Fluid Mechanics Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB)

Question 1Question →

Four identical airtight containers, each with a total volume of $1 \text{ m}^3$ , are placed in different situations. Arrange these situations in order from the GREATEST buoyant force acting on the container to the LEAST buoyant force.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from greatest to least buoyant force is: fully submerged in saltwater, fully submerged in freshwater, half-submerged in freshwater, and sitting on dry land.

Buoyant force is determined by the weight of the displaced fluid. A fully submerged object in saltwater displaces the heaviest amount of fluid. A fully submerged object in freshwater displaces slightly less weight because freshwater is less dense. A half-submerged object displaces only half the volume of water, yielding an even smaller force. An object on dry land displaces no water, experiencing zero buoyant force.

Step-by-Step Solution

1

Recall Archimedes' Principle.

The buoyant force is equal to the weight of the displaced fluid (

F_b = \text{fluid density} \times \text{volume displaced} \times \text{gravity}

).

This formula identifies the two main variables that affect buoyant force: the density of the fluid and the volume of fluid displaced.

2

Compare the volume of fluid displaced in each scenario.

Both completely submerged scenarios displace the full

1 \text{ m}^3

volume. The half-submerged floating scenario displaces

0.5 \text{ m}^3

. The dry land scenario displaces

0 \text{ m}^3

.

Volume displaced is the primary factor in determining the magnitude of buoyant force.

3

Compare the fluid densities for the fully submerged scenarios.

Saltwater has a higher density than standard freshwater. Therefore,

1 \text{ m}^3

of saltwater weighs more than

1 \text{ m}^3

of freshwater.

When displaced volumes are equal, the fluid with the higher density exerts a greater buoyant force.

4

Order the scenarios based on the weight of the displaced fluid.

Submerged in Saltwater > Submerged in Freshwater > Half-submerged in Freshwater > Dry land.

This sequence correctly ranks the situations from greatest to least buoyant force.

Key Concept

According to Archimedes' Principle, buoyant force is directly proportional to both the volume of fluid displaced and the density of that fluid.

Question 2Question →

A cylindrical water storage tank is filled to a depth of $4\text{ meters}$ , creating a hydrostatic pressure of $40,000\text{ Pascals}$ at the tank's flat bottom. If water is drained from the tank until the depth is exactly $1\text{ meter}$ , what is the new hydrostatic pressure at the bottom in Pascals? (Assume the density of water and gravity remain constant.)

Show answer & explanation

Answer: 10000

Answer

10000

Because hydrostatic pressure is directly proportional to depth (

P = \rho g h

), reducing the depth to one-fourth of its original value (

1\text{ meter}

instead of

4\text{ meters}

) also reduces the pressure exactly to one-fourth of its original value (

40,000 / 4 = 10,000

).

Step-by-Step Solution

1

Identify the relationship between hydrostatic pressure and fluid depth.

Hydrostatic pressure is directly proportional to depth (

P \propto h

) when density and gravity are constant.

This fundamental principle allows us to solve the problem using a simple ratio.

2

Calculate the pressure per unit of depth.

\frac{40,000\text{ Pascals}}{4\text{ meters}} = 10,000\text{ Pascals per meter}

.

Finding the rate of pressure change per meter simplifies the calculation for the new depth.

3

Multiply the pressure per meter by the new depth.

10,000\text{ Pascals/meter} \times 1\text{ meter} = 10,000\text{ Pascals}

.

This yields the final pressure at the reduced depth.

Key Concept

Hydrostatic pressure is directly proportional to the depth of the fluid.

Question 3Question →

In a hydraulic car lift system, a mechanic applies a downward force of $150\text{ N}$ on a small input piston with a cross-sectional area of $5\text{ cm}^2$ . This force is transmitted through hydraulic fluid to a larger output piston that has a cross-sectional area of $100\text{ cm}^2$ . Assuming ideal conditions, what is the maximum upward force exerted by the larger piston?

7.5 N

750 N

3,000 N

15,000 N

Show answer & explanation

Answer: 3,000 N

Answer

3,000 N

The correct answer is

3,000\text{ N}

. According to Pascal's Principle, the fluid pressure in the system remains constant (

P = F_1/A_1 = 150/5 = 30\text{ N/cm}^2

). Multiplying this pressure by the area of the larger piston (

30 \times 100

) yields an upward force of

3,000\text{ N}

. Alternatively, the mechanical advantage is the ratio of the two areas (

100/5 = 20

). Multiplying the input force by this mechanical advantage (

150 \times 20

) directly gives

3,000\text{ N}

.

