.

2

Determine the rotational speed of Gear C.

160\text{ RPM}

Gears mounted on the exact same solid shaft rotate at the same speed, regardless of their size.

3

Determine the rotational speed of Gear D (the rotor).

720\text{ RPM}

Gear C (

45

teeth) drives Gear D (

10

teeth). The speed multiplier is

45 / 10 = 4.5

. Therefore, Gear D spins at

160 \times 4.5 = 720\text{ RPM}

.

Key Concept

Compound Gear Train Speed Ratios

Question 66Question →

A depot worker must slide a $1200\text{ lb}$ steel safe up a straight ramp to a storage platform that is $3\text{ feet}$ above the main floor. The worker uses a mechanical winch that can pull with a maximum force of $240\text{ lbs}$ . Assuming a frictionless surface, what is the minimum ramp length required to move the safe?

Show answer & explanation

Answer: 15

Answer

15

The mechanical advantage required to move a

1200\text{ lb}

load using only

240\text{ lbs}

of force is

1200 / 240 = 5

. Because the Ideal Mechanical Advantage of an inclined plane is its length divided by its height, a ramp with an IMA of

5

lifting a load

3\text{ feet}

vertically must be

5 \times 3 = 15\text{ feet}

long.

Step-by-Step Solution

1

Calculate the required Ideal Mechanical Advantage (IMA).

IMA = 5

The mechanical advantage is the ratio of the load force to the maximum effort force (

1200 / 240 = 5

).

2

Set up the equation for the mechanical advantage of an inclined plane.

5 = Length / 3

For an inclined plane, IMA is equal to the length of the slope divided by the vertical height.

3

Solve for the ramp length.

Length = 15 feet

Multiplying the IMA by the height (

5 \times 3

) gives the minimum required ramp length to achieve that mechanical advantage.

Key Concept

Ideal Mechanical Advantage (IMA) of an Inclined Plane

Question 67Question →

Based on fundamental mechanical principles, match each basic gear configuration with its resulting effect on the output gear.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings connect gear size relationships to their speed/torque trade-offs, and gear train arrangements to their rotational direction outcomes.

The correct pairings follow the fundamental laws of gear mechanics: size differences dictate speed and torque tradeoffs (larger driven gears increase torque, smaller driven gears increase speed), while the number of gears determines rotation direction (even numbers reverse direction, odd numbers maintain it).

Step-by-Step Solution

1

Analyze the size relationship between the driving and driven gears in the first two configurations.

A smaller driver yields lower speed but higher torque. A larger driver yields higher speed but lower torque.

Torque and speed have an inverse relationship determined by the gear ratio.

2

Evaluate the effect of meshing identical gears.

Identical gears do not change speed or torque, but they do reverse the direction of rotation.

Adjacent meshing gears must always spin in opposite directions, and a 1:1 size ratio means no mechanical advantage is gained or lost.

3

Determine the impact of an idler gear.

An idler gear ensures the output gear spins in the same direction as the input gear.

The idler acts as a middle step, reversing the opposite rotation back to the original direction.

Key Concept

Basic Gear Relationships and Mechanical Advantage

Estimated Time:1m 0s

Question 68Question →

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}

Show answer & explanation

Answer: $20\text{ lbs}$

Answer

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}

).

Step-by-Step Solution

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.

Key Concept

Torque Equilibrium and Lever Mechanics

Alternative Method

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}

.

Estimated Time:1m 30s

Question 69Question →

A stonemason uses a steel wedge to fracture a granite boulder. The wedge has a penetration length of $160\text{ mm}$ and tapers to a maximum width (thickness) of $20\text{ mm}$ . Assuming an ideal mechanical system with zero friction, if the mason applies a downward effort force of $450\text{ N}$ to the head of the wedge, what is the maximum lateral outward force, in Newtons, that the wedge exerts on the stone?

Show answer & explanation

Answer: 3600

Answer

3600

The correct lateral outward force is 3600 N. This is found by first determining the ideal mechanical advantage (IMA) of the wedge, which is its length divided by its thickness (160 / 20 = 8). Multiplying this IMA by the input effort force (450 N) gives the total ideal output force exerted on the stone.

