All practice questions

Question 1641Question →

A jackscrew system has a stationary base containing a standard right-hand threaded nut. The screw, which has 4 threads per inch (TPI), supports a platform on its top. A technician uses a 1.25-foot long handle to turn the screw and lift a $720 \text{ lb}$ load exactly $2 \text{ inches}$ . Assuming $\pi \approx 3$ to simplify calculations and $100\%$ mechanical efficiency, which of the following correctly identifies the required rotation direction, effort force, and total work done?

Counterclockwise rotation; 2 lbs effort force; 1440 in-lbs total work

Clockwise rotation; 2 lbs effort force; 1440 in-lbs total work

Counterclockwise rotation; 24 lbs effort force; 1440 in-lbs total work

Counterclockwise rotation; 2 lbs effort force; 4 in-lbs total work

Show answer & explanation

Answer: Counterclockwise rotation; 2 lbs effort force; 1440 in-lbs total work

Answer

Counterclockwise rotation; 2 lbs effort force; 1440 in-lbs total work

Lifting the load requires backing the right-hand screw out of its nut, which means turning it counterclockwise. The IMA is calculated by matching units:

R = 15 \text{ inches}

and Pitch

= 0.25 \text{ inches}

. Using

\text{IMA} = \frac{2\pi R}{Pitch}

, the IMA is

\frac{90}{0.25} = 360

. The required effort force is the load divided by the IMA (

720 / 360 = 2 \text{ lbs}

). Finally, simple machines do not change total work; Work Out = Load

\times

Distance (

720 \text{ lbs} \times 2 \text{ inches} = 1440 \text{ in-lbs}

).

Step-by-Step Solution

1

Determine the required rotation direction for the screw.

Counterclockwise rotation.

A standard right-hand thread advances (moves inward/downward) when turned clockwise. To lift a load, the screw must back out of the base, which requires counterclockwise rotation.

2

Convert the handle length to inches and calculate the Ideal Mechanical Advantage (IMA).

R = 1.25 \text{ ft} = 15 \text{ inches}

. Pitch (

P

) =

\frac{1}{4} = 0.25 \text{ inches}

.

\text{IMA} = \frac{2\pi R}{P} = \frac{2(3)(15)}{0.25} = 360

.

Units must match (both in inches) to accurately calculate the mechanical advantage. IMA relates the circumference traveled by the effort to the linear pitch.

3

Calculate the required effort force.

\text{Effort Force} = \frac{\text{Load}}{\text{IMA}} = \frac{720 \text{ lbs}}{360} = 2 \text{ lbs}

.

The machine multiplies the effort force by a factor equal to the IMA (assuming

100\%

efficiency).

4

Calculate the total work done.

\text{Work} = \text{Load} \times \text{Vertical Distance} = 720 \text{ lbs} \times 2 \text{ inches} = 1440 \text{ in-lbs}

.

In any ideal machine, Work In equals Work Out. The total work performed against gravity depends only on the load weight and the distance it is lifted, regardless of the mechanical advantage.

Key Concept

Mechanics of Screws: Pitch, Mechanical Advantage, and Conservation of Work

Estimated Time:2m 0s

Question 1642Question →

Arrange the following stages of a skydiver's jump in chronological order, starting from the moment they step out of a hovering helicopter until just before they touch the ground.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence follows the progression of forces: maximum initial acceleration, decreasing acceleration as drag builds, first terminal velocity, rapid deceleration upon parachute deployment, and finally a second, slower terminal velocity.

A falling object initially accelerates under gravity alone. As its velocity increases, air resistance grows until it perfectly opposes gravity, resulting in a constant terminal velocity. Deploying a parachute drastically increases the surface area, creating a temporary net upward force that slows the fall until a new, slower terminal velocity is established for a safe landing.

Step-by-Step Solution

1

Analyze the initial phase of the jump.

Identify that at the moment of the jump, velocity is zero.

Without velocity, there is no air resistance. Gravity is the only force, causing maximum acceleration.

2

Evaluate the effect of increasing speed.

Determine that accelerating downwards causes air resistance to grow.

Air drag is proportional to velocity. As drag increases, it opposes gravity, reducing the net downward acceleration.

3

Identify the first equilibrium state.

Recognize that terminal velocity is reached when air resistance equals gravity.

When the net force is zero, Newton's first law dictates that acceleration stops and velocity becomes constant.

4

Assess the impact of parachute deployment.

Identify that the parachute causes a massive, sudden increase in air drag.

The drag force temporarily exceeds the gravitational force, creating an upward net force that rapidly slows the downward fall.