Step-by-Step Solution

1

Calculate the pressure exerted by the input piston on the hydraulic fluid.

P = \frac{F_1}{A_1} = \frac{150\text{ N}}{5\text{ cm}^2} = 30\text{ N/cm}^2

Pascal's Principle states that pressure applied to a confined fluid is transmitted undiminished in all directions.

2

Calculate the resulting upward force on the larger output piston.

F_2 = P \times A_2 = 30\text{ N/cm}^2 \times 100\text{ cm}^2 = 3,000\text{ N}

The uniform pressure of

30\text{ N/cm}^2

acts over the entire

100\text{ cm}^2

area of the larger piston, multiplying the original force.

Key Concept

Pascal's Principle and Hydraulic Mechanical Advantage

Question 4Question →

A hydraulic lift system has a small input piston with an area of $2 \text{ in}^2$ and a large output piston with an area of $10 \text{ in}^2$ . If a downward force of $20 \text{ lbs}$ is applied to the small piston, what is the resulting upward force exerted by the large piston?

4 \text{ lbs}

40 \text{ lbs}

100 \text{ lbs}

200 \text{ lbs}

Show answer & explanation

Answer: $100 \text{ lbs}$

Answer

The upward force exerted by the large piston is

100 \text{ lbs}

.

According to Pascal's Principle, pressure is uniform throughout a confined fluid system. The pressure at the input piston is

10 \text{ psi}

(

20 \text{ lbs}

divided by

2 \text{ in}^2

). When this

10 \text{ psi}

pressure acts upward across the

10 \text{ in}^2

area of the output piston, it generates an output force of

100 \text{ lbs}

(

10 \times 10

). Alternatively, you can recognize that the area ratio is

5

to

1

(

10 \div 2 = 5

), which means the input force of

20 \text{ lbs}

is multiplied by

5

to equal

100 \text{ lbs}

.

Step-by-Step Solution

1

Calculate the pressure exerted by the small piston.

P = \frac{20 \text{ lbs}}{2 \text{ in}^2} = 10 \text{ psi}

Pascal's Principle states that pressure applied to a confined fluid is transmitted equally in all directions. Pressure is equal to Force divided by Area.

2

Calculate the output force on the large piston using the uniform pressure.

F_{output} = 10 \text{ psi} \times 10 \text{ in}^2 = 100 \text{ lbs}

Force is equal to pressure multiplied by the area of the piston.

Key Concept

Pascal's Principle and Hydraulic Multiplication

Estimated Time:45s

Question 5Question →

A mechanic uses a hydraulic jack to lift a 2,000-pound engine. The output piston supporting the engine has a diameter that is exactly 4 times larger than the diameter of the input piston. Assuming an ideal system, if the mechanic pushes the input piston down a total distance of 16 inches, what is the required input force, and how far will the engine be lifted?

125 pounds of force; the engine is lifted 1 inch

500 pounds of force; the engine is lifted 4 inches

125 pounds of force; the engine is lifted 256 inches

32,000 pounds of force; the engine is lifted 1 inch

Show answer & explanation

Answer: 125 pounds of force; the engine is lifted 1 inch

Answer

125 pounds of force; the engine is lifted 1 inch

In a hydraulic system, the mechanical advantage is determined by the ratio of the piston areas. Since the output piston's diameter is 4 times the input piston's diameter, its area is

4^2

, or 16 times larger. This provides a mechanical advantage of 16. To lift a 2,000-pound engine, the required input force is

2,000 \div 16 = 125

pounds. Furthermore, because work is conserved (

F_{in} \times d_{in} = F_{out} \times d_{out}

), the input piston must move 16 times farther than the output piston. Therefore, pushing the input piston down 16 inches will lift the engine

16 \div 16 = 1

inch.

Step-by-Step Solution

1

Determine the area ratio between the two pistons.

Area Ratio =

4^2 = 16

.

In hydraulics, mechanical advantage depends on the cross-sectional area of the pistons. Because area scales with the square of the diameter (

A = \pi r^2

), a piston with 4 times the diameter has

4^2

, or 16 times, the area.

2

Calculate the required input force using Pascal's Principle.

Input Force =

2,000 \div 16 = 125

pounds.