Step-by-Step Solution

1

Calculate the Ideal Mechanical Advantage (IMA) of the wedge.

IMA = \frac{160\text{ mm}}{20\text{ mm}} = 8

The mechanical advantage of a wedge is determined by the ratio of its penetration length to its maximum thickness.

2

Calculate the output resistance force.

Output\ Force = 450\text{ N} \times 8 = 3600\text{ N}

In an ideal machine, the output force is equal to the input effort force multiplied by the mechanical advantage.

Key Concept

Mechanical Advantage of a Wedge

Question 70Question →

In a simple mechanical drive system, a small driving gear with 10 teeth meshes directly with a larger driven gear that has 30 teeth. If the driving gear is spinning clockwise at $120\text{ RPM}$ , what is the rotational speed and direction of the driven gear?

40\text{ RPM}

, counterclockwise

360\text{ RPM}

, counterclockwise

40\text{ RPM}

, clockwise

360\text{ RPM}

, clockwise

Show answer & explanation

Answer: $40\text{ RPM}$ , counterclockwise

Answer

The driven gear will rotate at

40\text{ RPM}

in a counterclockwise direction.

The correct answer accurately applies both fundamental rules of simple two-gear trains: speed reduction and direction reversal. Because the driven gear has three times as many teeth (30 vs 10), it turns at one-third the speed (40 RPM). Because it meshes directly with the clockwise-turning driving gear, it must turn counterclockwise.

Step-by-Step Solution

1

Determine the direction of rotation for the driven gear.

Counterclockwise

When two external gears mesh directly, they rotate in opposite directions. Since the driving gear rotates clockwise, the driven gear must rotate counterclockwise.

2

Calculate the speed ratio and the output speed.

40\text{ RPM}

The speed of the driven gear is inversely proportional to its number of teeth. The ratio of driving teeth to driven teeth is 10/30 (or 1/3). Multiplying the input speed (

120\text{ RPM}

) by this ratio gives

120 \times (1/3) = 40\text{ RPM}

.

Key Concept

Two adjacent meshing gears always rotate in opposite directions, and a larger driven gear turns slower than a smaller driving gear in direct proportion to their tooth counts.

Estimated Time:45s

Question 71Question →

A combat engineer squad needs to push a $60\text{ kg}$ supply crate up a smooth, frictionless ramp onto a platform. The platform is $2\text{ meters}$ above the ground, and the ramp is $6\text{ meters}$ long. Assuming the acceleration due to gravity is $10\text{ m/s}^2$ , what is the ideal effort force required to push the crate up the ramp?

20\text{ N}

200\text{ N}

600\text{ N}

1200\text{ N}

Show answer & explanation

Answer: $200\text{ N}$

Answer

The required effort force is

200\text{ N}

.

The ideal mechanical advantage (IMA) of an inclined plane is calculated by dividing its length by its height (6 m / 2 m = 3). The crate's weight is its mass multiplied by gravity (60 kg × 10 m/s² = 600 N). The effort force required is the weight divided by the IMA (600 N / 3 = 200 N).

Step-by-Step Solution

1

Calculate the weight (gravitational force) of the crate.

Weight = 60\text{ kg} \times 10\text{ m/s}^2 = 600\text{ N}

Effort force must oppose the weight (force of gravity) pulling the object down, not just the mass.

2

Calculate the ideal mechanical advantage (IMA) of the ramp.

IMA = \frac{6\text{ m}}{2\text{ m}} = 3

IMA compares the length of the ramp (distance the effort is applied) to the vertical height the load is lifted.

3

Calculate the effort force.

Effort = \frac{600\text{ N}}{3} = 200\text{ N}

The inclined plane reduces the required force by a factor equal to its mechanical advantage.

Key Concept

Ideal Mechanical Advantage of Inclined Planes

Question 72Question →

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?

Show answer & explanation

Answer: 480

Answer

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.

Step-by-Step Solution

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.

Key Concept

Compound Lever Systems and Torque Balance

Estimated Time:1m 30s

Question 73Question →

A small water pump uses a direct gear drive system. The driving gear, attached to an electric motor, has $20$ teeth and spins at $300\text{ rpm}$ . It meshes directly with a driven gear on the pump shaft that has $60$ teeth. What is the rotational speed of the driven gear in rpm?

