Simple Machines Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 2

Question 21Question →

Calculate the total effort force required for each scenario and arrange the following inclined plane configurations in order from the GREATEST required effort force to the LEAST required effort force.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from greatest to least effort force is: 240 lb crate (90 lb effort), 160 lb crate (80 lb effort), 300 lb crate (75 lb effort), and 450 lb crate (70 lb effort).

The total effort force required is the sum of the ideal effort (weight divided by IMA) and the friction force. For the 240 lb crate, the total effort is 90 lb. For the 160 lb crate, it is 80 lb. For the 300 lb crate, it is 75 lb. For the 450 lb crate, it is 70 lb. Arranging these from greatest to least yields the sequence of 90 lb, 80 lb, 75 lb, and 70 lb.

Step-by-Step Solution

1

Calculate the Ideal Mechanical Advantage (IMA) for each ramp.

Item 1: 5. Item 2: 2. Item 3: 10. Item 4: 4.

IMA is determined by the ratio of the ramp's length to its vertical height (

IMA = \frac{L}{h}

).

2

Determine the ideal effort force required to move each load without friction.

Item 1: 60 lb. Item 2: 80 lb. Item 3: 45 lb. Item 4: 60 lb.

The ideal effort force is the weight of the load divided by the IMA (

F_{\text{ideal}} = \frac{W}{IMA}

).

3

Add the constant friction force to the ideal effort force to find the total required effort.

Item 1: 75 lb. Item 2: 80 lb. Item 3: 70 lb. Item 4: 90 lb.

Friction acts against the movement up the ramp, so it must be added to the ideal effort force to find the true required effort.

4

Arrange the final total effort values in descending order.

90 lb, 80 lb, 75 lb, 70 lb.

The question asks for the order from greatest required effort force to least.

Key Concept

Calculating total required effort force on an inclined plane by combining Ideal Mechanical Advantage (IMA) and frictional forces.

Question 22Question →

A construction team uses various temporary ramps to roll heavy equipment platforms to different elevated levels. Assuming friction is negligible, arrange the following ramp configurations in order from the LEAST effort force required to the GREATEST effort force required.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from least effort force to greatest effort force is: the 25 ft ramp, followed by the 20 ft ramp, then the 18 ft ramp, and finally the 14 ft ramp.

The effort force required to move an object up a frictionless inclined plane is determined by the ratio of the height to the length of the plane. A lower height-to-length ratio means a higher mechanical advantage and less required effort force. By calculating this ratio for each ramp, we find the 25 ft ramp requires

\frac{1}{5}

(20%) of the load's weight, the 20 ft ramp requires

\frac{1}{4}

(25%), the 18 ft ramp requires

\frac{1}{3}

(33.3%), and the 14 ft ramp requires

\frac{1}{2}

(50%). Therefore, the sequence is 25 ft ramp, 20 ft ramp, 18 ft ramp, and 14 ft ramp.

Step-by-Step Solution

1

Identify the relationship between effort force, load, ramp length, and height.

\text{Effort Force} = \text{Load} \times \left(\frac{\text{height}}{\text{length}}\right)

.

The effort force is inversely proportional to the ideal mechanical advantage (

\text{IMA} = \frac{\text{length}}{\text{height}}

).

2

Calculate the effort force fraction for each ramp configuration.

25 ft ramp:

\frac{5}{25} = \frac{1}{5}

; 20 ft ramp:

\frac{5}{20} = \frac{1}{4}

; 18 ft ramp:

\frac{6}{18} = \frac{1}{3}

; 14 ft ramp:

\frac{7}{14} = \frac{1}{2}

.

Comparing these fractions allows us to determine the relative force required for an identical load.

3

Order the calculated fractions from lowest to highest.

\frac{1}{5} < \frac{1}{4} < \frac{1}{3} < \frac{1}{2}

.

The prompt asks to arrange the setups from the least effort force to the greatest effort force.

Key Concept

Ideal Mechanical Advantage and Effort Force on an Inclined Plane

Question 23Question →

Rank the following inclined plane configurations in order from the LEAST effort force required to the GREATEST effort force required. Assume ideal conditions with no friction.

