Simple Machines Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 3

Question 41Question →

A mechanic is using a C-clamp with a threaded screw that has $16$ threads per inch (TPI). The clamp is currently open to a width of $5.0$ inches. If the mechanic needs to tightly secure two steel plates that have a combined thickness of $1.5$ inches, how many full revolutions of the screw are required to close the clamp to the exact thickness of the plates?

Show answer & explanation

Answer: 56

Answer

56

The screw must travel a linear distance of

3.5

inches to close the gap from

5.0

inches to

1.5

inches. With

16

threads per inch, each inch of linear travel requires

16

full revolutions. Multiplying the

3.5

inches of travel by

16

revolutions per inch yields exactly

56

revolutions.

Step-by-Step Solution

1

Calculate the linear distance the clamp's jaw needs to travel.

5.0 - 1.5 = 3.5

inches

The screw only needs to cover the empty space between the current opening and the target thickness of the plates.

2

Calculate the total number of revolutions based on the travel distance and TPI.

3.5 \times 16 = 56

revolutions

TPI (threads per inch) indicates that one full revolution moves the screw

1/16

of an inch. Therefore, moving it

1

inch requires

16

revolutions.

Key Concept

Linear Travel and Threads Per Inch (TPI)

Question 42Question →

A heavy-duty steel wedge is used to split a solid masonry block. The wedge has a penetration length of $24\text{ cm}$ and a maximum width (thickness) of $4\text{ cm}$ at its head. A pneumatic tool applies a constant downward driving effort force of $850\text{ N}$ to the head of the wedge. Due to high surface resistance, the wedge operates with a mechanical efficiency of $60\%$ . What is the total lateral separating force exerted on the masonry block by the wedge? (Enter your answer as a number in Newtons)

Show answer & explanation

Answer: 3060

Answer

3060

The correct lateral separating force is found by determining the wedge's Ideal Mechanical Advantage (

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

), adjusting for energy lost to friction by applying the

60\%

efficiency (

6 \times 0.60 = 3.6\text{ AMA}

), and multiplying this actual advantage by the input force (

850\text{ N} \times 3.6 = 3060\text{ N}

).

Step-by-Step Solution

1

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

IMA = 6

The IMA of a wedge is defined as the ratio of its penetration length to its maximum thickness (

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

).

2

Determine the Actual Mechanical Advantage (AMA).

AMA = 3.6

Because the wedge operates at

60\%

efficiency due to friction, the actual mechanical advantage is the IMA multiplied by the efficiency (

6 \times 0.60

).

3

Calculate the total lateral separating force.

3060\text{ N}

The output force is the product of the input effort force and the Actual Mechanical Advantage (

850\text{ N} \times 3.6

).

Key Concept

Mechanical Advantage and Efficiency of Wedges

Question 43Question →

A woodworker uses a steel splitting wedge to split a heavy log. The wedge has a length of $8$ inches and a maximum thickness of $2$ inches. If a lateral splitting force of $400$ lbs is needed to split the log, what ideal effort force must be applied to the top of the wedge? (Assume no friction)

400

lbs

800

lbs

100

lbs

1600

lbs

Show answer & explanation

Answer: $100$ lbs

Answer

The ideal effort force is

100

lbs.

The correct answer is found by first calculating the ideal mechanical advantage (IMA) of the wedge, which is its length divided by its thickness (

8 / 2 = 4

). Because the wedge multiplies the input force by a factor of

4

, the required effort force is the splitting force divided by the IMA (

400 \text{ lbs} / 4 = 100 \text{ lbs}

).

Step-by-Step Solution

1

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

IMA = \frac{\text{Length}}{\text{Thickness}} = \frac{8}{2} = 4

The mechanical advantage of a wedge depends on the ratio of its length to its maximum thickness.

2

Calculate the required ideal effort force using the IMA.

\text{Effort Force} = \frac{\text{Load Force}}{IMA} = \frac{400}{4} = 100 \text{ lbs}

A wedge multiplies the effort force by its mechanical advantage to produce the required load force.

Key Concept

Wedges provide a mechanical advantage equal to their length divided by their maximum thickness, which reduces the effort force required to split an object.

Question 44Question →

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?

Show answer & explanation

Answer: 350

Answer

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.

Step-by-Step Solution

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.