5

Determine the final state before landing.

Conclude that a new equilibrium is reached at a much slower speed.

As the skydiver slows down, the drag on the parachute decreases until it balances gravity once more.

Key Concept

The relationship between gravity, air resistance, and terminal velocity during free fall.

Question 1643Question →

A solid drive shaft transmits rotational power to a heavy gear. To lock them together and prevent the gear from slipping, a rectangular steel key is fitted into a shared horizontal slot between the shaft and the gear's inner hub. When the motor engages and the shaft forcefully rotates while the heavy gear resists, the shaft pushes against one side of the key's bottom half, while the gear's hub pushes back against the opposite side of the key's top half. What primary type of mechanical stress does the steel key experience in this scenario?

Shear stress

Torsional stress

Tensile stress

Compressive stress

Show answer & explanation

Answer: Shear stress

Answer

Shear stress

The correct answer accurately identifies that the key experiences two opposing forces acting on parallel, offset planes. The shaft pushes one half, and the gear pushes the other. This creates a cutting or sliding effect along the horizontal plane between the two halves, which is the exact definition of shear stress.

Step-by-Step Solution

1

Analyze the forces acting on the specific component in question (the steel key).

The shaft pushes the bottom half of the key in the direction of rotation, while the gear hub resists, pushing the top half of the key in the opposite direction.

Accurate force identification is the first step in determining internal stress.

2

Evaluate the alignment of these opposing forces.

The forces act parallel to each other but on slightly offset planes (top half vs. bottom half).

The alignment of the load dictates how the material will tend to deform.

3

Match the observed force pattern to standard mechanical stress definitions.

Opposing forces acting on offset, parallel planes attempt to slide one portion of the material over another, which defines shear stress.

This confirms the correct classification of the mechanical load.

Key Concept

Identifying Shear Stress in Fasteners and Load-Bearing Joints

Question 1644Question →

A uniform $10$ -foot steel pry bar weighing $40$ pounds is used to lift a rock. A fulcrum is placed exactly $2$ feet from the load end of the bar. A $480$ -pound downward force from the rock acts exactly at the load end. How much downward effort force must be applied at the opposite end of the pry bar to keep it perfectly balanced and horizontal?

84 lbs

105 lbs

120 lbs

135 lbs

Show answer & explanation

Answer: 105 lbs

Answer

105 lbs

The system is balanced when clockwise and counter-clockwise torques are equal. The rock creates

960 \text{ lb-ft}

of counter-clockwise torque (

480 \times 2

). The bar's weight creates

120 \text{ lb-ft}

of clockwise torque (

40 \times 3

, since its center is

3 \text{ ft}

from the fulcrum). The remaining clockwise torque must be provided by the effort force applied at

8 \text{ ft}

from the fulcrum. Therefore,

960 = 120 + 8E

, which solves to

E = 105 \text{ lbs}

.

Step-by-Step Solution

1

Calculate the counter-clockwise torque created by the rock.

480 \text{ lbs} \times 2 \text{ ft} = 960 \text{ lb-ft}

Torque is force multiplied by the distance to the fulcrum. The rock acts

2

feet from the fulcrum.

2

Determine the location of the pry bar's center of gravity relative to the fulcrum.

The center is

5 \text{ ft}

from either end. The fulcrum is at

2 \text{ ft}

. The distance from the fulcrum to the center of gravity is

5 - 2 = 3 \text{ ft}

on the effort side.

A uniform bar's weight acts at its exact center point.

3

Calculate the clockwise torque created by the pry bar's own weight.

40 \text{ lbs} \times 3 \text{ ft} = 120 \text{ lb-ft}

The bar's weight assists the effort force because its center of gravity is on the same side of the fulcrum as the applied effort.

4

Set up the torque balance equation to find the required effort force.

960 \text{ lb-ft} = 120 \text{ lb-ft} + (E \times 8 \text{ ft})

For the lever to be balanced, total counter-clockwise torque must equal total clockwise torque. The effort is applied

10 - 2 = 8 \text{ ft}

from the fulcrum.

5

Solve for the effort force (E).

8E = 960 - 120 \Rightarrow 8E = 840 \Rightarrow E = 105 \text{ lbs}

Isolating E gives the exact downward force required at the end of the bar.

Key Concept

To balance a lever, the sum of all clockwise torques must equal the sum of all counter-clockwise torques. For heavy levers, the object's own weight acting at its center of gravity must be included in the calculation.