Drag items to arrange them in the correct order

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Answer

The correct sequence from least to greatest effort force is: the 500 lb load on a 15 ft ramp (100 lbs), followed by the 600 lb load on a 12 ft ramp (150 lbs), then the 400 lb load on an 8 ft ramp (200 lbs), and finally the 750 lb load on a 15 ft ramp (250 lbs).

The effort force required for an inclined plane is calculated by dividing the weight of the load by the Ideal Mechanical Advantage (IMA = Length / Height). Calculating this for each scenario yields: 100 lbs, 150 lbs, 200 lbs, and 250 lbs respectively. Organizing these from least to greatest provides the correct sequence.

Step-by-Step Solution

1

Calculate the effort force for the

750\text{ lb}

cargo.

Force =

750 \times (5 / 15) = 750 \times (1 / 3) = 250\text{ lbs}

.

To determine the actual force needed, divide the load by the Ideal Mechanical Advantage (IMA = Length / Height).

2

Calculate the effort force for the

400\text{ lb}

cargo.

Force =

400 \times (4 / 8) = 400 \times 0.5 = 200\text{ lbs}

.

The ramp's IMA is 2, meaning it halves the required lifting force.

3

Calculate the effort force for the

600\text{ lb}

cargo.

Force =

600 \times (3 / 12) = 600 \times 0.25 = 150\text{ lbs}

.

The ramp has an IMA of 4, which divides the

600\text{ lb}

load by 4.

4

Calculate the effort force for the

500\text{ lb}

cargo.

Force =

500 \times (3 / 15) = 500 \times 0.2 = 100\text{ lbs}

.

This configuration provides an IMA of 5, which significantly reduces the force.

5

Order the calculated effort forces from least to greatest.

The correct order is

100\text{ lbs}

,

150\text{ lbs}

,

200\text{ lbs}

,

250\text{ lbs}

.

The question specifically requests the sequence from least effort to greatest effort.

Key Concept

Ideal Mechanical Advantage and Effort Force in Inclined Planes

Estimated Time:2m 0s

Question 24Question →

Rank the following jackscrew configurations in order from the highest Ideal Mechanical Advantage (IMA) to the lowest IMA.

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Answer

The correct order from highest to lowest Ideal Mechanical Advantage is: the 8-inch handle with 8 TPI, followed by the 10-inch handle with 6 TPI, then the 12-inch handle with 4 TPI, and finally the 15-inch handle with 3 TPI.

The Ideal Mechanical Advantage of a jackscrew is determined by multiplying the circumference of the effort circle (

2 \times \pi \times L

) by the Threads Per Inch (TPI). Because

2\pi

is constant for all options, we can simply compare the product of the handle length (

L

) and the TPI. The 8-inch/8 TPI setup yields the highest proportional factor (

8 \times 8 = 64

), followed by the 10-inch/6 TPI setup (

60

), the 12-inch/4 TPI setup (

48

), and finally the 15-inch/3 TPI setup (

45

).

Step-by-Step Solution

1

Identify the formula for the Ideal Mechanical Advantage (IMA) of a jackscrew.

IMA = (2 × π × Handle Length) / Pitch. Since Pitch = 1 / TPI, this simplifies to IMA = 2 × π × Handle Length × TPI.

To find the relative IMA, we need to compare the product of the handle length and TPI for each configuration.

2

Calculate the proportional IMA factor (Handle Length × TPI) for each jackscrew configuration.

8-inch handle, 8 TPI: 8 × 8 = 64. 10-inch handle, 6 TPI: 10 × 6 = 60. 12-inch handle, 4 TPI: 12 × 4 = 48. 15-inch handle, 3 TPI: 15 × 3 = 45.

This provides a comparative numerical value to rank the mechanical advantages without needing to calculate the exact factor of 2π.

3

Order the configurations based on the calculated factors.

64 > 60 > 48 > 45.