Show answer & explanation

Answer: 100

Answer

100 rpm

Because the driven gear has three times as many teeth as the driving gear (

60

compared to

20

), it requires three times the effort to turn. Therefore, it will rotate at exactly one-third the speed of the driving gear. Dividing the driving speed of

300\text{ rpm}

by

3

yields

100\text{ rpm}

.

Step-by-Step Solution

1

Recall the fundamental gear relationship formula.

S_1 \times T_1 = S_2 \times T_2

The product of the speed and number of teeth of the driving gear equals the product of the speed and number of teeth of the driven gear.

2

Substitute the given values into the formula.

300 \times 20 = S_2 \times 60

We are given the driving gear speed (

300\text{ rpm}

) and teeth (

20

), along with the driven gear teeth (

60

).

3

Solve for the driven gear speed (

S_2

).

S_2 = \frac{6000}{60} = 100\text{ rpm}

Isolating

S_2

provides the final rotational speed of the driven gear.

Key Concept

The speed of a driven gear is inversely proportional to its number of teeth compared to the driving gear.

Question 74Question →

A heavy machinery technician needs to slightly lift a generator weighing $1000\text{ lbs}$ to slide a permanent leveling mount underneath it. To do this, the technician drives a solid steel wedge under the edge of the generator. If the technician can apply a maximum driving effort of $200\text{ lbs}$ and the wedge has a penetration length of $20\text{ inches}$ , what is the maximum thickness the wedge can be to successfully lift the load? (Assume an ideal system with no friction.)

100\text{ inches}

4\text{ inches}

5\text{ inches}

0.25\text{ inches}

Show answer & explanation

Answer: $4\text{ inches}$

Answer

The maximum thickness the wedge can be is

4\text{ inches}

.

To find the maximum thickness, first determine the required mechanical advantage (MA) by dividing the load force (

1000\text{ lbs}

) by the effort force (

200\text{ lbs}

), which equals 5. Because the MA of an ideal wedge is equal to its length divided by its thickness, you divide the length (

20\text{ inches}

) by the MA (

5

). This calculation yields a maximum thickness of

4\text{ inches}

.

Step-by-Step Solution

1

Calculate the required Mechanical Advantage (MA) for the task.

MA = \frac{\text{Load Force}}{\text{Effort Force}} = \frac{1000\text{ lbs}}{200\text{ lbs}} = 5

The mechanical advantage determines how much the simple machine must multiply the input force to overcome the load.

2

Relate the mechanical advantage to the geometry of the wedge.

MA = \frac{\text{Length}}{\text{Thickness}} \Rightarrow 5 = \frac{20\text{ inches}}{\text{Thickness}}

In an ideal wedge, the ratio of the input distance (penetration length) to the output distance (vertical lift or thickness) equals the mechanical advantage.

3

Solve the equation for the maximum thickness.

\text{Thickness} = \frac{20\text{ inches}}{5} = 4\text{ inches}

Rearranging the formula isolates the unknown variable. A wedge thicker than

4\text{ inches}

would provide an MA of less than 5, making the

200\text{ lbs}

effort insufficient to lift the generator.

Key Concept

Mechanical Advantage of Wedges

Question 75Question →

Arrange the following inclined planes in order from the LEAST ideal mechanical advantage (IMA) to the GREATEST ideal mechanical advantage.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from least to greatest mechanical advantage is: the 10-foot ramp (IMA=2), the 12-foot ramp (IMA=4), the 15-foot ramp (IMA=5), and the 16-foot ramp (IMA=8).

The ideal mechanical advantage (IMA) of an inclined plane is calculated by dividing the length of the slope by its vertical height. A longer slope relative to its height yields a higher mechanical advantage, meaning less effort force is required. Ordering the ramps by their calculated IMA values gives 2 (least), 4, 5, and 8 (greatest).

Step-by-Step Solution

1

Recall the formula for the ideal mechanical advantage (IMA) of an inclined plane.

\text{IMA} = \frac{\text{Length of Incline}}{\text{Vertical Height}}

The mechanical advantage quantifies how much the inclined plane multiplies the input force or reduces the required effort.

