All practice questions

Question 1621Question →

A simple wall-mounted jib crane is used to lift a heavy engine block. The crane consists of a rigid horizontal beam attached to a vertical wall at one end, and a diagonal support cable connecting the outer end of the beam to a point higher up on the wall. The engine block hangs directly from the outer end of the horizontal beam. Based on this configuration, which of the following correctly identifies the primary type of stress experienced by the diagonal cable and the horizontal beam?

The cable is in tension; the beam is in compression.

The cable is in compression; the beam is in tension.

Both the cable and the beam are in tension.

The cable is in tension; the beam is in shear.

Show answer & explanation

Answer: The cable is in tension; the beam is in compression.

Answer

The diagonal cable experiences tension, and the horizontal beam experiences compression.

The heavy load pulls down on the outer connection point. The diagonal cable resists this by pulling up and in toward the wall, putting the cable in tension. To counteract the inward pull of the cable, the horizontal beam is squeezed against the wall, putting the beam in compression.

Step-by-Step Solution

1

Analyze the forces acting on the outer end of the horizontal beam.

The suspended load creates a downward force at the connection point.

Identifying the primary external force is necessary to determine internal member stresses.

2

Determine the stress on the diagonal support cable.

To prevent the beam from falling, the cable must pull upward and backward toward the wall. This stretching force means the cable is in tension.

Cables can only support pulling forces (tension); they would buckle under pushing forces.

3

Determine the stress on the horizontal beam.

Because the cable is pulling backward toward the wall, the horizontal beam must push outward to keep the system in equilibrium. This squeezing force means the beam is in compression.

Structural members that are squeezed between opposing forces undergo compressive stress.

Key Concept

Tension and Compression in Structural Members

Estimated Time:45s

Question 1622Question →

A jackscrew used for lifting heavy loads has a thread pitch of $\frac{1}{8}$ inch. How far will the load be lifted if the screw is turned exactly 4 complete revolutions?

\frac{1}{2}

inch

\frac{1}{8}

inch

4 inches

32 inches

Show answer & explanation

Answer: $\frac{1}{2}$ inch

Answer

The load will be lifted

\frac{1}{2}

inch.

The pitch of a screw indicates how far it will move linearly with one full revolution. To find the total distance the load is lifted, multiply the pitch by the number of complete revolutions. In this case, multiplying

\frac{1}{8}

inch by 4 revolutions equals

\frac{4}{8}

inch, which simplifies to

\frac{1}{2}

inch.

Step-by-Step Solution

1

Identify the distance the screw travels in one revolution.

The screw travels

\frac{1}{8}

inch per revolution.

The pitch of a screw defines its linear travel distance for exactly one

360

-degree turn.

2

Calculate the total distance for the given number of revolutions.

Multiply the pitch (

\frac{1}{8}

inch) by the number of revolutions (4).

The total linear distance is the product of the pitch and the total number of revolutions.

3

Solve the multiplication.

4 \times \frac{1}{8} = \frac{4}{8} = \frac{1}{2}

inch.

Simplifying the fraction gives the final total travel distance.

Key Concept

Screw Pitch and Linear Travel

Question 1623Question →

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 1624Question →

A spacecraft hovers above a moon with no atmosphere. An astronaut drops a $5\text{ kg}$ tool from rest and observes that it takes exactly $4\text{ seconds}$ to fall $32\text{ meters}$ to the surface. Later, the astronaut drops a $10\text{ kg}$ piece of equipment from the exact same height. What is the weight of the $10\text{ kg}$ equipment on this moon, and how long will it take to reach the surface?

Weight is

40\text{ N}

; fall time is

4\text{ s}

Weight is

10\text{ kg}

; fall time is

4\text{ s}

Weight is

100\text{ N}

; fall time is

4\text{ s}

Weight is

40\text{ N}

; fall time is

2\text{ s}

Show answer & explanation

Answer: Weight is $40\text{ N}$ ; fall time is $4\text{ s}$

Answer

The weight of the equipment is

40\text{ N}

, and it will take

4\text{ s}

to fall.