The prompt requests ranking from the highest IMA to the lowest IMA.

Key Concept

Ideal Mechanical Advantage of a Screw

Question 25Question →

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

Click a left item, then click its matching right item

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Answer

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).

Step-by-Step Solution

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.

Key Concept

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).

Estimated Time:2m 0s

Question 26Question →

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

Click a left item, then click its matching right item

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Answer

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).

Step-by-Step Solution

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.

Key Concept

Lever Classes, Torque, and Mechanical Advantage

Question 27Question →

A technician is selecting a bench vise to clamp heavy metal parts securely. They want to use the vise that provides the greatest mechanical advantage, meaning it requires the least amount of effort force on the handle to generate a strong clamping force. If all available vises have the exact same handle length, which of the following thread designs should the technician choose?

A design with a smaller thread pitch (more threads per inch)

A design with a larger thread pitch (fewer threads per inch)

A design that uses left-handed threads instead of right-handed threads

A design with a thicker solid central threaded shaft

Show answer & explanation

Answer: A design with a smaller thread pitch (more threads per inch)

Answer

A design with a smaller thread pitch (more threads per inch)

A screw is essentially an inclined plane wrapped around a central cylinder. The thread pitch is the linear distance between adjacent threads. A smaller thread pitch represents a less steep inclined plane, meaning the vise jaw advances a shorter distance for each full revolution of the handle. Because the jaw moves less distance per turn, the effort force required to turn the handle is reduced, resulting in a higher ideal mechanical advantage and a stronger clamping force.

Step-by-Step Solution

1

Identify the relationship between a screw's threads and an inclined plane.

A screw is essentially an inclined plane wrapped around a cylinder.

Understanding the underlying simple machine explains how force is multiplied.

2

Determine how thread pitch affects the steepness of the inclined plane.

A smaller thread pitch (threads are closer together) creates a longer, less steep inclined plane over the same linear distance.

Pitch determines the linear distance the screw travels per revolution.

3

Relate the steepness to mechanical advantage and effort force.

A less steep inclined plane provides a higher mechanical advantage, meaning less effort force on the handle is required to generate the clamping force.

Mechanical advantage is the ratio of effort distance to load distance; a smaller pitch increases the effort distance per unit of clamping movement.

Key Concept

Mechanical Advantage of Screws

Estimated Time:45s

Question 28Question →

A civil engineer is designing a road to reach the summit of a steep mountain. Instead of building a road that goes straight up the slope, the engineer designs a series of gradual switchbacks (zigzags). Ignoring friction and air resistance, how does the switchback design affect the physical task for a vehicle climbing the mountain compared to a direct, steep route?

The design reduces the total amount of work the vehicle must perform because the slope is not as steep.

The design reduces the actual weight of the vehicle against gravity, which reduces the required engine force.

The design requires less force from the vehicle's engine, but the total amount of work done to reach the summit is unchanged.

The design requires less force from the vehicle's engine, which directly decreases the total amount of work performed.

Show answer & explanation

Answer: The design requires less force from the vehicle's engine, but the total amount of work done to reach the summit is unchanged.

Answer

The design requires less force from the vehicle's engine, but the total amount of work done to reach the summit is unchanged.

A switchback acts as an inclined plane. By making the path longer and more gradual, the mechanical advantage increases. This means the vehicle's engine does not have to provide as much force at any given moment to push the vehicle upward. However, because this smaller force must be applied over a much longer distance, the total amount of work done against gravity (Force × Distance) remains exactly the same as taking a shorter, steeper route.

Step-by-Step Solution

1

Identify the simple machine principle at work.

A switchback road acts as an inclined plane, stretching the vertical climb over a much longer physical distance.

To determine which mechanical principles govern the vehicle's climb.

2

Analyze how the inclined plane affects the required force.

Because the slope is more gradual, the mechanical advantage increases, meaning the engine has to provide less effort force to push the vehicle upward at any given moment.

To understand the immediate benefit of using a ramp or switchback.

3

Apply the principle of conservation of work.