Key Concept

Torque Equilibrium and Composite Loads in Second-Class Levers

Question 45Question →

A vintage printing press utilizes a compound gear train to feed paper. Shaft 1 is the input and is driven at $120$ RPM in a counter-clockwise direction. Gear A ( $40$ teeth) is mounted on Shaft 1 and meshes directly with Gear B ( $20$ teeth), which is mounted on Shaft 2. Also rigidly mounted on Shaft 2 is Gear C ( $50$ teeth). Gear C meshes directly with Gear D ( $25$ teeth) mounted on Shaft 3. Based on this configuration, what is the rotational speed and direction of Shaft 3?

480

RPM, counter-clockwise

480

RPM, clockwise

30

RPM, counter-clockwise

120

RPM, counter-clockwise

Show answer & explanation

Answer: $480$ RPM, counter-clockwise

Answer

The output shaft rotates at

480

RPM in a counter-clockwise direction.

In a compound gear train, mechanical advantage and speed ratios are multiplied across stages. In the first stage (Gears A to B), a

40

-tooth gear drives a

20

-tooth gear, doubling the speed to

240

RPM and reversing the direction to clockwise. Both gears on Shaft 2 rotate together. In the second stage (Gears C to D), a

50

-tooth gear drives a

25

-tooth gear, doubling the speed again to

480

RPM and reversing the direction back to counter-clockwise.

Step-by-Step Solution

1

Calculate the rotational speed of Shaft 2.

Speed =

120 \times (40 / 20) = 240

RPM.

The speed ratio between meshing gears is inversely proportional to their tooth count. A smaller driven gear rotates faster than the driving gear.

2

Determine the rotational direction of Shaft 2.

Clockwise.

When two external gears mesh, they rotate in opposite directions. Since Shaft 1 is counter-clockwise, Shaft 2 must be clockwise. Because Gears B and C are on the same shaft, Gear C also rotates at

240

RPM clockwise.

3

Calculate the rotational speed of Shaft 3.

Speed =

240 \times (50 / 25) = 480

RPM.

Applying the speed ratio principle again from Gear C (driving) to Gear D (driven).

4

Determine the rotational direction of Shaft 3.

Counter-clockwise.

Gear C (clockwise) meshes with Gear D, forcing Gear D to rotate in the opposite direction.

Key Concept

Compound Gear Trains and Speed Ratios

Question 46Question →

A manual boat winch is used to pull a small boat onto a trailer. The crank handle travels in a circle with a radius of $15 \text{ inches}$ , turning a central spool (the axle) with a radius of $3 \text{ inches}$ .

What is the Ideal Mechanical Advantage (IMA) of this winch?

Show answer & explanation

Answer: 5

Answer

The Ideal Mechanical Advantage is 5.

The Ideal Mechanical Advantage of a wheel and axle is calculated by dividing the radius of the wheel by the radius of the axle. Here, the wheel's radius is

15 \text{ inches}

and the axle's radius is

3 \text{ inches}

, which gives

15 \div 3 = 5

.

Step-by-Step Solution

1

Identify the given dimensions of the wheel and the axle.

R_{wheel} = 15 \text{ inches}

,

R_{axle} = 3 \text{ inches}

These are the required variables for the IMA equation.

2

Apply the Ideal Mechanical Advantage formula for a wheel and axle.

IMA = \frac{R_{wheel}}{R_{axle}}

IMA represents the ratio of the effort distance to the resistance distance.

3

Calculate the final value.

IMA = \frac{15}{3} = 5

Dividing the wheel radius by the axle radius yields the mechanical advantage.

Key Concept

Ideal Mechanical Advantage (IMA) of a Wheel and Axle

Question 47Question →

A flight line technician must load a cargo pallet with a mass of $240\text{ kg}$ into a transport aircraft. The aircraft's cargo deck is $1.5\text{ meters}$ above the tarmac. To assist with loading, the technician deploys a straight, $6.0\text{-meter}$ long rigid ramp equipped with frictionless rollers. Assuming the acceleration due to gravity is $10\text{ m/s}^2$ , which of the following identifies the ideal effort force required to push the pallet up the ramp, and correctly compares the total work performed to lifting the pallet vertically?

600\text{ N}

; the total work performed is the same as lifting the pallet vertically.

600\text{ N}

; the total work performed is less than lifting the pallet vertically.

60\text{ N}

; the total work performed is the same as lifting the pallet vertically.

60\text{ N}

; the total work performed is less than lifting the pallet vertically.

