All practice questions

Question 1661Question →

In a small vehicle's hydraulic braking system, a driver applies an input force of $30 \text{ lbs}$ to the master cylinder piston. The master cylinder piston has a cross-sectional area of $1 \text{ in}^2$ . If the brake caliper piston has a cross-sectional area of $4 \text{ in}^2$ , what is the resulting output force applied by the brake caliper piston?

Show answer & explanation

Answer: 120

Answer

120

According to Pascal's Principle, pressure is transmitted equally throughout an enclosed fluid. Because the brake caliper piston has an area 4 times larger than the master cylinder piston, it will multiply the input force by a factor of 4, resulting in 120 lbs of output force.

Step-by-Step Solution

1

Calculate the fluid pressure created by the master cylinder.

P = \frac{30 \text{ lbs}}{1 \text{ in}^2} = 30 \text{ psi}

Pressure is defined as force divided by area.

2

Calculate the output force at the brake caliper using Pascal's Principle.

F_2 = 30 \text{ psi} \times 4 \text{ in}^2 = 120 \text{ lbs}

Pascal's Principle states that the 30 psi pressure is transmitted undiminished to the 4 sq in output piston.

Key Concept

Pascal's Principle and Mechanical Advantage in Hydraulics

Estimated Time:45s

Question 1662Question →

A stationary drone drops a $5\text{ kg}$ supply package from a high altitude. Ignoring air resistance and assuming the acceleration due to gravity is $10\text{ m/s}^2$ , what is the downward velocity of the package after it has been falling for exactly $4\text{ seconds}$ ?

80\text{ m/s}

50\text{ m/s}

40\text{ m/s}

8\text{ m/s}

Show answer & explanation

Answer: $40\text{ m/s}$

Answer

The downward velocity of the package is

40\text{ m/s}

.

In free fall without air resistance, velocity increases at a constant rate equal to gravitational acceleration, regardless of mass. Multiplying the acceleration (

10\text{ m/s}^2

) by the time (

4\text{ seconds}

) gives the correct velocity of

40\text{ m/s}

.

Step-by-Step Solution

1

Identify the given values from the problem statement.

Initial velocity (

v_0

) =

0\text{ m/s}

, Acceleration (

g

) =

10\text{ m/s}^2

, Time (

t

) =

4\text{ s}

, Mass (

m

) =

5\text{ kg}

.

Extracting variables is the necessary first step to choosing the correct kinematic formula.

2

Select the correct formula to find the final velocity.

Use the formula

v = v_0 + gt

.

This formula relates velocity, acceleration, and time without needing the fall distance.

3

Substitute the values into the formula and solve.

v = 0 + (10 \times 4) = 40\text{ m/s}

.

Performing the calculation yields the final velocity. The mass of

5\text{ kg}

is irrelevant because all objects in free fall accelerate at the same rate in the absence of air resistance.

Key Concept

In free fall with no air resistance, the velocity of an object increases linearly with time (

v = gt

), completely independent of the object's mass.

Question 1663Question →

An industrial hydraulic press uses two pistons connected by an enclosed, incompressible fluid. The input piston has a cross-sectional area of $3 \text{ in}^2$ , and the output piston has a cross-sectional area of $30 \text{ in}^2$ . If an operator applies a constant force of $50 \text{ lbs}$ to the input piston, what is the maximum weight the press can lift, and how far must the input piston travel in total to raise that weight by exactly $5 \text{ inches}$ ?

500 \text{ lbs}

maximum weight;

50 \text{ inches}

of input travel

5 \text{ lbs}

maximum weight;

50 \text{ inches}

of input travel

500 \text{ lbs}

maximum weight;

0.5 \text{ inches}

of input travel

5 \text{ lbs}

maximum weight;

0.5 \text{ inches}

of input travel

Show answer & explanation

Answer: $500 \text{ lbs}$ maximum weight; $50 \text{ inches}$ of input travel

Answer

The press can lift a maximum weight of 500 lbs, and the input piston must travel 50 inches.

The correct answer correctly applies the area ratio of 10 (

30/3

). The output force is multiplied by this ratio (

50 \times 10 = 500 \text{ lbs}

), and the input distance must also be 10 times the output distance (

5 \times 10 = 50 \text{ inches}

) to displace enough fluid volume.

Step-by-Step Solution

1

Calculate the area ratio between the output and input pistons.

Area Ratio =

30 \text{ in}^2 / 3 \text{ in}^2 = 10

.

This ratio represents the mechanical advantage of the hydraulic system.

2

Calculate the maximum weight (output force) the press can lift.

F_{output} = 50 \text{ lbs} \times 10 = 500 \text{ lbs}

.

According to Pascal's Principle, pressure is transmitted equally. Therefore, the output force is the input force multiplied by the area ratio.