Work is equal to Force multiplied by Distance. The switchback decreases the force but increases the distance by the exact same proportion, meaning the total work done to lift the vehicle to the summit remains exactly the same.

To complete the comparison between the switchback and the direct route.

Key Concept

Ideal Mechanical Advantage and Work on an Inclined Plane

Question 29Question →

A mechanical winch uses a compound gear train to lift heavy loads. The input motor drives Gear A (30 teeth) clockwise at $300$ RPM. Gear A meshes directly with Gear B (45 teeth). Gear C (15 teeth) is fixed to the same shaft as Gear B, meaning they rotate together. Gear C meshes with an idler Gear D (30 teeth), which in turn meshes with the output Gear E (60 teeth).

Based on this configuration, what is the rotational speed and direction of the output Gear E?

50 RPM, counter-clockwise

50 RPM, clockwise

150 RPM, clockwise

1800 RPM, counter-clockwise

Show answer & explanation

Answer: 50 RPM, counter-clockwise

Answer

The correct answer is 50 RPM, counter-clockwise.

The correct answer is calculated by tracking the speed reduction and direction reversal across each mesh. Shaft 2 rotates at 200 RPM CCW (

300 \times \frac{30}{45}

). Shaft 3 (idler) rotates at 100 RPM CW (

200 \times \frac{15}{30}

). The final output shaft rotates at 50 RPM CCW (

100 \times \frac{30}{60}

).

Step-by-Step Solution

1

Calculate the speed and direction of the second shaft (Gears B and C).

Gear B speed =

300 \times \frac{30}{45} = 200

RPM. Direction = counter-clockwise.

When meshing, speed is inversely proportional to tooth count. Meshing gears rotate in opposite directions. Because Gears B and C share a shaft, Gear C also rotates at 200 RPM counter-clockwise.

2

Calculate the speed and direction of the idler Gear D.

Gear D speed =

200 \times \frac{15}{30} = 100

RPM. Direction = clockwise.

Gear C drives Gear D. The ratio is

\frac{15}{30}

, and the direction reverses from counter-clockwise to clockwise.

3

Calculate the final speed and direction of output Gear E.

Gear E speed =

100 \times \frac{30}{60} = 50

RPM. Direction = counter-clockwise.

Gear D drives Gear E. The ratio is

\frac{30}{60}

, and the direction reverses from clockwise to counter-clockwise.

Key Concept

Compound Gear Trains and Idler Mechanics

Question 30Question →

A heavy transport truck with a mass of $10\text{ metric tons}$ ( $1\text{ metric ton} = 1000\text{ kg}$ ) needs to deliver supplies to a mountain facility located $500\text{ m}$ vertically above its current position. A direct, steep road up the mountainside to the facility is $1000\text{ m}$ long. Instead, the truck takes a winding switchback road that is $4000\text{ m}$ long to reach the exact same facility. Assuming an ideal system with no friction and using $g = 10\text{ m/s}^2$ , what is the required effort force on the switchback road, and how does the total work done compare to using the direct road?

The effort force is

12.5\text{ N}

, and the total work done is the same as the direct road.

The effort force is

12,500\text{ N}

, and the total work done is one-fourth of the work required on the direct road.

The effort force is

12,500\text{ N}

, and the total work done is the same as the direct road.

The effort force is

1,250\text{ N}

, and the total work done is the same as the direct road.

Show answer & explanation

Answer: The effort force is $12,500\text{ N}$ , and the total work done is the same as the direct road.

Answer

The effort force is

12,500\text{ N}

, and the total work done is the same as the direct road.

The correct answer accurately calculates the truck's weight (

100,000\text{ N}

), finds the ideal mechanical advantage of the switchback road (

4000\text{ m} / 500\text{ m} = 8

), and divides the weight by the IMA to find the effort force (

12,500\text{ N}

). It also correctly identifies that in an ideal system, the total work done depends only on the vertical height and load, remaining constant regardless of the path taken.

Step-by-Step Solution

1

Calculate the total weight (load force) of the truck.