The hydraulic system provides a mechanical advantage of 16. Therefore, the input force required is

1/16

th of the output load.

3

Calculate the distance the output piston (and engine) will move.

Output Distance =

16 \text{ inches} \div 16 = 1 \text{ inch}

.

By the principle of conservation of work (Work In = Work Out), if force is multiplied by 16, the distance traveled must be divided by 16 to keep the total volume of fluid displaced equal.

Key Concept

Pascal's Principle and Mechanical Advantage

Alternative Method

You can verify your answers using the principle of conservation of energy (Work In = Work Out). The work done by the input piston is

125 \text{ lbs} \times 16 \text{ inches} = 2,000 \text{ in-lbs}

. The work done on the engine is

2,000 \text{ lbs} \times 1 \text{ inch} = 2,000 \text{ in-lbs}

. Since the work values match, the force and distance calculations are correct.

Estimated Time:1m 30s

Question 6Question →

A solid metal sphere is suspended from a digital spring scale and lowered completely into four different storage tanks, each containing a different liquid. The true weight of the sphere in the air remains constant.

The liquids and their approximate densities are:
- Gasoline: $0.75 \text{ g/cm}^3$
- Glycerin: $1.26 \text{ g/cm}^3$
- Freshwater: $1.00 \text{ g/cm}^3$
- Ocean water: $1.03 \text{ g/cm}^3$

Arrange the liquids in order based on the apparent weight shown on the spring scale when the sphere is completely submerged, from the GREATEST apparent weight to the LEAST apparent weight.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

Gasoline, Freshwater, Ocean water, Glycerin

The correct order follows the fluid densities from lowest to highest. A fluid with a lower density displaces less mass for a given volume, exerting a smaller upward buoyant force. Because Apparent Weight = True Weight - Buoyant Force, the smallest buoyant force results in the largest apparent weight. Therefore, gasoline (lowest density) yields the greatest apparent weight, and glycerin (highest density) yields the least apparent weight.

Step-by-Step Solution

1

Identify the relationship between true weight, apparent weight, and buoyant force.

Apparent Weight = True Weight - Buoyant Force

A submerged object experiences an upward buoyant force that opposes gravity, reducing the measured weight on the scale.

2

Determine how the buoyant force relates to the different liquids.

Buoyant Force is directly proportional to the density of the liquid.

According to Archimedes' principle, buoyant force equals the weight of the displaced fluid. Since the same sphere is completely submerged every time, the displaced volume is constant, making buoyant force depend entirely on fluid density.

3

Correlate fluid density with apparent weight.

Lower fluid density produces a smaller buoyant force, resulting in a greater apparent weight.

Since apparent weight is calculated by subtracting the buoyant force from the constant true weight, subtracting a smaller value leaves a larger apparent weight.

4

Order the liquids from lowest density to highest density.

Gasoline (0.75), Freshwater (1.00), Ocean water (1.03), Glycerin (1.26).

This sequence represents the progression from the smallest buoyant force (greatest apparent weight) to the largest buoyant force (least apparent weight).

Key Concept

The apparent weight of a submerged object is its true weight minus the buoyant force. For a constant volume, a lower density fluid exerts a smaller buoyant force, resulting in a greater apparent weight.

Estimated Time:1m 0s

Question 7Question →

A heavy-duty hydraulic lifting system features an input piston with a cross-sectional area of $2\text{ in}^2$ and an output piston with an area of $40\text{ in}^2$ . The input piston is driven by a lever that provides a $4:1$ mechanical advantage. If the operator pushes the lever handle down a distance of $16\text{ inches}$ for each stroke, how far will the output piston move upward after $10$ complete strokes? (Assume an incompressible fluid and zero system losses.)

8 inches

32 inches

2 inches

800 inches

Show answer & explanation

Answer: 2 inches

Answer

The output piston will move upward by 2 inches.

In any hydraulic system, the volume of fluid displaced by the input piston must equal the volume received by the output piston. Due to the

4:1

mechanical advantage of the lever, the input piston only moves

1/4

of the handle's distance per stroke (

16 / 4 = 4\text{ inches}

). Displacing a

2\text{ in}^2

piston by

4\text{ inches}

pushes

8\text{ in}^3

of fluid per stroke. Over

10

strokes,

80\text{ in}^3

of fluid is moved. To accommodate this volume, the larger

40\text{ in}^2

output piston will move upward by

80 / 40 = 2\text{ inches}

.