The correct answer accurately calculates the moon's gravity as

4\text{ m/s}^2

using the first object's fall data. It then applies this gravity to find the second object's weight (

10\text{ kg} \times 4\text{ m/s}^2 = 40\text{ N}

) and correctly applies the principle that in a vacuum, mass does not affect fall time, resulting in the same

4\text{ s}

duration.

Step-by-Step Solution

1

Calculate the acceleration due to gravity on the moon using the kinematic equation

d = \frac{1}{2}at^2

.

32 = 0.5 \times a \times (4^2) \rightarrow 32 = 8a \rightarrow a = 4\text{ m/s}^2

.

The local acceleration must be known to determine the weight of the second object.

2

Determine the weight of the

10\text{ kg}

equipment on the moon.

W = m \times a = 10 \times 4 = 40\text{ N}

.

Weight is the force of gravity exerted on a mass in a specific gravitational field.

3

Determine the fall time for the

10\text{ kg}

equipment.

The fall time is

4\text{ s}

.

In a vacuum (no atmosphere), all objects accelerate downward at the exact same rate regardless of their mass.

Key Concept

Free fall kinematics and the distinction between mass and weight

Estimated Time:2m 0s

Question 1625Question →

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 1626Question →

A worker must load a $500\text{ lb}$ crate into a truck bed that is $4\text{ ft}$ off the ground. He can choose between two ramps. Ramp X is $8\text{ ft}$ long and operates as an ideal, frictionless machine. Ramp Y is $16\text{ ft}$ long, but has a rough surface that reduces its mechanical efficiency to $50\%$ . How do the required effort force and total work input compare when using Ramp Y instead of Ramp X?

Ramp Y requires the exact same effort force but twice the total work.

Ramp Y requires half the effort force and the exact same total work.

Ramp Y requires half the effort force but twice the total work.

Ramp Y requires twice the effort force and twice the total work.

Show answer & explanation

Answer: Ramp Y requires the exact same effort force but twice the total work.

Answer

Ramp Y requires the exact same effort force but twice the total work.

Although Ramp Y is twice as long as Ramp X (which ideally cuts the required force in half), its 50% efficiency means half of the worker's effort is wasted fighting friction. These two factors perfectly cancel each other out, resulting in the exact same effort force of 250 lbs. However, because the worker is applying that same 250 lbs of force over twice the distance (16 ft instead of 8 ft), the total work performed is doubled.

Step-by-Step Solution

1

Calculate the useful work output required to lift the crate.

\text{Work}_{\text{out}} = \text{Load} \times \text{Height} = 500\text{ lb} \times 4\text{ ft} = 2,000\text{ ft-lb}

.

This baseline value is the minimum energy required to achieve the lift regardless of the ramp used.

2

Evaluate the effort force and work input for Ramp X.

\text{IMA}_{\text{X}} = \frac{8\text{ ft}}{4\text{ ft}} = 2

.

\text{Effort}_{\text{X}} = \frac{500\text{ lb}}{2} = 250\text{ lb}

.

\text{Work}_{\text{in X}} = 250\text{ lb} \times 8\text{ ft} = 2,000\text{ ft-lb}

.

Because Ramp X is frictionless (100% efficient), work input equals work output, and the force is simply the load divided by the IMA.

3

Calculate the total work input required for Ramp Y.

\text{Efficiency} = \frac{\text{Work}_{\text{out}}}{\text{Work}_{\text{in}}}

. So,

0.50 = \frac{2,000}{\text{Work}_{\text{in}}}

, resulting in

\text{Work}_{\text{in Y}} = 4,000\text{ ft-lb}

.

Friction wastes energy as heat. A 50% efficiency means twice as much energy must be put into the system to get the required useful work out.

4

Calculate the effort force required for Ramp Y.

\text{Work}_{\text{in}} = \text{Effort} \times \text{Distance}

.

4,000\text{ ft-lb} = \text{Effort}_{\text{Y}} \times 16\text{ ft}

.

\text{Effort}_{\text{Y}} = 250\text{ lb}

.

Force can be derived by dividing the total actual work input by the total distance the force is applied.

5

Compare the force and work values between the two ramps.