Weight =

10,000\text{ kg} \times 10\text{ m/s}^2 = 100,000\text{ N}

.

The mass must be converted to standard SI units (kilograms) and multiplied by gravity to find the downward force required to lift it.

2

Calculate the Ideal Mechanical Advantage (IMA) of the switchback road.

\text{IMA} = \frac{\text{Length}}{\text{Height}} = \frac{4000\text{ m}}{500\text{ m}} = 8

.

The mechanical advantage of an inclined plane is the ratio of its slope length to its vertical height.

3

Calculate the required effort force on the switchback road.

\text{Effort Force} = \frac{\text{Load}}{\text{IMA}} = \frac{100,000\text{ N}}{8} = 12,500\text{ N}

.

In an ideal system, the required effort force is equal to the load divided by the mechanical advantage.

4

Compare the total work done between the two paths.

The total work remains the same for both the direct road and the switchback road.

According to the conservation of energy, an inclined plane reduces the effort force by increasing the distance traveled, but the total work done (Force

\times

Distance) remains constant in an ideal system.

Key Concept

Ideal Mechanical Advantage of Inclined Planes and Conservation of Work

Question 31Question →

A facility operator needs to open a large industrial gate valve to allow water flow. The valve's stem features a standard right-hand thread with $5$ threads per inch (TPI). The valve is currently fully closed. To allow maximum water flow, the internal gate must be lifted exactly $3\text{ inches}$ . What action must the operator take to fully open the valve?

Turn the handwheel counterclockwise for

15\text{ full revolutions}

.

Turn the handwheel clockwise for

15\text{ full revolutions}

.

Turn the handwheel counterclockwise for

0.6\text{ of a revolution}

.

Turn the handwheel clockwise for

0.6\text{ of a revolution}

.

Show answer & explanation

Answer: Turn the handwheel counterclockwise for $15\text{ full revolutions}$ .

Answer

Turn the handwheel counterclockwise for

15\text{ full revolutions}

.

For a standard right-hand thread, counterclockwise rotation is required to withdraw or lift the mechanism. Since the stem has 5 threads per inch, it takes 5 full revolutions to move the gate 1 inch. To move the gate 3 inches, the operator must turn the handwheel

3 \times 5 = 15\text{ full revolutions}

.

Step-by-Step Solution

1

Determine the required direction of rotation.

Counterclockwise rotation.

Standard right-hand threads advance (close or move away) when turned clockwise. To lift the gate (open or move toward the operator), the handwheel must be turned counterclockwise.

2

Determine the linear distance traveled per revolution (pitch).

Pitch =

1 / 5\text{ inch}

per revolution.

TPI (threads per inch) tells us that

5\text{ threads}

fit in

1\text{ inch}

. The distance of a single thread (one revolution) is the inverse of the TPI.

3

Calculate the total revolutions needed to travel the target distance.

3\text{ inches} \times 5\text{ revolutions/inch} = 15\text{ revolutions}

.

Multiplying the total required distance by the number of threads per inch gives the total number of revolutions required.

Key Concept

Screw Pitch, TPI, and Thread Direction

Estimated Time:1m 0s

Question 32Question →

A mechanic needs to move a $150\text{ kg}$ engine block onto a storage shelf that is $120\text{ cm}$ above the floor. He uses a frictionless inclined plane and applies a constant effort force of $300\text{ N}$ parallel to the ramp's surface to push the block at a constant speed. Assuming the acceleration due to gravity is $10\text{ m/s}^2$ , what is the required length of the inclined plane, and how does the total work done using the ramp compare to the work required to lift the engine vertically?

6.0\text{ m}

; total work done is equal to lifting it vertically

6.0\text{ m}

; total work done is less than lifting it vertically

0.6\text{ m}

; total work done is less than lifting it vertically

600\text{ m}

; total work done is equal to lifting it vertically

Show answer & explanation

Answer: $6.0\text{ m}$ ; total work done is equal to lifting it vertically

Answer

The required length is

6.0\text{ m}

, and the total work done is equal to the work required to lift the engine vertically.