2

Calculate the IMA for each given ramp.

10-foot ramp:

10 \div 5 = 2

. 12-foot ramp:

12 \div 3 = 4

. 16-foot ramp:

16 \div 2 = 8

. 15-foot ramp:

15 \div 3 = 5

.

Calculating the individual values allows for direct comparison.

3

Order the calculated values from least to greatest.

The values in ascending order are 2, 4, 5, and 8.

The question specifically asks for the arrangement starting from the lowest mechanical advantage to the highest.

Key Concept

The ideal mechanical advantage of a ramp is determined by the ratio of its length to its vertical height.

Question 76Question →

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

Show answer & explanation

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

Answer

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}

.

Step-by-Step Solution

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.

Key Concept

Torque balance and Third-class levers

Question 77Question →

Match each wedge geometry or characteristic with its corresponding mechanical property or practical application.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches align the high length-to-thickness wedge with high mechanical advantage, the low length-to-thickness wedge with low mechanical advantage, the single-beveled wedge with one-directional pushing, and the double-beveled wedge with symmetrical splitting.

Each wedge geometry dictates a specific mechanical trade-off. Long and narrow profiles maximize mechanical advantage to reduce required effort, while short and wide profiles sacrifice force multiplication for rapid, wide separation. Furthermore, the bevel symmetry determines the specific direction of the applied lateral splitting forces.

Step-by-Step Solution

1

Evaluate the length-to-thickness ratios for mechanical advantage.

A high ratio means a long, thin wedge (high MA), while a low ratio means a short, thick wedge (low MA).

The ideal mechanical advantage for a wedge is calculated as the length of the wedge divided by its maximum thickness.

2

Evaluate the structural symmetry (bevel shapes).

A single-bevel focuses material displacement in one direction, while a double-bevel splits material evenly on both sides.

The physical symmetry of the wedge's inclined planes determines the direction of the output forces relative to the driving force.

3

Match the calculated geometries to their practical tool applications.

High ratio corresponds to high MA; low ratio corresponds to low MA; single-bevel corresponds to flush chiseling; double-bevel corresponds to log splitting.

Connecting theoretical physics principles to practical tool designs verifies the correct logical pairings.

Key Concept

Wedge Geometry and Mechanical Advantage

Question 78Question →

A shipwright uses a steel wedge to precisely align a heavy hull plate. The wedge has a length of $24\text{ cm}$ and a maximum heel thickness of $4\text{ cm}$ . If the shipwright strikes the heel, applying an average effort force of $600\text{ N}$ , what is the theoretical maximum lateral lifting force exerted by the wedge on the hull plate?

2400\text{ N}

100\text{ N}

3600\text{ N}

600\text{ N}

Show answer & explanation

Answer: $3600\text{ N}$

Answer

3600\text{ N}

The mechanical advantage (IMA) of a wedge is calculated by dividing its length by its maximum thickness. In this case,

24\text{ cm} / 4\text{ cm} = 6

. The theoretical maximum lateral force is the effort force multiplied by the IMA, which is

600\text{ N} \times 6 = 3600\text{ N}

.

Step-by-Step Solution

1

Calculate the ideal mechanical advantage (IMA) of the wedge.

IMA = \frac{24\text{ cm}}{4\text{ cm}} = 6

The IMA of a wedge is the ratio of its penetration length to its maximum thickness.

2

Calculate the theoretical maximum lateral force.

F_{\text{out}} = 6 \times 600\text{ N} = 3600\text{ N}

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

Key Concept

The ideal mechanical advantage of a wedge is determined by the ratio of its length to its thickness.

Question 79Question →

A theater stage crew needs to roll a heavy $400$ -pound lighting rig up to a stage platform that is $5$ feet high. They can build either a steep ramp that is $10$ feet long or a gentle ramp that is $20$ feet long. Assuming the ramp surfaces and wheels are perfectly frictionless, which of the following statements correctly compares the two ramp options?

The

20

-foot ramp reduces both the required effort force and the total work done by half compared to the

10

-foot ramp.

The

10

-foot ramp requires less total work than the

20

-foot ramp because the lighting rig is moved across a shorter total distance.