Effort force is identical (

250\text{ lb}

for both). Total work for Ramp Y (

4,000\text{ ft-lb}

) is twice that of Ramp X (

2,000\text{ ft-lb}

).

This establishes the final correct comparison.

Key Concept

Actual Mechanical Advantage and Efficiency

Question 1627Question →

A $50\text{-kg}$ solid iron block and a $10\text{-kg}$ solid iron block are dropped simultaneously from a height of $100\text{ m}$ inside a large vacuum chamber. Which of the following statements correctly describes the acceleration of the two blocks as they fall?

Both blocks will experience an identical downward acceleration of approximately

9.8 \text{ m/s}^2

.

The

50\text{-kg}

block will accelerate five times faster than the

10\text{-kg}

block due to its greater mass.

Both blocks will experience an identical downward acceleration of approximately

9.8 \text{ km/h}^2

.

The

50\text{-kg}

block will accelerate at

490 \text{ m/s}^2

and the

10\text{-kg}

block at

98 \text{ m/s}^2

.

Show answer & explanation

Answer: Both blocks will experience an identical downward acceleration of approximately $9.8 \text{ m/s}^2$ .

Answer

Both blocks will experience an identical downward acceleration of approximately

9.8 \text{ m/s}^2

.

In the absence of air resistance, all objects near Earth's surface fall with the same constant acceleration due to gravity, which is roughly

9.8 \text{ m/s}^2

. Although the

50\text{-kg}

block experiences a stronger gravitational force (weight), it also has proportionally more inertia resisting motion. These two factors perfectly balance each other out, resulting in an identical rate of acceleration for both blocks.

Step-by-Step Solution

1

Identify the forces acting on the objects during their fall in the vacuum chamber.

The only force acting on both blocks is the gravitational pull of the Earth (their weight).

Because the chamber is a vacuum, air resistance is completely eliminated.

2

Apply Newton's Second Law of Motion (

F = ma

) to determine the rate of acceleration.

Acceleration (

a

) is equal to the force of gravity divided by the mass of the object (

a = \frac{F}{m}

).

This formula shows the relationship between the downward force, the object's inertia, and its resulting acceleration.

3

Calculate the theoretical acceleration for both masses.

Since the force of gravity (

F

) is equal to mass times gravitational acceleration (

F = m \times g

), the formula simplifies to

a = \frac{m \times g}{m}

, which means

a = g

.

The mass variable cancels itself out mathematically, proving that acceleration due to gravity is independent of an object's mass.

Key Concept

In a vacuum, all objects experience free fall at the exact same rate of gravitational acceleration, roughly

9.8 \text{ m/s}^2

, because their proportionally greater weight is perfectly offset by their proportionally greater inertia.

Question 1628Question →

A technician is comparing two threaded machine bolts of the same diameter. Bolt X has 10 threads per inch (TPI) and Bolt Y has 16 TPI. If both bolts are driven into a metal block using the exact same wrench, which of the following statements correctly describes how Bolt Y compares to Bolt X?

Bolt Y provides a greater mechanical advantage but advances a shorter linear distance per revolution.

Bolt X provides a greater mechanical advantage and advances a greater linear distance per revolution.

Bolt Y advances a greater linear distance per revolution because it has a higher thread count.

Both bolts provide the same mechanical advantage, but Bolt X advances faster.

Show answer & explanation

Answer: Bolt Y provides a greater mechanical advantage but advances a shorter linear distance per revolution.

Answer

Bolt Y provides a greater mechanical advantage but advances a shorter linear distance per revolution.

The correct answer accurately reflects the physics of a screw. A higher threads-per-inch (TPI) count (16 TPI for Bolt Y vs. 10 TPI for Bolt X) means the threads are closer together. This smaller pitch results in a shorter linear distance traveled per revolution. Because the effort force is applied over the same rotational distance (using the same wrench) but results in a smaller linear output distance, the mechanical advantage is greater. You trade distance for increased force multiplication.

Step-by-Step Solution

1

Determine the relationship between threads per inch (TPI) and pitch.