The weight of the engine block is

1500\text{ N}

. To lift it

1.2\text{ m}

vertically requires

1800\text{ J}

of work. Using a frictionless ramp, the exact same

1800\text{ J}

of work is performed by applying a smaller

300\text{ N}

force over a longer distance (

6.0\text{ m}

). Simple machines change the ratio of force and distance but do not alter the total work done.

Step-by-Step Solution

1

Calculate the weight of the engine block and convert the height to meters.

Weight =

150\text{ kg} \times 10\text{ m/s}^2 = 1500\text{ N}

. Height =

1.2\text{ m}

.

Standard SI units (meters) and forces (Newtons) are required for accurate work and energy calculations.

2

Apply the principle of conservation of work (Work Input = Work Output) to solve for the length.

300\text{ N} \times L = 1500\text{ N} \times 1.2\text{ m} \rightarrow 300L = 1800 \rightarrow L = 6.0\text{ m}

.

In a frictionless system, the input work done by the effort force over the ramp's length equals the output work needed to lift the object vertically.

3

Compare the total work done in both scenarios.

Work on ramp (

300\text{ N} \times 6.0\text{ m} = 1800\text{ J}

) equals Work lifted vertically (

1500\text{ N} \times 1.2\text{ m} = 1800\text{ J}

).

Simple machines like inclined planes provide a mechanical advantage by spreading the same amount of work over a longer distance, reducing the required force.

Key Concept

Ideal Mechanical Advantage and Work Conservation on an Inclined Plane

Question 33Question →

A lead screw on a lathe has 8 threads per inch (TPI). How far, in inches, will the carriage move if the screw is turned exactly 12 full revolutions?

Show answer & explanation

Answer: 1.5

Answer

1.5

Because the screw has 8 threads per inch, completing one full revolution advances the screw exactly 1/8 of an inch (0.125 inches). Turning the screw 12 times results in a total linear movement of 12 * 0.125 = 1.5 inches.

Step-by-Step Solution

1

Determine the pitch of the screw.

Pitch = 1 / 8 = 0.125 inches per revolution.

Pitch is the linear distance a screw travels in one complete revolution, which is the inverse of the TPI (Threads Per Inch).

2

Calculate the total linear distance traveled.

Distance = 0.125 inches/revolution * 12 revolutions = 1.5 inches.

Total distance is the product of the distance per revolution and the number of full revolutions.

Key Concept

The relationship between Threads Per Inch (TPI), pitch, and linear travel distance of a screw mechanism.

Alternative Method

Set up a proportion based on the definition of TPI: If 8 revolutions move the carriage 1 inch (since there are 8 threads in one inch), then 12 revolutions will move it 12 / 8 = 1.5 inches.

Estimated Time:45s

Question 34Question →

A delivery crew needs to load a $600\text{ lb}$ safe into the back of a moving truck that sits $4\text{ ft}$ above the ground. They deploy a straight, uniform ramp that is $16\text{ ft}$ long. Assuming friction is negligible, what is the ideal effort force, in pounds, required to push the safe up the ramp?

Show answer & explanation

Answer: 150

Answer

150

By using a ramp, the distance over which the force is applied increases, which proportionally decreases the required force. A 16-foot ramp lifting an object 4 feet provides a mechanical advantage of 4 (16 divided by 4). Therefore, pushing a 600-pound load requires only one-quarter of the weight in force, which is 150 pounds.

Step-by-Step Solution

1

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

IMA = 16 / 4 = 4

The mechanical advantage of an inclined plane is the ratio of its length to its vertical height.

2

Calculate the ideal effort force required to move the load.

Effort Force = 600 / 4 = 150 lbs

The required effort force is the total load divided by the mechanical advantage.

Key Concept

Ideal Mechanical Advantage of an Inclined Plane

Question 35Question →

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}

Show answer & explanation

Answer: $50\text{ lbs}$

Answer

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

).

Step-by-Step Solution

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.

Key Concept

Torque equilibrium in levers accounting for uniform mass distribution.