Step-by-Step Solution

1

Calculate the distance the input piston moves during a single stroke.

d_1 = 16\text{ inches} / 4 = 4\text{ inches}

Because the lever provides a

4:1

mechanical advantage, the input piston moves one-fourth of the distance the handle moves.

2

Calculate the volume of fluid displaced in one stroke.

V_{\text{stroke}} = 2\text{ in}^2 \times 4\text{ inches} = 8\text{ in}^3

The volume of a cylinder is its cross-sectional area multiplied by the distance the piston travels.

3

Determine the total volume of fluid displaced after 10 strokes.

V_{\text{total}} = 8\text{ in}^3 \times 10 = 80\text{ in}^3

Each stroke moves the same volume of fluid, so we multiply the single-stroke volume by the total number of strokes.

4

Calculate the upward distance the output piston moves.

d_2 = 80\text{ in}^3 / 40\text{ in}^2 = 2\text{ inches}

Since liquids are incompressible, the volume entering the large cylinder equals the total volume displaced by the input piston.

Key Concept

Pascal's Principle and Conservation of Volume in Hydraulics

Question 8Question →

Match each component or characteristic of a basic hydraulic system to its correct mechanical behavior or property based on Pascal's Principle.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Small input piston -> Requires less mechanical force but must move a longer distance; Large output piston -> Exerts a greater mechanical force but moves a shorter distance; Incompressible fluid -> Transmits force instantly without absorbing it through compression; Hydraulic pressure -> Remains constant and uniform throughout the entire enclosed system.

In a hydraulic system, Pascal's Principle dictates that pressure is uniform. An incompressible liquid ensures energy is transferred efficiently. To conserve work, a smaller piston requires less force but moves further, while the displaced fluid causes the larger piston to move a shorter distance with greater force.

Step-by-Step Solution

1

Recall the fundamental rule of Pascal's Principle regarding pressure.

Pressure is constant throughout the system.

This allows matching 'Hydraulic pressure' with 'Remains constant and uniform'.

2

Analyze the trade-off between force and distance in a hydraulic system.

Work input equals work output. Small area means small force but long distance; large area means large force but short distance.

This correctly pairs the small input piston and the large output piston with their respective distance and force behaviors.

3

Identify the property of the fluid used in the system.

The fluid must not compress so power isn't lost.

This connects 'Incompressible fluid' to 'Transmits force instantly'.

Key Concept

Pascal's Principle states that pressure in a confined, incompressible fluid is transmitted equally in all directions, allowing a system to multiply force at the direct expense of travel distance.

Question 9Question →

In an ideal hydraulic lift system, an input piston applies force to an enclosed, incompressible fluid to lift a heavy load on an output piston. Match each modification made to the system on the left with its corresponding physical result on the right. Assume only the stated modification occurs and input force remains constant unless otherwise noted.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings link output area changes to inverse distance effects, input area reductions to increased system pressure, stroke length changes to proportional output distance changes, and fluid density changes to having no effect on theoretical mechanical advantage.

The correct pairings correctly apply the two governing rules of ideal hydraulics:

P = F/A

(which determines force multiplication) and

V = A \times d

(which determines piston travel distance). Every change to an area variable inversely affects the distance variable.

Step-by-Step Solution

1

Analyze the relationship for output force based on Pascal's Principle.

The formula is

F_{out} = F_{in} \times \frac{A_{out}}{A_{in}}

.

This establishes how changes in either piston area affect the maximum lifting force.

2

Analyze the relationship for piston travel distances based on conservation of fluid volume.

The formula is

d_{out} = d_{in} \times \frac{A_{in}}{A_{out}}

.

This establishes the trade-off between force amplification and distance; mechanical advantage always comes at the cost of requiring greater input distance.

3

Evaluate each modification against both formulas simultaneously.

Doubling

A_{out}

doubles

F_{out}

but halves

d_{out}

. Halving

A_{in}

doubles pressure and

F_{out}

, but halves the volume displaced per inch of stroke, requiring twice the

d_{in}

.

Hydraulic systems must obey both the pressure equation and the conservation of volume simultaneously.

Key Concept

Pascal's Principle and Conservation of Volume in Hydraulics

Estimated Time:2m 0s

Question 10Question →

Match each change made to a closed hydraulic system with its resulting effect on the output piston. Assume all other initial parameters remain constant.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The input area affects both pressure (inversely) and volume pushed (directly). The output area affects output force (directly) and output distance (inversely). Input force only affects pressure, and input distance only affects volume.