3

Calculate the required input travel distance.

d_{input} = 5 \text{ inches} \times 10 = 50 \text{ inches}

.

To conserve work and fluid volume, the input piston must move proportionally farther than the output piston by a factor equal to the area ratio.

Key Concept

In a hydraulic system, a larger output piston multiplies the input force by the ratio of their areas, but the input piston must travel proportionally farther to conserve fluid volume and work.

Estimated Time:1m 30s

Question 1664Question →

Match each mechanical design feature or procedure with its underlying thermal expansion principle or purpose.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The heated pulley matches with interference fit upon cooling; the pipeline loops match with preventing structural buckling; the riveted metal strips match with unequal expansion for movement; and the hydraulic accumulator matches with volumetric expansion of fluids.

Each distinct mechanical design feature intentionally harnesses or mitigates the physical size changes that occur when solid materials and fluids undergo temperature variations.

Step-by-Step Solution

1

Analyze the thermal behavior of a heated metal pulley placed on a cold shaft.

The heated pulley has an temporarily enlarged bore; upon cooling, thermal contraction reduces the bore size.

This physical reduction in size locks the pulley onto the shaft securely without fasteners.

2

Evaluate the mechanical purpose of U-shaped loops in long metal pipelines.

The loops provide a flexible section that can compress or stretch slightly.

Metal expands linearly when heated; without flexibility, the expansion would cause the pipe to warp or break under massive internal stress.

3

Determine the effect of heating two dissimilar metals riveted together (a bimetallic strip).

The metal with the higher coefficient of expansion grows more, forcing the entire bonded assembly to curl or bend.

This predictable bending motion is intentionally utilized to flip switches or move dials in temperature-sensing devices.

4

Assess why a closed fluid system requires an accumulator or overflow tank.

The tank provides an empty or gas-filled space for hydraulic fluid to enter as the system warms up.

Fluids expand when heated but are virtually incompressible; without extra volume, the expanding fluid would rupture the rigid lines.

Key Concept

Thermal Expansion and Contraction

Question 1665Question →

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

Show answer & explanation

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 1666Question →

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 1667Question →

A heavy shipping container is suspended steadily in the air by a single vertical steel cable attached to a helicopter. Which of the following types of stress is the steel cable primarily experiencing?

Tension

Compression

Shear

Torsion

Show answer & explanation

Answer: Tension

Answer

Tension

The correct answer is tension because the cable is subjected to opposing forces pulling outward from its center (the lifting force of the helicopter above and the weight of the container below). This pulling action tends to stretch or elongate the material, which is the exact definition of tensile stress.

Step-by-Step Solution

1

Identify the direction of the forces acting on the cable.

The helicopter exerts an upward force on the top of the cable, while gravity acting on the shipping container exerts a downward force on the bottom of the cable.

Understanding the direction of applied forces is required to determine the type of stress.

2

Determine how these forces affect the material of the cable.

The opposing outward forces pull the cable from both ends, attempting to stretch or elongate it.

Matching the physical effect to the definition of stress types reveals the answer.

3

Match the stretching effect to the correct stress terminology.

A pulling force that stretches a material is defined as tension.

Tension is the specific mechanical term for this type of load.

Key Concept

Types of Mechanical Stress (Tension)

Question 1668Question →

Match each structural scenario or mechanical component to the primary type of internal stress it experiences under load.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Concrete pillar matches to Compressive stress; crane cable matches to Tensile stress; drive shaft matches to Torsional stress; bolt holding overlapping plates matches to Shear stress.

Structural and mechanical components are designed to withstand specific forces based on their application. Pillars support weight by resisting being crushed (compression). Cables lift loads by resisting being stretched (tension). Drive shafts transmit torque by resisting being twisted (torsion). Fasteners like bolts hold sliding members together by resisting being cut in half (shear).

Step-by-Step Solution

1

Analyze the force acting on the concrete foundation pillar.

The pillar is being pushed down upon, which squeezes the material.

Pushing forces that attempt to shorten or crush a material create compressive stress.

2

Analyze the force acting on the crane cable.

The cable is being pulled from both ends by the crane mechanism and the load.

Pulling forces that attempt to stretch or elongate a material create tensile stress.

3

Analyze the force acting on the drive shaft.

The shaft is being rotated rapidly to transmit power, creating a twisting motion.

Twisting forces applied to an object create torsional stress.

4

Analyze the force acting on the structural bolt.

The overlapping plates pull in opposite directions, acting like scissors across the middle of the bolt.

Forces acting parallel to a cross-section, attempting to slice or cut the material, create shear stress.

Key Concept

Different types of external loads applied to structural components result in specific types of internal stresses: tension (pulling apart), compression (pushing together), torsion (twisting), and shear (slicing across).