Show answer & explanation

Answer: $600\text{ N}$ ; the total work performed is the same as lifting the pallet vertically.

Answer

The effort force is

600\text{ N}

, and the total work performed is exactly the same as lifting the pallet vertically.

The correct answer accurately calculates both the required force and applies the principle of work conservation. First, weight is determined by multiplying mass by gravity (

240\text{ kg} \times 10\text{ m/s}^2 = 2400\text{ N}

). The ideal mechanical advantage (IMA) of the ramp is its length divided by its height (

6.0\text{ m} / 1.5\text{ m} = 4

). The effort force is the weight divided by the IMA (

2400\text{ N} / 4 = 600\text{ N}

). Finally, while the ramp reduces the effort force, it requires moving the object over a longer distance, meaning the total work performed (

600\text{ N} \times 6.0\text{ m} = 3600\text{ Joules}

) is identical to lifting it straight up (

2400\text{ N} \times 1.5\text{ m} = 3600\text{ Joules}

).

Step-by-Step Solution

1

Calculate the weight (load force) of the cargo pallet.

W = 240\text{ kg} \times 10\text{ m/s}^2 = 2400\text{ N}

Mass must be converted to weight (force) before mechanical advantage formulas can be properly applied.

2

Determine the Ideal Mechanical Advantage (IMA) of the inclined plane.

\text{IMA} = \frac{6.0\text{ m}}{1.5\text{ m}} = 4

IMA for an inclined plane is the ratio of the effort distance (ramp length) to the load distance (vertical height).

3

Calculate the ideal effort force.

\text{Effort Force} = \frac{2400\text{ N}}{4} = 600\text{ N}

The effort force is equal to the total load force divided by the mechanical advantage of the machine.

4

Evaluate the total work performed in both scenarios.

Work via ramp:

600\text{ N} \times 6.0\text{ m} = 3600\text{ J}

. Work vertical:

2400\text{ N} \times 1.5\text{ m} = 3600\text{ J}

. Work is equal.

According to the principle of conservation of energy, simple machines trade increased distance for reduced effort force, leaving the total work unchanged.

Key Concept

Mechanical advantage reduces effort force by increasing distance, but total work is conserved.

Question 48Question →

A construction worker needs to push a heavy wheelbarrow up to a raised foundation that is $2\text{ m}$ above the ground. The worker has three different sturdy wooden planks available to use as a ramp. Arrange the following planks in order from the LEAST amount of effort force required to the MOST amount of effort force required.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The planks should be ordered from longest (

10\text{ m}

) to shortest (

4\text{ m}

). The

10\text{ m}

plank requires the least effort, the

8\text{ m}

plank requires a moderate amount, and the

4\text{ m}

plank requires the most effort.

The effort force needed to move an object up a ramp is inversely proportional to the ramp's length when the vertical height is fixed. The

10\text{ m}

plank provides the most gradual slope (highest mechanical advantage) and therefore requires the least effort force. As the plank length decreases to

8\text{ m}

and then

4\text{ m}

, the slope becomes steeper, reducing the mechanical advantage and requiring increasingly more effort force. Thus, ordering from least to most effort force is synonymous with ordering from longest to shortest plank.

Step-by-Step Solution

1

Recall the formula for the Ideal Mechanical Advantage (IMA) of an inclined plane.

\text{IMA} = \frac{\text{Length of the slope}}{\text{Height of the incline}}

This formula helps us determine how much a ramp multiplies the input force.

2

Determine the relationship between length, IMA, and effort force.

Since the height (

2\text{ m}

) is constant for all three planks, a longer plank results in a higher IMA. A higher IMA means less effort force is required to push the load.

Understanding this inverse relationship allows us to rank the effort force without doing complex calculations.

3

Rank the planks based on length to determine the required effort force.

Longest plank (

10\text{ m}

) = Highest IMA = Least effort force. Shortest plank (

4\text{ m}

) = Lowest IMA = Most effort force.

The question asks to order from LEAST effort force to MOST effort force, which corresponds to ordering from longest ramp to shortest ramp.

Key Concept

For an inclined plane of a given height, increasing the length of the plane increases its mechanical advantage, which proportionally decreases the effort force required to move a load up the plane.

Estimated Time:45s

Question 49Question →

Match each type of wedge or wedge application with its correct mechanical characteristic.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches are: Long, thin wedge corresponds to high mechanical advantage and less effort; Short, thick wedge corresponds to lower mechanical advantage but rapid splitting; Doorstop corresponds to securing objects using friction.