Pitch is the inverse of TPI. Bolt X has a pitch of 1/10 inch, while Bolt Y has a pitch of 1/16 inch.

Pitch defines the linear distance the bolt travels in one complete revolution.

2

Compare the linear advancement of both bolts.

Since 1/16 inch is less than 1/10 inch, Bolt Y advances a shorter distance per revolution than Bolt X.

A smaller pitch means less linear travel for each rotation.

3

Evaluate the mechanical advantage based on the pitch.

Bolt Y has a greater mechanical advantage than Bolt X.

A screw is essentially an inclined plane wrapped around a cylinder. A smaller pitch is equivalent to a shallower incline, which requires less effort force to move the load, thereby providing a higher mechanical advantage.

Key Concept

Screws and Threads

Estimated Time:1m 0s

Question 1629Question →

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 1630Question →

Arrange the following stages of a skydiver's descent from an aircraft to the ground in the correct chronological order.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence starts with initial downward acceleration, followed by increasing air resistance, reaching a constant terminal velocity, and finally parachute deployment to decelerate.

The sequence accurately describes the physical forces during an atmospheric free fall. Initially, gravity causes acceleration. As speed increases, so does opposing air resistance. Eventually, air resistance equals the gravitational force, resulting in a constant speed known as terminal velocity. Finally, deploying a parachute increases surface area, creating excess drag to rapidly decelerate the skydiver.

Step-by-Step Solution

1

Identify the initial state of the fall.

The skydiver steps out and accelerates downward.

At the start of the fall, speed is very low, so gravity is the dominant force causing an initial downward acceleration.

2

Determine what happens as the skydiver's downward speed increases.

Upward air resistance grows.

Air resistance (drag) increases proportionally as the speed of the falling object moving through the atmosphere increases.

3

Identify the state where the forces eventually balance.

Air resistance balances gravity, leading to a constant speed.

This state is known as terminal velocity. The net force becomes zero, meaning acceleration stops and the skydiver falls at a steady rate.

4

Identify the final stage required before safely reaching the ground.

The parachute is deployed to slow the fall.

Deploying the parachute vastly increases the surface area, which temporarily makes air resistance much greater than gravity, safely decelerating the descent.

Key Concept

Gravity, Air Resistance, and Terminal Velocity

Question 1631Question →

A building's concrete foundation pillars, which support the heavy weight of the entire structure above them, are primarily subjected to tension stress.

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because foundation pillars are pushed down upon by the immense weight of the building they support. This squeezing action produces compression stress. Tension stress only occurs when a material is being pulled or stretched.

Step-by-Step Solution

1

Analyze the forces acting on the foundation pillars.

The pillars support the weight of the building pushing down on them from above.

Identifying the direction and nature of the force is necessary to determine the specific type of stress.

2

Determine the type of stress caused by this specific force.

The downward force of the building squeezes the pillars against the ground. This squeezing action is defined as compression.

Tension is a pulling force that stretches a material, while compression is a pushing force that squeezes or compacts it.

Key Concept

Compression vs. Tension Stress

Estimated Time:30s

Question 1632Question →

A worker needs to move a heavy crate into a truck bed that is $3$ feet above the ground. The worker is deciding whether to use an $8$ -foot ramp or a $12$ -foot ramp. Assuming friction is negligible, which of the following statements correctly describes the effect of using the $12$ -foot ramp instead of the $8$ -foot ramp?

It requires the same effort force, but increases the total work performed due to the longer distance.

It decreases both the required effort force and the total work performed.

It decreases the required effort force, but the total work performed remains the same.

It increases the required effort force because the crate must be pushed over a greater distance.

Show answer & explanation

Answer: It decreases the required effort force, but the total work performed remains the same.

Answer

The

12

-foot ramp decreases the required effort force, but the total work performed remains the same.

A longer inclined plane provides a greater Ideal Mechanical Advantage, which means the effort force is reduced. However, because this smaller force must be applied over a proportionally longer distance, the total work done (

W = F \times d

) remains constant. In the absence of friction, the work done equals the work required to lift the load straight up.

Step-by-Step Solution

1

Determine the relationship between ramp length and Ideal Mechanical Advantage (IMA).