Question 1669Question →

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 1670Question →

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 1671Question →

A construction crew uses a pile driver machine that drops a solid steel hammer freely onto a concrete pillar. The operator replaces the original $500\text{-kg}$ hammer with a $1,000\text{-kg}$ hammer and drops it from the exact same height.

Ignoring air resistance and guide rail friction, which of the following statements correctly describes how the final free-fall speed of the heavier hammer compares to the original lighter hammer?

The speeds will be exactly the same, because gravitational acceleration is independent of an object's mass.

The heavier hammer will fall faster, because a larger mass results in a greater gravitational acceleration.

The heavier hammer will fall slower, because its increased mass creates more inertia to resist the downward pull.

The heavier hammer will fall faster, because mass and weight are identical properties that directly dictate an object's falling speed.

Show answer & explanation

Answer: The speeds will be exactly the same, because gravitational acceleration is independent of an object's mass.

Answer

The final speeds will be exactly the same because the acceleration due to gravity is independent of the falling object's mass.

In the absence of air resistance and friction, all objects in free fall experience the exact same uniform acceleration due to gravity, regardless of their mass. Because both hammers are dropped from the same height and accelerate at the same rate, they will achieve the exact same speed just before impact.

Step-by-Step Solution

1

Identify the forces acting on both hammers during their descent.

Since air resistance and friction are ignored, the only force acting on the hammers is the downward pull of gravity.

Establishing that the objects are in a state of free fall is required to determine their rate of acceleration.

2

Relate gravitational force to acceleration using Newton's laws.

The downward force is weight (

F = mg

), and Newton's second law is

F = ma

. Equating them gives

mg = ma

, which simplifies to

a = g

.

This mathematical relationship proves that the mass (

m

) cancels out, meaning acceleration depends only on gravity (

g

).

3

Compare the final velocities of the two hammers.

Both hammers drop from the exact same height with the same constant acceleration. Therefore, their final speeds just before impact must be equal.

According to kinematic equations, identical initial conditions (starting from rest) and identical acceleration over the same distance result in the exact same final velocity.

Key Concept

Independence of Mass in Free Fall

Question 1672Question →

A hydraulic press is used to stamp metal parts. The system features an input piston with a cross-sectional area of $5 \text{ cm}^2$ and an output piston with a cross-sectional area of $50 \text{ cm}^2$ . If a worker applies a downward force of $20 \text{ N}$ to the input piston, what is the resulting upward force exerted by the output piston?

2 \text{ N}

100 \text{ N}

200 \text{ N}

1000 \text{ N}

Show answer & explanation

Answer: $200 \text{ N}$

Answer

The resulting upward force is

200 \text{ N}

.

In a closed hydraulic system, pressure is distributed equally throughout the fluid. Because the output piston's area (

50 \text{ cm}^2

) is 10 times larger than the input piston's area (

5 \text{ cm}^2

), it produces 10 times the force. Therefore, a

20 \text{ N}

input force is multiplied by 10, resulting in a

200 \text{ N}

output force.

Step-by-Step Solution

1

Identify the given values from the problem description.

Input Force =

20 \text{ N}

, Input Area =

5 \text{ cm}^2

, Output Area =

50 \text{ cm}^2

.

These are the necessary parameters to apply Pascal's Principle.

2

Calculate the area ratio, which represents the mechanical advantage, between the output and input pistons.

Area Ratio =

50 / 5 = 10

.

The mechanical advantage in a hydraulic system is directly proportional to the ratio of the piston areas.

3

Multiply the input force by the mechanical advantage to find the final output force.

Output Force =

20 \times 10 = 200 \text{ N}

.

Because pressure is transmitted equally, the force is multiplied by the same factor as the area.

Key Concept

Pascal's Principle and Hydraulic Mechanical Advantage

Question 1673Question →

A hiker accidentally knocks a rock off the edge of a steep cliff. Assuming there is no air resistance and the acceleration due to gravity is $10\text{ m/s}^2$ , how fast is the rock traveling exactly $3$ seconds after it begins to fall?

Show answer & explanation

Answer: 30

Answer

30

In free fall with no air resistance, an object's speed increases by the acceleration of gravity each second. After

3

seconds of accelerating at

10\text{ m/s}^2

, the speed is exactly

30\text{ m/s}

.

Step-by-Step Solution

1

Identify the relevant formula

v = gt

The rock starts from rest, so its speed depends only on gravitational acceleration multiplied by the elapsed time.

2

Substitute the given values into the formula

v = 10 \times 3 = 30

Gravity accelerates the object by

10\text{ m/s}^2

for every second it falls.

Key Concept

Velocity of an object in free fall increases linearly with time based on gravitational acceleration.