Each relationship is governed by two fundamental principles of hydraulics. Pascal's Principle dictates that

F_{out} = F_{in} \times (A_{out}/A_{in})

. The conservation of volume dictates that

d_{out} = d_{in} \times (A_{in}/A_{out})

. Applying a factor of 2 to any single variable in these equations yields the correct changes to the output force and distance.

Step-by-Step Solution

1

Analyze the formula for mechanical advantage and output force.

F_{out} = F_{in} \times \frac{A_{out}}{A_{in}}

Pascal's Principle states that pressure is transmitted equally (

P = F/A

), allowing us to solve for the force tradeoff.

2

Analyze the formula for output travel distance.

d_{out} = d_{in} \times \frac{A_{in}}{A_{out}}

Because liquids are virtually incompressible, the volume of fluid pushed by the input piston must equal the volume moving the output piston (

V = A \times d

).

3

Evaluate the effect of doubling each parameter independently.

Doubling

A_{in}

halves

F_{out}

and doubles

d_{out}

. Doubling

A_{out}

doubles

F_{out}

and halves

d_{out}

.

The direct and inverse proportions in the formulas dictate the final force multiplication and travel distance outcomes.

Key Concept

Hydraulic Mechanical Advantage and Conservation of Work

Question 11Question →

A barber uses a hydraulic system to lift a customer in a chair. The output piston beneath the chair has a cross-sectional area that is exactly 10 times larger than the input piston connected to the foot pedal. If the barber applies a downward force of $25 \text{ lbs}$ on the input piston, how much upward force does the output piston generate?

2.5 \text{ lbs}

25 \text{ lbs}

250 \text{ lbs}

2,500 \text{ lbs}

Show answer & explanation

Answer: $250 \text{ lbs}$

Answer

The output piston generates

250 \text{ lbs}

of upward force.

Based on Pascal's Principle, hydraulic systems multiply force proportionally to the increase in the cross-sectional area of the pistons. Because the output piston's area is 10 times larger than the input piston's area, the output force is 10 times the input force. Multiplying the

25 \text{ lbs}

of input force by 10 gives

250 \text{ lbs}

of output force.

Step-by-Step Solution

1

Identify the mechanical advantage (MA) based on the area ratio.

The area ratio is given as 10 to 1, meaning the MA is 10.

Pascal's Principle states that pressure is transmitted equally, so force is multiplied by the same factor as the area increases.

2

Calculate the output force by multiplying the input force by the mechanical advantage.

25 \text{ lbs} \times 10 = 250 \text{ lbs}

.

The output force is directly proportional to the increase in area.

Key Concept

Pascal's Principle and Hydraulic Force Multiplication

Question 12Question →

A hydraulic rescue tool, often used by emergency responders to pry open crushed vehicles, operates based on Pascal's Principle. The tool's pump applies $200\text{ lbs}$ of force to an input piston that has a cross-sectional area of $0.5\text{ in}^2$ . In order to generate $4,000\text{ lbs}$ of spreading force at the output piston, what must be the cross-sectional area of the output piston?

10\text{ in}^2

20\text{ in}^2

40\text{ in}^2

0.025\text{ in}^2

Show answer & explanation

Answer: $10\text{ in}^2$

Answer

The cross-sectional area of the output piston must be

10\text{ in}^2

.

In a hydraulic system, pressure is transmitted equally in all directions based on Pascal's Principle. First, determine the system pressure generated by the input piston by dividing the applied force by the area:

200\text{ lbs} / 0.5\text{ in}^2 = 400\text{ psi}

. Since this same pressure acts on the output piston, the required area to produce

4,000\text{ lbs}

of lifting force is found by dividing the output force by the pressure:

4,000\text{ lbs} / 400\text{ psi} = 10\text{ in}^2

. Alternatively, you can identify that the output force is 20 times the input force, meaning the output area must be 20 times the input area:

0.5\text{ in}^2 \times 20 = 10\text{ in}^2

.

Step-by-Step Solution

1

Calculate the fluid pressure generated by the input piston using the formula Pressure = Force / Area.

P = \frac{200\text{ lbs}}{0.5\text{ in}^2} = 400\text{ psi}

Pascal's Principle states that pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid, so the output piston experiences this same pressure.