Question 36Question →

A logistics team needs to roll a heavy generator weighing $900\text{ lbs}$ from the ground onto a raised platform that is $5\text{ feet}$ high. If the team can apply a maximum combined push force of $150\text{ lbs}$ parallel to the ramp surface, what is the minimum ramp length required to accomplish this task? Assume ideal conditions with zero friction.

Show answer & explanation

Answer: 30

Answer

The minimum ramp length required is

30\text{ feet}

.

In an ideal frictionless system, the ratio of the load force to the effort force equals the ratio of the ramp length to its height. A load of

900\text{ lbs}

requiring only

150\text{ lbs}

of effort means the system needs a mechanical advantage of

6

. Since the height is

5\text{ feet}

, the ramp must be

6

times longer than the height, which is

30\text{ feet}

.

Step-by-Step Solution

1

Calculate the required Ideal Mechanical Advantage (IMA).

IMA = \frac{\text{Load}}{\text{Effort}} = \frac{900\text{ lbs}}{150\text{ lbs}} = 6

The mechanical advantage indicates how many times the machine multiplies the effort force. We need a mechanical advantage of 6 to move the

900\text{ lbs}

load with only

150\text{ lbs}

of force.

2

Calculate the required ramp length using the IMA and the given height.

IMA = \frac{\text{Length}}{\text{Height}} \Rightarrow 6 = \frac{\text{Length}}{5\text{ feet}} \Rightarrow \text{Length} = 6 \times 5 = 30\text{ feet}

For an inclined plane in ideal conditions, the mechanical advantage is also the ratio of the ramp's length to its vertical height.

Key Concept

The relationship between effort force, load, ramp length, and vertical height on an ideal inclined plane (Ideal Mechanical Advantage).

Estimated Time:1m 30s

Question 37Question →

A hydraulic log splitter is designed to drive a steel wedge into a large piece of timber. The wedge has a length of $20\text{ cm}$ and a maximum thickness of $5\text{ cm}$ . Assuming an ideal system with no friction, if the hydraulic cylinder applies a downward effort force of $2,000\text{ N}$ to the wedge, which of the following statements correctly identifies the theoretical lateral splitting force exerted on the timber and the underlying physical principle regarding the work done?

The lateral splitting force is

8,000\text{ N}

; the wedge multiplies the effort force, but the total work output cannot exceed the total work input from the cylinder.

The lateral splitting force is

8,000\text{ N}

; the wedge multiplies the effort force and therefore decreases the total amount of work required to split the timber.

The lateral splitting force is

500\text{ N}

; the mechanical advantage is determined by dividing the thickness by the length, which divides the effort force.

The lateral splitting force is

8,000\text{ N}

; the force is multiplied because the larger surface area of the wedge decreases the frictional force opposing the motion.

Show answer & explanation

Answer: The lateral splitting force is $8,000\text{ N}$ ; the wedge multiplies the effort force, but the total work output cannot exceed the total work input from the cylinder.

Answer

The correct answer states that the theoretical lateral splitting force is

8,000\text{ N}

and that the total work output cannot exceed the work input.

The correct answer accurately calculates the Ideal Mechanical Advantage by dividing the length by the thickness (

20 / 5 = 4

), resulting in an output force of

4 \times 2,000 = 8,000\text{ N}

. It also correctly identifies that while the wedge multiplies the effort force, it does not reduce the total amount of work done, adhering to the conservation of energy.

Step-by-Step Solution

1

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

IMA = \frac{\text{Length}}{\text{Thickness}} = \frac{20\text{ cm}}{5\text{ cm}} = 4

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

2

Determine the theoretical lateral splitting force (Output Force).

F_{out} = F_{in} \times IMA = 2,000\text{ N} \times 4 = 8,000\text{ N}

In an ideal machine, the output force is the product of the input effort force and the mechanical advantage.

3

Evaluate the principles of work and energy.

Total work input

\geq

Total work output. Simple machines never reduce the amount of work required.

The Law of Conservation of Energy dictates that while a simple machine can reduce the required effort force by increasing the distance over which it is applied, it cannot reduce the total work done.