Question 1645Question →

A mechanic has four different bolts. Arrange the bolts in order from the LEAST linear distance traveled per full revolution to the GREATEST linear distance traveled per full revolution.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from least to greatest linear travel is: 32 TPI, 20 TPI, 16 TPI, and 10 TPI.

The linear distance a screw travels in one full revolution is its pitch, which is calculated as 1 divided by the Threads Per Inch (TPI). A higher TPI means the threads are closer together, so the screw advances a shorter distance per turn. Therefore, to order them from least to greatest distance traveled, you must arrange them from highest TPI to lowest TPI: 32 TPI (1/32 inch), 20 TPI (1/20 inch), 16 TPI (1/16 inch), and 10 TPI (1/10 inch).

Step-by-Step Solution

1

Understand the relationship between threads per inch (TPI) and linear travel.

Linear travel per revolution, also known as pitch, is equal to 1 divided by the TPI.

Each full turn advances the screw by one thread pitch.

2

Determine the linear travel for each bolt.

32 TPI = 1/32 inch, 20 TPI = 1/20 inch, 16 TPI = 1/16 inch, and 10 TPI = 1/10 inch.

Applying the relationship pitch = 1 / TPI.

3

Order the calculated values from least to greatest.

1/32 < 1/20 < 1/16 < 1/10.

The question asks for the order starting with the least linear distance traveled.

Key Concept

The linear distance a screw travels in one revolution is its pitch. For standard screws, pitch is the reciprocal of the threads per inch (TPI). Therefore, a higher TPI results in a smaller linear travel per revolution.

Estimated Time:45s

Question 1646Question →

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 1647Question →

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 1648Question →

Two overlapping steel plates are fastened together by a single rivet. If opposing lateral forces pull the two plates in opposite directions away from each other, what type of internal stress is primarily experienced by the rivet?

Shear stress

Tensile stress

Compressive stress

Torsional stress

Show answer & explanation

Answer: Shear stress

Answer

Shear stress

The forces from the plates act in parallel but opposite directions across the cross-section of the rivet. This attempts to slide the top half of the rivet off the bottom half, which perfectly describes shear stress (similar to how a pair of shears cuts paper).

Step-by-Step Solution

1

Analyze the direction of the external forces acting on the joint.

The plates are being pulled in opposite directions, parallel to their mating surfaces.

Identifying the force vectors is necessary to determine how they interact with the fastener.

2

Determine how these forces translate to the rivet.

The top plate pushes against the top half of the rivet in one direction, while the bottom plate pushes against the bottom half in the opposite direction.

This cross-sectional cutting action defines the type of stress on the internal component.

3

Match the physical action to the definition of structural stresses.

Forces causing internal layers of a material to slide past each other create shear stress.

Applying the standard definition of shear stress in mechanical systems yields the final answer.

Key Concept

Types of Stress and Strain

Estimated Time:30s

Question 1649Question →

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 1650Question →

A woodworker is selecting a replacement spindle for a workbench vise. They have four options with identical handle lengths and spindle diameters. Arrange the vise spindles in order from the one that applies the MAXIMUM clamping force for a given amount of effort, to the one that applies the MINIMUM clamping force.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence is: 12 TPI, 8 TPI, 5 TPI, and 3 TPI.

The mechanical advantage of a screw determines how much clamping force is produced from a given effort force. It is inversely proportional to its pitch (the linear distance between adjacent threads). Threads per inch (TPI) is the inverse of pitch, meaning a higher TPI results in a smaller pitch. A smaller pitch creates a shallower thread angle, which maximizes the mechanical advantage and produces the greatest clamping force. Therefore, the spindle with 12 TPI applies the maximum force, while the one with 3 TPI applies the minimum force.

Step-by-Step Solution

1

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

Pitch is the inverse of TPI. A higher TPI translates to a smaller pitch (distance between threads).

Understanding pitch is necessary to determine the mechanical advantage of the screw.

2

Relate pitch to mechanical advantage and clamping force.

A smaller pitch means the screw advances less linear distance per revolution, acting like a shallower inclined plane. This increases mechanical advantage, resulting in greater output clamping force for the same effort.

The question asks to order the spindles by the amount of clamping force applied.

3

Rank the spindles from maximum to minimum clamping force.

Maximum clamping force corresponds to the highest TPI. The correct order is 12 TPI, 8 TPI, 5 TPI, and 3 TPI.

This establishes the final required arrangement.

Key Concept

Mechanical Advantage of Screws Based on TPI

Question 1651Question →

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

Drag items to arrange them in the correct order

Show answer & explanation

Answer

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

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

Step-by-Step Solution

1

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

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

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

IMA = \frac{L}{h}

).