IMA is calculated as ramp length divided by vertical height. The

12

-foot ramp has a greater IMA than the

8

-foot ramp since both reach the same height of

3

feet.

To establish how the machine alters the required input force.

2

Determine the effect of a greater IMA on effort force.

A higher IMA means less effort force is required to move the same load. Therefore, the

12

-foot ramp requires less effort force.

Effort force is inversely proportional to mechanical advantage.

3

Evaluate the effect of ramp length on total work performed.

Work is the product of force and distance (

W = F \times d

). The smaller force applied over the longer

12

-foot distance equals the larger force applied over the

8

-foot distance. Total work remains unchanged.

Simple machines conserve energy; they cannot reduce the total amount of work required.

Key Concept

Work and Force Relationship on Inclined Planes

Estimated Time:1m 0s

Question 1633Question →

A mechanic uses a 60-inch straight pry bar to lift a 300 lb machinery crate. Initially, they place a pivot block 12 inches from the crate, pressing down on the opposite end to lift the load. To lift the crate higher on the next attempt, they move the pivot block so it is 20 inches from the crate, again pressing down on the opposite end. By how much does the required effort force change, and what is the new mechanical advantage of the lever?

The effort force increases by 75 lbs; the new mechanical advantage is 2.

The effort force increases by 40 lbs; the new mechanical advantage is 3.

The effort force increases by 150 lbs; the new mechanical advantage is 0.5.

The effort force increases by 50 lbs; the new mechanical advantage is 2.4.

Show answer & explanation

Answer: The effort force increases by 75 lbs; the new mechanical advantage is 2.

Answer

The effort force increases by 75 lbs; the new mechanical advantage is 2.

The correct answer accurately calculates that moving the pivot increases the load arm to 20 inches and shrinks the effort arm to 40 inches. This cuts the mechanical advantage in half (from 4 to 2), doubling the required effort force from 75 lbs to 150 lbs, which is an increase of 75 lbs.

Step-by-Step Solution

1

Identify the lever class and the formula for torque balance.

This is a first-class lever. The formula is: Effort Force × Effort Arm = Load Force × Load Arm. The Mechanical Advantage (MA) is Effort Arm / Load Arm.

The fulcrum (pivot block) is located between the applied effort and the load.

2

Calculate the initial effort force and mechanical advantage (State 1).

Load Arm = 12 inches. Effort Arm = 60 - 12 = 48 inches. MA = 48 / 12 = 4. Initial Effort Force = 300 / 4 = 75 lbs.

The total length of the bar is 60 inches, so the effort arm is the remainder of the bar after subtracting the load arm.

3

Calculate the new effort force and mechanical advantage (State 2).

New Load Arm = 20 inches. New Effort Arm = 60 - 20 = 40 inches. New MA = 40 / 20 = 2. New Effort Force = 300 / 2 = 150 lbs.

Moving the pivot 8 inches away from the load increases the load arm and correspondingly decreases the effort arm.

4

Determine the change in the required effort force.

150 lbs (New) - 75 lbs (Initial) = 75 lbs increase.

The question asks for the difference between the two required forces.

Key Concept

Torque balance and changing fulcrum positions on a first-class lever.

Estimated Time:2m 0s

Question 1634Question →

A cargo aircraft accelerates from rest to a final velocity of $90 \text{ m/s}$ in exactly $3 \text{ minutes}$ . What is the average acceleration of the aircraft in $\text{m/s}^2$ ?

30 \text{ m/s}^2

2.0 \text{ m/s}^2

0.5 \text{ m/s}^2

270 \text{ m/s}^2

Show answer & explanation

Answer: $0.5 \text{ m/s}^2$

Answer

The average acceleration of the aircraft is

0.5 \text{ m/s}^2

.

The correct average acceleration is found by first converting the given time of 3 minutes into 180 seconds. Then, using the kinematic formula for acceleration (change in velocity divided by time), dividing the final velocity of

90 \text{ m/s}

by 180 seconds results in

0.5 \text{ m/s}^2

.

Step-by-Step Solution

1

Convert the time from minutes to seconds.