The correct pairings are based on the principle that the ideal mechanical advantage (IMA) of a wedge is its length divided by its thickness. A long, thin wedge has a high mechanical advantage and therefore requires less effort force to penetrate a material. Conversely, a short, thick wedge has a lower mechanical advantage, meaning it takes more effort force, but it successfully splits the object wider in a shorter distance. A doorstop is a specific application that uses the wedge shape to generate a strong normal force and resulting friction between the door and the floor, holding the object stationary.

Step-by-Step Solution

1

Analyze the mechanical advantage of a long, thin wedge.

Matches with 'Provides high mechanical advantage and requires less applied effort force.'

The mechanical advantage of a wedge is its length divided by its thickness. A larger ratio means less effort is needed to drive the wedge.

2

Analyze the mechanical advantage of a short, thick wedge.

Matches with 'Provides lower mechanical advantage but splits material apart rapidly over a short distance.'

A smaller length-to-thickness ratio results in a lower mechanical advantage, requiring more force, but the increased thickness spreads the material faster.

3

Analyze the practical application of a doorstop.

Matches with 'Secures objects in place primarily by utilizing friction against a surface.'

Instead of splitting material, a doorstop is driven under a door to increase normal force and create high friction, keeping it wedged in place.

Key Concept

The mechanical advantage of a wedge is determined by the ratio of its length to its thickness, governing the trade-off between the effort force required and the distance the material is separated.

Question 50Question →

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.*

Show answer & explanation

Answer: 975

Answer

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.

Step-by-Step Solution

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).

Key Concept

Torque Equilibrium in Compound and Uniform Levers

Question 51Question →

A mechanic needs to lift a heavy engine using a screw jack. They can equip the jack with one of two interchangeable threaded rods: Rod X has a coarse thread with $4$ threads per inch (TPI), and Rod Y has a fine thread with $8$ TPI.

Assuming the jack's handle length is identical for both setups and friction is negligible, which of the following statements accurately describes the required effort force and total work done if the mechanic uses Rod Y instead of Rod X to lift the engine by exactly $10$ inches?

Rod Y requires half the effort force on the handle, but the total work done to lift the engine remains the same.

Rod Y requires half the effort force on the handle, and it reduces the total work done to lift the engine by half.

Rod Y requires twice the effort force on the handle, but the total work done to lift the engine remains the same.

Rod Y requires twice the effort force on the handle, and it doubles the total work done to lift the engine.

Show answer & explanation

Answer: Rod Y requires half the effort force on the handle, but the total work done to lift the engine remains the same.

Answer

Rod Y requires half the effort force on the handle, but the total work done to lift the engine remains the same.

A screw is essentially an inclined plane wrapped around a central cylinder. Increasing the Threads Per Inch from

4

to

8

halves the pitch of the thread, making the 'incline' less steep. This doubles the mechanical advantage of the jack, cutting the required effort force on the handle in half. However, the fundamental principle of all simple machines is that they do not reduce the total amount of work required to move a load; they merely trade increased input distance (more rotations of the handle) for decreased input force. Because the engine is still being lifted exactly

10

inches, the total output work remains identical.

Step-by-Step Solution

1

Analyze the relationship between Threads Per Inch (TPI) and pitch.

Rod Y (

8

TPI) has half the pitch of Rod X (

4

TPI), meaning it advances half the linear distance per full rotation.

Pitch is the mathematical inverse of TPI (

1/\text{TPI}

). It determines the steepness of the screw's thread.

2

Determine the effect of a smaller pitch on Mechanical Advantage and effort force.

Because Rod Y advances half as far per rotation, its Ideal Mechanical Advantage is doubled. Thus, it requires half the effort force to lift the same load.

A screw is an inclined plane wrapped around a cylinder. A finer thread equates to a shallower incline, which trades distance for force.

3

Evaluate the total work done by the system in both scenarios.

The total work done remains identical regardless of the rod used.

Total output work is calculated as Load Force

\times

Lift Distance. Since the engine's weight and the

10

-inch lift distance do not change, the total work is conserved.

Key Concept

Mechanical Advantage of Screws and Conservation of Work

Estimated Time:1m 0s

Question 52Question →

A manual winch is used to lift a $240 \text{ lb}$ engine block exactly $2 \text{ feet}$ vertically. The winch consists of a steering-wheel style control with a diameter of $3 \text{ feet}$ , which is rigidly attached to an axle drum with a radius of $3 \text{ inches}$ . Assuming an ideal system with no friction, what continuous effort force must be applied to the outer edge of the wheel, and what total distance must the effort force move along the wheel's rotation?