The

20

-foot ramp requires half the effort force of the

10

-foot ramp, but the total work done remains exactly the same for both.

The

20

-foot ramp requires less effort force because placing the rig on a shallower angle effectively reduces the mass of the rig.

Show answer & explanation

Answer: The $20$ -foot ramp requires half the effort force of the $10$ -foot ramp, but the total work done remains exactly the same for both.

Answer

The

20

-foot ramp requires half the effort force of the

10

-foot ramp, but the total work done remains exactly the same for both.

The Ideal Mechanical Advantage (IMA) of an inclined plane is its length divided by its height. The

20

-foot ramp has an IMA of

4

(

20/5

), requiring

100

lbs of effort force. The

10

-foot ramp has an IMA of

2

(

10/5

), requiring

200

lbs of effort force. Therefore, the longer ramp requires half the effort force. However, because simple machines do not change the total amount of work done (

\text{Work} = \text{Force} \times \text{Distance}

), both ramps require

2000

ft-lbs of total work to lift the rig

5

feet vertically.

Step-by-Step Solution

1

Calculate the Ideal Mechanical Advantage (IMA) for both ramps using the formula

IMA = \text{Length} / \text{Height}

.

Steep ramp IMA =

10 / 5 = 2

. Gentle ramp IMA =

20 / 5 = 4

.

IMA determines how much the inclined plane multiplies the input force.

2

Determine the required effort force for each ramp using the formula

\text{Effort} = \text{Load} / IMA

.

Steep ramp effort =

400 / 2 = 200

lbs. Gentle ramp effort =

400 / 4 = 100

lbs.

This mathematical relationship shows that the

20

-foot ramp requires exactly half the effort force of the

10

-foot ramp.

3

Calculate the total work done for both scenarios using

\text{Work} = \text{Force} \times \text{Distance}

(along the ramp).

Work for steep ramp =

200 \text{ lbs} \times 10 \text{ ft} = 2000 \text{ ft-lbs}

. Work for gentle ramp =

100 \text{ lbs} \times 20 \text{ ft} = 2000 \text{ ft-lbs}

.

This confirms the principle of conservation of energy: while simple machines alter the force required, they do not change the total work done against gravity.

Key Concept

Ideal Mechanical Advantage and the Work-Force Tradeoff

Question 80Question →

A bucket is being raised from a well using a windlass mechanism (a type of wheel and axle). The cylindrical drum (axle) around which the rope winds has a diameter of $8 \text{ inches}$ . The crank handle used to turn the drum has a length (radius) from the center of rotation of $20 \text{ inches}$ . If the bucket is raised vertically by exactly $15 \text{ feet}$ , what is the total distance, in feet, that the crank handle must travel along its circular path? (Assume the rope does not overlap itself on the drum).

Show answer & explanation

Answer: 75

Answer

75

In any simple machine, assuming 100% efficiency, the work put in equals the work put out. The Ideal Mechanical Advantage (IMA) for a wheel and axle is the ratio of the effort radius to the load radius. Here, the load radius (the drum) is

4 \text{ inches}

(half of the

8 \text{ inch}

diameter) and the effort radius (the crank) is

20 \text{ inches}

. Thus, the IMA is

20 / 4 = 5

. This geometric relationship dictates that the effort force must travel

5

times further than the load to conserve work. If the load is lifted

15 \text{ feet}

, the effort distance is

15 \times 5 = 75 \text{ feet}

.

Step-by-Step Solution

1

Calculate the drum's radius.

4 \text{ inches}

The Ideal Mechanical Advantage compares radii, but the drum's dimension was provided as a diameter.

2

Calculate the Ideal Mechanical Advantage (IMA).

IMA =

5

Dividing the effort radius (

20 \text{ inches}

) by the load radius (

4 \text{ inches}

) gives the mechanical advantage of the system.

3

Calculate the total distance traveled by the crank handle.

75 \text{ feet}

By the principle of conservation of energy, the effort distance must be equal to the load distance (

15 \text{ feet}

) multiplied by the IMA (

5

).

Key Concept

Ideal Mechanical Advantage (IMA) and Force-Distance Tradeoff

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