2

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

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

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

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

).

3

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

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

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

4

Arrange the final total effort values in descending order.

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

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

Key Concept

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

Question 1652Question →

A skydiver jumps from a stationary hot air balloon. Arrange the following moments of the jump in order of the skydiver's acceleration, starting from the greatest downward acceleration and ending with the greatest upward acceleration.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence tracks the decreasing downward acceleration: immediately after jumping (maximum downward), followed by several seconds into the fall (reduced downward), then falling at terminal velocity (zero acceleration), and finally the moment the parachute inflates (strong upward acceleration).

The correct sequence tracks the net force acting on the skydiver. Initially, only gravity acts, causing maximum downward acceleration. As speed increases, upward air drag grows, reducing the net downward force and acceleration. At terminal velocity, drag equals weight, making net force and acceleration zero. When the parachute opens, drag vastly exceeds weight, creating a net upward force and thus an upward acceleration.

Step-by-Step Solution

1

Analyze the forces acting immediately after the jump.

Velocity is zero, so air drag is zero. The only force is gravity, resulting in the maximum downward acceleration of approximately 9.8 m/s².

Air drag depends on velocity. Without velocity, there is no drag to oppose gravity.

2

Evaluate the forces during the middle of the fall before terminal velocity.

As the skydiver speeds up, upward air drag increases. The net downward force becomes less than gravity, so downward acceleration is less than 9.8 m/s².

Net force equals weight minus air drag. As drag increases, net force decreases, reducing acceleration according to Newton's Second Law.

3

Identify the conditions at terminal velocity.

Upward air drag exactly equals downward weight. The net force is zero, meaning acceleration is exactly 0 m/s².

Terminal velocity is reached when opposing forces balance out completely, ending any change in speed.

4

Assess the moment the parachute inflates.

The huge surface area creates an upward air drag much greater than the downward weight. The net force points upward, creating a strong upward acceleration.

A net upward force is required to rapidly decrease the downward velocity to a safe landing speed.

Key Concept

The relationship between gravity, air resistance, and net acceleration during free fall through an atmosphere.

Question 1653Question →

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

Drag items to arrange them in the correct order

Show answer & explanation

Answer

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

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

\frac{1}{5}

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

\frac{1}{4}

(25%), the 18 ft ramp requires

\frac{1}{3}

(33.3%), and the 14 ft ramp requires

\frac{1}{2}

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

Step-by-Step Solution

1

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

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

.

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

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

).

2

Calculate the effort force fraction for each ramp configuration.

25 ft ramp:

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

; 20 ft ramp:

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

; 18 ft ramp:

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

; 14 ft ramp:

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

.

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

3

Order the calculated fractions from lowest to highest.

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

.

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

Key Concept

Ideal Mechanical Advantage and Effort Force on an Inclined Plane

Question 1654Question →

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 1655Question →

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

Drag items to arrange them in the correct order

Show answer & explanation

Answer

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

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

Step-by-Step Solution

1

Calculate the effort force for the

750\text{ lb}

cargo.

Force =

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

.

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

2

Calculate the effort force for the

400\text{ lb}

cargo.

Force =

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

.

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

3

Calculate the effort force for the

600\text{ lb}

cargo.

Force =

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

.

The ramp has an IMA of 4, which divides the

600\text{ lb}

load by 4.

4

Calculate the effort force for the

500\text{ lb}

cargo.

Force =

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

.

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

5

Order the calculated effort forces from least to greatest.

The correct order is

100\text{ lbs}

,

150\text{ lbs}

,

200\text{ lbs}

,

250\text{ lbs}

.

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

Key Concept

Ideal Mechanical Advantage and Effort Force in Inclined Planes

Estimated Time:2m 0s

Question 1656Question →

Match each type of mechanical stress to the description of the forces that cause it.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Tension matches stretching forces, Compression matches squeezing forces, Shear matches unaligned parallel forces causing sliding, and Torsion matches twisting forces.

Each type of stress is strictly defined by the nature of the applied load: outward pulling causes tension, inward pushing causes compression, unaligned sliding forces cause shear, and twisting causes torsion.

Step-by-Step Solution

1

Analyze 'Tension'

Matches with pulling and stretching forces.

By definition, tensile stress elongates materials.

2

Analyze 'Compression'

Matches with pushing and squeezing forces.

By definition, compressive stress shortens materials.

3

Analyze 'Shear'

Matches with unaligned parallel forces.