20 \text{ lbs}

of force;

24 \text{ feet}

of distance

480 \text{ lbs}

of force;

1 \text{ foot}

of distance

40 \text{ lbs}

of force;

12 \text{ feet}

of distance

1440 \text{ lbs}

of force;

1/3 \text{ foot}

of distance

Show answer & explanation

Answer: $40 \text{ lbs}$ of force; $12 \text{ feet}$ of distance

Answer

The correct answer is

40 \text{ lbs}

of force and

12 \text{ feet}

of distance.

The system's Ideal Mechanical Advantage is determined by the ratio of the effort radius (18 inches, half of the 3-foot diameter) to the load radius (3 inches), which yields an IMA of 6. This means the required effort force is divided by 6 (

240 \text{ lbs} / 6 = 40 \text{ lbs}

), while the required effort distance is multiplied by 6 (

2 \text{ feet} \times 6 = 12 \text{ feet}

) to conserve work.

Step-by-Step Solution

1

Convert all measurements to the same unit (inches).

Wheel diameter =

3 \text{ feet} = 36 \text{ inches}

.

Units must be identical to calculate a valid, dimensionless mechanical advantage ratio.

2

Identify the effort radius and the load radius.

Wheel radius (

R

) =

18 \text{ inches}

. Axle drum radius (

r

) =

3 \text{ inches}

.

The lever arms in a wheel and axle system are the distances from the central axis (the radii), not the diameters.

3

Calculate the Ideal Mechanical Advantage (IMA).

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

.

This establishes that the system multiplies the user's force by a factor of 6.

4

Calculate the required effort force.

\text{Effort Force} = \frac{\text{Load}}{\text{IMA}} = \frac{240 \text{ lbs}}{6} = 40 \text{ lbs}

.

Because the machine provides a mechanical advantage of 6, the required input force is one-sixth of the load.

5

Calculate the required effort distance.

\text{Effort Distance} = \text{Load Distance} \times \text{IMA} = 2 \text{ feet} \times 6 = 12 \text{ feet}

.

According to the principle of work, whatever factor force is reduced by, distance must be multiplied by that same factor.

Key Concept

Ideal Mechanical Advantage (IMA) and Work in a Wheel and Axle

Estimated Time:1m 30s

Question 53Question →

Match each gear system configuration with its resulting mechanical properties.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The simple train multiplies torque by 4 and reverses direction. The idler train quadruples speed and keeps the same direction. The compound train provides a mechanical advantage of 6 and keeps the same direction. The rack and pinion converts rotational motion to linear motion.

Each specific gear arrangement produces distinct mechanical outcomes based on the ratio of teeth between the driving and driven gears, the number of intermediary gears, and the physical shape of the gears involved.

Step-by-Step Solution

1

Determine the mechanical properties of the simple train.

Matches with: Torque is multiplied by 4; rotational direction is reversed.

A 15-tooth gear driving a 60-tooth gear has a ratio of 4:1. The driven gear is 4 times larger, multiplying torque by 4. Two meshing gears turn in opposite directions.

2

Determine the mechanical properties of the idler train.

Matches with: Output speed is 4 times the input speed; rotational direction is unchanged.

The 25-tooth idler gear does not change the ratio. The 60-tooth drive gear directly relates to the 15-tooth driven gear, meaning the output is 1/4 the size and spins 4 times as fast. An odd number of gears (3) preserves the rotational direction.

3

Determine the mechanical properties of the compound train.

Matches with: Overall mechanical advantage is 6; final gear rotates in the same direction as the input gear.

Calculate the ratio of each pair: 40/20 = 2, and 30/10 = 3. Multiply them: 2 * 3 = 6. This is the mechanical advantage. There are two gear meshes, causing the direction to reverse twice, returning to the original input direction.

4

Determine the mechanical properties of the rack and pinion.

Matches with: Rotational motion is converted into linear motion.

A round pinion turning against a straight rack is the standard mechanism for converting rotational motion into linear motion.

Key Concept

Calculating gear ratios and tracking rotational direction across simple, idler, compound, and rack-and-pinion trains.