Question 1674Question →

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 1675Question →

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 1676Question →

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 1677Question →

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 1678Question →

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 1679Question →

A military recovery vehicle is using a winch to pull a disabled transport truck out of deep mud on a level surface. The disabled truck has a weight of $50,000 \text{ N}$ . The mud exerts a constant resistive drag force of $12,000 \text{ N}$ against the truck's motion. If the winch must accelerate the truck horizontally at a rate of $0.5 \text{ m/s}^2$ , how much tension force must the winch cable provide? (Assume gravitational acceleration $g = 10 \text{ m/s}^2$ )

22,000 \text{ N}

37,000 \text{ N}

14,500 \text{ N}

2,500 \text{ N}

Show answer & explanation

Answer: $14,500 \text{ N}$

Answer

The winch cable must provide

14,500 \text{ N}

of tension force.

The correct answer is found by first converting the truck's weight (

50,000 \text{ N}

) to its mass (

5,000 \text{ kg}

) using

g = 10 \text{ m/s}^2

. Next, the net force required to accelerate this mass at

0.5 \text{ m/s}^2

is calculated as

2,500 \text{ N}

using

F = ma

. Finally, because the mud exerts a

12,000 \text{ N}

opposing drag, the winch must pull with enough force to overcome the drag and still provide the

2,500 \text{ N}

net force. Therefore, the total required tension is

12,000 \text{ N} + 2,500 \text{ N} = 14,500 \text{ N}

.

Step-by-Step Solution

1

Determine the mass of the disabled truck from its given weight.

m = \frac{W}{g} = \frac{50,000 \text{ N}}{10 \text{ m/s}^2} = 5,000 \text{ kg}

Newton's second law (

F = ma

) requires mass, not weight.

2

Calculate the required net force to achieve the target acceleration.

F_{\text{net}} = ma = 5,000 \text{ kg} \times 0.5 \text{ m/s}^2 = 2,500 \text{ N}

This is the net force that must remain after all opposing forces are overcome.

3

Set up the force balance equation for the horizontal direction.

F_{\text{net}} = T_{\text{winch}} - F_{\text{drag}}

The net force is the vector sum of the applied tension pulling forward and the drag force resisting backward.

4

Solve the equation for the required winch tension (

T_{\text{winch}}

).

2,500 \text{ N} = T_{\text{winch}} - 12,000 \text{ N} \implies T_{\text{winch}} = 14,500 \text{ N}

Adding the drag force to the required net force gives the total applied force.

Key Concept

Newton's Second Law applied to multi-force systems.

Question 1680Question →

A U-shaped glass tube is open to the atmosphere at both ends. The left arm of the tube has a cross-sectional area of $10\text{ cm}^2$ , and the right arm has a cross-sectional area of $5\text{ cm}^2$ . The tube initially contains water with a density of $1.0\text{ g/cm}^3$ . A student carefully pours a column of oil measuring $20\text{ cm}$ in length, with a density of $0.8\text{ g/cm}^3$ , into the narrower right arm. Assuming the oil-water interface remains in the vertical section of the right arm, what is the vertical height difference between the top surface of the oil and the top surface of the water once the fluids come to rest?

16\text{ cm}

0\text{ cm}

4\text{ cm}

8\text{ cm}

Show answer & explanation

Answer: $4\text{ cm}$

Answer

4\text{ cm}

Hydrostatic pressure is determined solely by fluid density and depth (

P = \rho g h

), entirely independent of the container's cross-sectional area. Equating the pressures at the fluid interface gives a water column height of

16\text{ cm}

. Subtracting this from the

20\text{ cm}

oil column reveals that the oil surface rests

4\text{ cm}

higher than the water surface.

Step-by-Step Solution

1

Identify the reference point of equal pressure.

The horizontal plane at the oil-water interface in the right arm represents the point where pressure from both sides must be equal.

In a static fluid system, pressure at any given horizontal depth within the same continuous fluid (water) is uniform.

2

Set up the hydrostatic pressure equation.

\rho_{\text{water}} \times g \times h_{\text{water}} = \rho_{\text{oil}} \times g \times h_{\text{oil}}

. The cross-sectional areas (

10\text{ cm}^2

and

5\text{ cm}^2

) are irrelevant and ignored.

Hydrostatic pressure (

P = \rho g h

) depends only on fluid density and depth, not on the container's shape or the total volume of fluid.

3

Calculate the required height of the water column.

1.0\text{ g/cm}^3 \times h_{\text{water}} = 0.8\text{ g/cm}^3 \times 20\text{ cm}

, which simplifies to

h_{\text{water}} = 16\text{ cm}

.

This establishes how high the water surface in the left arm must rise above the interface to balance the downward pressure of the oil.

4

Calculate the difference in surface heights.

20\text{ cm} - 16\text{ cm} = 4\text{ cm}

.

The question asks for the vertical difference between the top of the oil and the top of the water, not just the absolute height of the water column.

Key Concept

Hydrostatic pressure is independent of container shape or cross-sectional area, depending solely on fluid density and vertical column height.

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