2

Use the calculated pressure to find the required area of the output piston using the rearranged formula Area = Force / Pressure.

A_2 = \frac{4,000\text{ lbs}}{400\text{ psi}} = 10\text{ in}^2

Dividing the desired output force by the uniform system pressure yields the necessary cross-sectional area to achieve that force.

Key Concept

Pascal's Principle (Hydraulic Multiplication)

Question 13Question →

A heavy machinery hydraulic lift is being redesigned to alter its performance characteristics. Match each proposed modification to the hydraulic pistons with its resulting effect on the output force and output piston travel distance. (Assume the input force and input piston stroke distance remain constant in all scenarios).

Click a left item, then click its matching right item

Show answer & explanation

Answer

Match each structural modification by first converting any diameter changes to area changes (by squaring the factor), then calculating the new mechanical advantage ratio, and applying the inverse factor to the stroke distance to conserve work.

Each modification changes the ratio of the output piston area to the input piston area, which dictates the mechanical advantage (MA). Because hydraulic systems conserve work, any factor that multiplies the output force must inversely scale the output distance. Crucially, while area scales directly, changes given as diameter must be squared to find the effect on area.

Step-by-Step Solution

1

Analyze the relationship between diameter and cross-sectional area for circular pistons.

Area is proportional to the square of the diameter (

A \propto D^2

). Doubling a diameter quadruples the area, and halving a diameter reduces the area to one-fourth.

To correctly determine how geometric modifications affect the actual area used in Pascal's principle.

2

Calculate the new mechanical advantage (MA) for the first modification (decreasing input diameter to half).

Input area becomes

1/4

of the original.

MA = \frac{A_{out}}{A_{in}}

, so the new MA is

\frac{1}{1/4} = 4

times the original. Output force becomes

4\times

.

To match the first modification to its force multiplication effect.

3

Apply the principle of conservation of work to determine the effect on output distance for the first modification.

Since Work In = Work Out (

F_{in} \times d_{in} = F_{out} \times d_{out}

), an increase in output force by a factor of 4 must result in a proportional decrease in output distance to

1/4

.

Pascal's principle dictates an inverse relationship between force multiplication and distance traveled.

4

Evaluate the remaining modifications using the same area scaling and work principles.

Modification 2 yields

4/2 = 2\times

MA. Modification 3 yields

9/9 = 1\times

MA. Modification 4 yields

1/4 = 1/4\times

MA. Output distances scale inversely for each.

To correctly pair all remaining items based on their calculated force and distance multipliers.

Key Concept

Pascal's Principle, Mechanical Advantage, and Work Conservation

Question 14Question →

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 15Question →

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 16Question →

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 17Question →

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.

Question 18Question →

A maintenance barge is floating in a closed canal lock filled with water. The barge carries a heavy load of solid steel beams. If the crew unloads the steel beams directly into the water, causing them to sink to the bottom of the lock, what happens to the overall water level in the lock, and why?

The water level rises, because the submerged steel now displaces a volume of water equal to its heavy weight.

The water level remains the same, because the total mass of the barge, steel, and water in the lock system has not changed.

The water level falls, because the sunken steel displaces less water than it did when its weight was supported by the floating barge.

The water level rises, because the steel beams displace more water when resting on the bottom than when sitting on the barge.

Show answer & explanation

Answer: The water level falls, because the sunken steel displaces less water than it did when its weight was supported by the floating barge.

Answer

The water level falls, because the sunken steel displaces less water than it did when its weight was supported by the floating barge.

When the steel beams are on the barge, they must displace a volume of water equal to their weight in order to float. Because steel is denser than water, this displaced volume of water is larger than the actual physical volume of the steel beams. When the beams are thrown overboard and sink, they only displace a volume of water equal to their own physical volume. Since they are now displacing a smaller volume of water than before, the overall water level in the lock falls.

Step-by-Step Solution

1

Determine the volume of water displaced by the steel beams when they are floating on the barge.

The steel beams are part of a floating system, so they displace a weight of water equal to their own weight.

According to Archimedes' principle, a floating object displaces a volume of fluid whose weight equals the object's weight. The displaced volume is

V_{floating} = \frac{W_{steel}}{d_{water}}

where

d

is density.

2

Determine the volume of water displaced by the steel beams when they sink to the bottom of the lock.

The steel beams are completely submerged, so they displace a volume of water equal to their own geometric volume.