Question 54Question →

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}

Show answer & explanation

Answer: $70\text{ lbs}$

Answer

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}

.

Step-by-Step Solution

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.

Key Concept

Torque Equilibrium Involving Lever Mass

Question 55Question →

Arrange the following material handling tasks in order from the LEAST amount of effort force required to the GREATEST amount of effort force required. (Assume frictionless ramps)

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from least to greatest effort force is: Task A (

30\text{ lbs}

), Task D (

40\text{ lbs}

), Task C (

50\text{ lbs}

), Task B (

60\text{ lbs}

).

Task A requires

30\text{ lbs}

of force, Task D requires

40\text{ lbs}

, Task C requires

50\text{ lbs}

, and Task B requires

60\text{ lbs}

. Ordering these mathematically from lowest to highest yields Task A, Task D, Task C, and Task B.

Step-by-Step Solution

1

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

Task A:

12/3 = 4

. Task B:

6/2 = 3

. Task C:

8/4 = 2

. Task D:

5/1 = 5

.

The IMA determines how much the ramp multiplies the applied force, calculated as the length of the ramp divided by its height.

2

Calculate the required effort force for each task.

Task A:

120/4 = 30\text{ lbs}

. Task B:

180/3 = 60\text{ lbs}

. Task C:

100/2 = 50\text{ lbs}

. Task D:

200/5 = 40\text{ lbs}

.

Effort force equals the load's total weight divided by the IMA of the ramp.

3

Order the tasks from least effort to greatest effort.

Task A (

30\text{ lbs}

) < Task D (

40\text{ lbs}

) < Task C (

50\text{ lbs}

) < Task B (

60\text{ lbs}

).

The question specifically asks for the sequence starting with the lowest force and ending with the highest.

Key Concept

Ideal Mechanical Advantage and Effort Force on an Inclined Plane

Question 56Question →

A homeowner needs to move a $300\text{ lb}$ refrigerator onto a porch that is $2\text{ feet}$ above ground level. They place a straight, frictionless wooden ramp that is $6\text{ feet}$ long from the ground to the porch edge. How much effort force is required to push the refrigerator up the ramp?

150\text{ lbs}

300\text{ lbs}

100\text{ lbs}

900\text{ lbs}

Show answer & explanation

Answer: $100\text{ lbs}$

Answer

100\text{ lbs}

To find the force required to push an object up a frictionless ramp, first find the ramp's Ideal Mechanical Advantage (IMA). The IMA is the ramp's length divided by its vertical height (

6\text{ ft} / 2\text{ ft} = 3

). The effort force needed is the object's weight divided by this mechanical advantage (

300\text{ lbs} / 3 = 100\text{ lbs}

).

Step-by-Step Solution

1

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

\text{IMA} = \frac{\text{Length}}{\text{Height}} = \frac{6\text{ ft}}{2\text{ ft}} = 3

The IMA tells us how many times easier the ramp makes the task compared to lifting the object straight up.

2

Calculate the required effort force.

\text{Effort Force} = \frac{\text{Weight}}{\text{IMA}} = \frac{300\text{ lbs}}{3} = 100\text{ lbs}

An inclined plane reduces the necessary force by a factor equal to its mechanical advantage.

Key Concept

Ideal Mechanical Advantage and Effort Force of an Inclined Plane

Estimated Time:1m 0s

Question 57Question →

A worker uses a steel wedge to split a large block of ice. If the worker switches to a longer wedge that has the exact same maximum thickness, which of the following statements correctly describes the result?

It decreases the effort force required to split the ice.

It decreases the total amount of work required to split the ice.

It increases the total frictional force resisting the wedge due to the larger surface area.

It requires a greater effort force because the wedge must be driven a longer distance.

Show answer & explanation

Answer: It decreases the effort force required to split the ice.

Answer

It decreases the effort force required to split the ice.

The ideal mechanical advantage of a wedge is its length divided by its thickness. By increasing the length while keeping the thickness the same, the mechanical advantage increases, which reduces the effort force needed to do the job.

Step-by-Step Solution

1

Recall the formula for the ideal mechanical advantage (IMA) of a wedge.

IMA = Length / Thickness

This relationship determines how much a wedge multiplies the applied force.

2

Analyze the change in the wedge's dimensions described in the scenario.

The length increases, while the thickness remains exactly the same.

Understanding the variables helps determine the new mechanical advantage.

3

Determine the effect on the effort force.