Key Concept

Mechanical Advantage of Wedges and Conservation of Work

Question 38Question →

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

Show answer & explanation

Answer: 800 pounds; Third-class lever

Answer

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}

.

Step-by-Step Solution

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.

Key Concept

Torque Balance and Third-Class Levers

Question 39Question →

A motorized winch is pulling a $1,500\text{ lb}$ boat and trailer assembly up a concrete boat ramp at a constant speed. The ramp has a total length of $40\text{ ft}$ and a vertical drop of $8\text{ ft}$ from the parking lot to the water level. If the mechanical efficiency of the ramp system is $80\%$ (due to friction in the wheel bearings and surface resistance), what is the actual effort force, in pounds, the winch must apply to pull the assembly up the ramp?

Show answer & explanation

Answer: 375

Answer

The winch must apply an actual effort force of 375 pounds.

To solve this problem, you must integrate the concepts of Ideal Mechanical Advantage (IMA) and mechanical efficiency. First, find the IMA by dividing the ramp's length by its vertical drop (

40 / 8 = 5

). In an ideal, frictionless system, the effort force required would be the load divided by the IMA (

1500 / 5 = 300\text{ lb}

). However, the problem states the system is only

80\%

efficient due to friction. Efficiency is the ratio of ideal effort to actual effort. To find the actual effort force, divide the ideal effort by the efficiency (

300 / 0.80 = 375\text{ lb}

).

Step-by-Step Solution

1

Calculate the Ideal Mechanical Advantage (IMA).

\text{IMA} = 5

The IMA of an inclined plane is found by dividing the length of the slope by the vertical height (

40 / 8 = 5

).

2

Calculate the theoretical effort force required under ideal, frictionless conditions.

\text{Theoretical Effort} = 300\text{ lb}

Theoretical effort force is the load divided by the IMA (

1500 / 5 = 300\text{ lb}

).

3

Calculate the actual effort force by accounting for the mechanical efficiency.

\text{Actual Effort} = 375\text{ lb}

Because the system is only

80\%

efficient, it requires more force than the ideal scenario. Divide the theoretical effort by the efficiency as a decimal (

300 / 0.80 = 375\text{ lb}

).

Key Concept

Actual Mechanical Advantage and Efficiency of Inclined Planes

Question 40Question →

A worker needs to push a supply cart weighing $150\text{ lbs}$ up to a loading dock. The dock is $3\text{ feet}$ above the ground, and the straight ramp leading up to it is $30\text{ feet}$ long. Assuming friction is negligible, what is the effort force required to push the cart up the ramp?

150\text{ lbs}

450\text{ lbs}

15\text{ lbs}

1,500\text{ lbs}

Show answer & explanation

Answer: $15\text{ lbs}$

Answer

The correct effort force required is

15\text{ lbs}

.

The effort force is found by dividing the total load by the Ideal Mechanical Advantage (IMA). The IMA of an inclined plane is the ratio of its length to its vertical height (

30\text{ feet} / 3\text{ feet} = 10

). Therefore, the required effort force is the load divided by the IMA:

150\text{ lbs} / 10 = 15\text{ lbs}

.

Step-by-Step Solution

1

Identify the known variables from the problem.

Load =

150\text{ lbs}

, Ramp Length =

30\text{ feet}

, Height =

3\text{ feet}

.

These values are required to calculate the mechanical advantage and the resulting effort force.

2

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

\text{IMA} = \frac{\text{Length}}{\text{Height}} = \frac{30}{3} = 10

The IMA tells us by what factor the machine multiplies the input force (or divides the required effort).

3

Calculate the required effort force.

\text{Effort Force} = \frac{\text{Load}}{\text{IMA}} = \frac{150}{10} = 15\text{ lbs}

Dividing the total weight of the object by the mechanical advantage gives the theoretical force needed to move it up the incline.

Key Concept

Ideal Mechanical Advantage (IMA) and Effort Force on an Inclined Plane

Estimated Time:45s

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