A completely submerged object displaces a volume of fluid exactly equal to its own volume, so

V_{submerged} = V_{steel}

.

3

Compare the two displaced volumes.

Since steel is denser than water (

d_{steel} > d_{water}

), the volume of the steel itself is much less than the volume of water that has the same weight. Therefore,

V_{submerged} < V_{floating}

.

Density is mass divided by volume. For a given mass (or weight), a material with a higher density occupies a smaller geometric volume.

4

Determine the effect on the overall water level in the lock.

Because the total volume of water displaced by the system decreases when the steel is dropped to the bottom, the overall water level falls.

The water level in a closed system depends directly on the total volume of water being displaced by objects in the water.

Key Concept

Flotation equilibrium vs. Submerged volume

Question 19Question →

A hydraulic theater stage lift is used to raise a heavy set piece. The output piston supporting the stage has a cross-sectional area of $120 \text{ in}^2$ , while the input piston connected to the control system has an area of $8 \text{ in}^2$ . If the set piece and stage exert a combined downward force of $2,400 \text{ lbs}$ on the output piston, how much input force, in pounds, must be applied to the input piston to hold the stage steady?

Show answer & explanation

Answer: 160

Answer

160

Based on Pascal's Principle, an enclosed fluid transmits pressure equally. The stage creates a pressure of

20 \text{ psi}

(

2,400 \text{ lbs} / 120 \text{ in}^2

). Because this

20 \text{ psi}

is transmitted to the input piston, which has an area of

8 \text{ in}^2

, the required force is

20 \text{ psi} \times 8 \text{ in}^2 = 160 \text{ lbs}

.

Step-by-Step Solution

1

State Pascal's Principle for the system.

P_{in} = P_{out}

, meaning

\frac{F_{in}}{A_{in}} = \frac{F_{out}}{A_{out}}

Pascal's Principle dictates that pressure applied to an enclosed fluid is transmitted equally in all directions.

2

Calculate the pressure at the output piston.

P = \frac{2,400 \text{ lbs}}{120 \text{ in}^2} = 20 \text{ psi}

This establishes the uniform pressure throughout the entire hydraulic system.

3

Calculate the required input force.

F_{in} = P \times A_{in} = 20 \text{ psi} \times 8 \text{ in}^2 = 160 \text{ lbs}

Applying the system pressure to the smaller area of the input piston yields the necessary input force.

Key Concept

Pascal's Principle and Mechanical Advantage

Question 20Question →

In a specialized heavy-duty hydraulic system, the master cylinder has a piston diameter of $2 \text{ cm}$ and the lifting cylinder has a piston diameter of $10 \text{ cm}$ . If the master cylinder piston is pushed through a total stroke distance of $75 \text{ cm}$ , how far, in cm, will the lifting cylinder piston move?

Show answer & explanation

Answer: 3

Answer

3

In a closed hydraulic system, the volume of fluid displaced by the input piston must equal the volume of fluid that moves the output piston. Because volume is the product of the piston's cross-sectional area and its stroke distance (

V = A \times d

), and area is proportional to the square of the diameter (

A \propto D^2

), the distance moved is inversely proportional to the square of the diameter. The lifting cylinder has a diameter

5

times larger than the master cylinder (

10 / 2 = 5

), which means its area is

25

times larger (

5^2 = 25

). Therefore, to conserve fluid volume, the lifting piston will move

1/25

of the distance of the master piston. Dividing the master piston's distance of

75 \text{ cm}

by

25

gives an output distance of

3 \text{ cm}

.

Step-by-Step Solution

1

State the principle of volume conservation for incompressible fluids.

V_{in} = V_{out}

In a closed hydraulic system, the volume of fluid pushed by the input piston must equal the volume displacing the output piston.

2

Express volume in terms of piston diameter and stroke distance.

V \propto D^2 \times d

Volume equals cross-sectional area times distance, and area is proportional to the square of the diameter.

3

Set up the equation using the given values.

2^2 \times 75 = 10^2 \times d_{out}

Substituting

D_{in} = 2

,

d_{in} = 75

, and

D_{out} = 10

allows us to solve for the unknown output distance.

4

Simplify and solve for the output distance.

300 = 100 \times d_{out} \rightarrow d_{out} = 3

Basic algebra isolates the variable to find the final distance.

Key Concept

Hydraulic volumetric conservation and area-distance tradeoffs

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