Since the numerator (length) is larger and the denominator (thickness) is constant, the IMA increases. A higher IMA means less effort force is needed.

An increased mechanical advantage trades a longer distance for a reduced effort force requirement.

Key Concept

Mechanical Advantage of Wedges

Question 58Question →

A technician is analyzing the thread specifications of four different fasteners to determine their linear travel. Assuming all fasteners have single-start threads, arrange them in order from the LEAST to the GREATEST linear distance they will travel into a fixed base for each full $360^\circ$ rotation.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from least to greatest linear travel per rotation is: the machine screw (20 TPI), the wood screw (1/16 inch pitch), the lag bolt (10 TPI), and finally the power screw (0.125 inch lead).

The sequence correctly calculates and compares the linear travel per rotation for each fastener. By converting all specifications to decimal inches (

0.05

in,

0.0625

in,

0.10

in, and

0.125

in), they can be accurately ordered from least to greatest.

Step-by-Step Solution

1

Define the relationship between threads per inch (TPI), pitch, lead, and linear travel.

For single-start threads, the linear travel per rotation equals the lead, which is equal to the pitch. Pitch can be calculated as

1 \div \text{TPI}

.

Establishing the mathematical formulas is necessary to compare different types of thread specifications on a single scale.

2

Calculate the decimal linear travel for the fasteners specified in TPI.

The machine screw travels

1 \div 20 = 0.05

inches per rotation. The lag bolt travels

1 \div 10 = 0.10

inches per rotation.

Converting TPI into decimal pitch allows for direct numerical comparison.

3

Convert the fractional pitch of the wood screw to a decimal.

The wood screw travels

1 \div 16 = 0.0625

inches per rotation. The power screw is already given as

0.125

inches.

Decimals provide a consistent format for sequential ordering.

4

Compare all decimal values and arrange them from least to greatest.

0.05

(machine screw)

< 0.0625

(wood screw)

< 0.10

(lag bolt)

< 0.125

(power screw).

Finalizes the sequence according to the explicit instructions in the stem.

Key Concept

The linear distance a screw travels in one full rotation is determined by its lead. For single-start threads, the lead equals the pitch, and the pitch is the reciprocal of the threads per inch (

P = 1 / \text{TPI}

).

Question 59Question →

A motorized mixer uses two meshing gears to adjust the mixing speed. The driving gear, which is attached to the motor, has $12$ teeth and spins at $150$ RPM. It meshes directly with a driven gear that has $36$ teeth. What is the rotational speed of the driven gear in RPM?

Show answer & explanation

Answer: 50

Answer

50

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

36

vs

12

). Because gears have an inverse relationship between size and speed, the larger driven gear will rotate at one-third the speed of the driving gear (

150 \div 3 = 50

RPM).

Step-by-Step Solution

1

State the gear speed formula

T_1 \times S_1 = T_2 \times S_2

To establish the inverse relationship between the number of gear teeth and rotational speed

2

Substitute the given values into the formula

12 \times 150 = 36 \times S_2

To set up the equation with the known tooth counts and driving speed

3

Solve for the unknown driven speed (

S_2

)

S_2 = \frac{1800}{36} = 50

To find the final RPM of the driven gear

Key Concept

The rotational speed of meshing gears is inversely proportional to their number of teeth.

Question 60Question →

A carpenter uses a wooden wedge as a doorstop to hold a heavy door open. The wedge has a length of $15\text{ cm}$ and a maximum thickness of $3\text{ cm}$ . What is the ideal mechanical advantage (IMA) of this wedge?

Show answer & explanation

Answer: 5

Answer

The ideal mechanical advantage is 5.

The ideal mechanical advantage of a wedge is found by dividing its length (the distance the wedge penetrates) by its maximum thickness (the distance the object is separated). For a wedge with a length of

15\text{ cm}

and a thickness of

3\text{ cm}

, the mechanical advantage is calculated as

15 / 3 = 5

.

Step-by-Step Solution

1

Identify the given dimensions of the wedge.

Length (

L

) =

15\text{ cm}

, Thickness (

t

) =

3\text{ cm}

.

These values are required to calculate the mechanical advantage.

2

Apply the formula for the ideal mechanical advantage of a wedge.

IMA = \frac{L}{t} = \frac{15}{3} = 5

.

The IMA is the ratio of the effort distance (length) to the resistance distance (thickness).

Key Concept

Ideal Mechanical Advantage of a Wedge

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