Shear forces act like scissors, pushing adjacent planes of material in opposite directions.

4

Analyze 'Torsion'

Matches with twisting forces.

Torsional stress is applied rotationally around an axis.

Key Concept

Identifying the primary types of mechanical stress based on the direction and application of applied forces.

Estimated Time:45s

Question 1657Question →

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).

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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 1658Question →

In mechanical engineering, understanding the dominant forces at contact interfaces is critical for system design. Match each practical mechanical scenario to the primary type of friction or resistance that governs its operation.

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Answer

The conveyor belt pairs with Static Friction; the skidding tire pairs with Kinetic (Sliding) Friction; the oil-supported crankshaft pairs with Viscous/Fluid Resistance; and the steel spheres pair with Rolling Resistance.

Each mechanical scenario operates under a distinct physical resistance mechanism. Static friction transfers torque when parts move together without slipping. Kinetic friction dominates during sliding events like skidding. Viscous resistance occurs when fluids separate moving parts, and rolling resistance occurs due to material deformation in rolling elements.

Step-by-Step Solution

1

Analyze the contact interface of the conveyor belt.

Identify that the belt does not slip against the drive pulley.

Since the surfaces are stationary relative to one another at the point of contact, the force transferring the torque is static friction.

2

Analyze the state of the skidding automotive tire.

Identify that the rubber is actively sliding across the asphalt.

Sliding motion between two dry solid surfaces is always characterized by kinetic (sliding) friction.

3

Analyze the hydrodynamic journal bearing supporting the crankshaft.

Identify that a pressurized oil film entirely separates the solid components.

Because there is no solid-to-solid contact, the resistance to rotation is purely the viscous drag of the fluid.

4

Analyze the spheres in the high-speed wheel hub.

Identify that the spheres roll along the raceway rather than sliding.

The continuous micro-deformation of the materials as they roll creates energy loss, which defines rolling resistance.

Key Concept

Identifying applied frictional forces in complex mechanical systems

Estimated Time:1m 30s

Question 1659Question →

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

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Answer

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

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

2 \times \pi \times L

) by the Threads Per Inch (TPI). Because

2\pi

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

L

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

8 \times 8 = 64

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

60

), the 12-inch/4 TPI setup (

48

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

45

).

Step-by-Step Solution

1

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

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

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

2

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

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

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

3

Order the configurations based on the calculated factors.

64 > 60 > 48 > 45.

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

Key Concept

Ideal Mechanical Advantage of a Screw

Question 1660Question →

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

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Answer

Pry bar matches Class 1 (MA > 1). Hand truck matches Class 2 (MA > 1). Forceps match Class 3 (MA < 1). Trebuchet matches Class 1 (MA < 1).

Each mechanism must be evaluated by locating the fulcrum, effort, and load. A pry bar and trebuchet both have the fulcrum in the middle (Class 1), but differ in their arm ratios, resulting in MA > 1 and MA < 1 respectively. A hand truck has the load in the middle (Class 2, MA > 1). Forceps have the effort in the middle (Class 3, MA < 1).

Step-by-Step Solution

1

Analyze the pry bar scenario.

Identified as a Class 1 lever with MA > 1.

The fulcrum is located between the effort (operator) and the load (crate). Because the fulcrum is placed closer to the load, the effort arm is longer than the load arm, creating a mechanical advantage greater than 1.

2

Analyze the hand truck scenario.

Identified as a Class 2 lever with MA > 1.

The wheels serve as the fulcrum at one end. The refrigerator's weight (load) acts in the middle, and the operator's hands apply effort at the opposite end. This configuration is a Class 2 lever, which inherently multiplies force (MA > 1).

3

Analyze the forceps scenario.

Identified as a Class 3 lever with MA < 1.

The joined end is the fulcrum. The surgeon applies effort to the middle of the tool, and the tissue is the load at the far end. Effort in the middle defines a Class 3 lever, which always acts to multiply distance and speed at the expense of force (MA < 1).

4

Analyze the trebuchet scenario.

Identified as a Class 1 lever with MA < 1.

The pivot is between the dropping counterweight (effort) and the projectile (load), making it a Class 1 lever. Because the effort arm is much shorter than the load arm, the mechanical advantage is less than 1, which serves to drastically increase the speed of the projectile.

Key Concept

Identifying the relative positions of the fulcrum, effort, and load to classify levers (Class 1, 2, or 3) and analyzing arm lengths to determine if the mechanism multiplies force (MA > 1) or speed (MA < 1).

Estimated Time:2m 0s

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