3 \text{ minutes} \times 60 \text{ seconds/minute} = 180 \text{ seconds}

.

Acceleration is requested in

\text{m/s}^2

, so the time variable must be in seconds.

2

Apply the formula for average acceleration:

a = \frac{v_f - v_i}{t}

.

a = \frac{90 \text{ m/s} - 0 \text{ m/s}}{180 \text{ s}}

.

This formula determines the rate of change of velocity over the given time period.

3

Calculate the final acceleration value.

a = 0.5 \text{ m/s}^2

.

Dividing

90

by

180

yields

0.5

.

Key Concept

Calculating average acceleration requiring unit conversion.

Question 1635Question →

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 1636Question →

A machinist tests two threaded shafts being driven into fixed, stationary bases. Shaft A is equipped with a $14\text{-inch}$ handle and standard right-hand threads at $4$ threads per inch (TPI). Shaft B is equipped with a $7\text{-inch}$ handle and left-hand threads at $8$ TPI. From a top-down view, the machinist turns the handle of both shafts exactly one full revolution clockwise. Assuming no friction, which of the following statements correctly describes the linear movement of the shafts and compares their Ideal Mechanical Advantage (IMA)?

Shaft A moves exactly

0.25

inches downward into the base, Shaft B moves exactly

0.125

inches upward out of the base, and both shafts possess the exact same IMA.

Both shafts move downward into their respective bases, but Shaft B possesses twice the IMA of Shaft A due to its finer thread pitch.

Shaft A moves exactly

0.25

inches upward out of the base, Shaft B moves exactly

0.125

inches downward into the base, and both shafts possess the exact same IMA.

Shaft A moves exactly

0.25

inches downward into the base, Shaft B moves exactly

0.125

inches upward out of the base, and Shaft A possesses four times the IMA of Shaft B.

Show answer & explanation

Answer: Shaft A moves exactly $0.25$ inches downward into the base, Shaft B moves exactly $0.125$ inches upward out of the base, and both shafts possess the exact same IMA.

Answer

Shaft A moves

0.25

inches downward, Shaft B moves

0.125

inches upward, and both shafts have the exact same Ideal Mechanical Advantage.

The correct answer accurately calculates that a clockwise turn drives a right-hand thread downward by its pitch (

1/4 = 0.25

inches) and a left-hand thread upward by its pitch (

1/8 = 0.125

inches). Furthermore, it recognizes that Ideal Mechanical Advantage (IMA) is the effort circumference (

2\pi R

) multiplied by the TPI. The larger

14\text{-inch}

handle on Shaft A exactly compensates for its coarser

4

TPI pitch, resulting in an identical IMA (

112\pi

) for both systems.

Step-by-Step Solution

1

Determine the direction of linear movement based on thread handedness.

Shaft A moves downward; Shaft B moves upward.

By convention, turning a standard right-hand thread clockwise drives it inward (downward into the base). Turning a left-hand thread clockwise drives it outward (upward out of the base).

2

Calculate the linear travel distance per revolution (pitch) for each shaft.

Shaft A travels

0.25

inches; Shaft B travels

0.125

inches.

Pitch is the reciprocal of Threads Per Inch (TPI). For Shaft A:

1 / 4 = 0.25

inches. For Shaft B:

1 / 8 = 0.125

inches.

3

Calculate the Ideal Mechanical Advantage (IMA) for Shaft A.

\text{IMA}_A = 112\pi

The IMA of a screw is the ratio of effort distance to load distance per revolution:

(2 \cdot \pi \cdot R) / \text{Pitch}

. For A:

(2 \cdot \pi \cdot 14) / 0.25 = 28\pi / 0.25 = 112\pi

.

4

Calculate the Ideal Mechanical Advantage (IMA) for Shaft B.

\text{IMA}_B = 112\pi

Applying the same formula for B:

(2 \cdot \pi \cdot 7) / 0.125 = 14\pi / 0.125 = 112\pi

. The doubling of the TPI exactly offsets the halving of the handle length, yielding identical IMA.

Key Concept

The IMA of a screw depends on both the length of the lever arm (handle) and the pitch of the threads, and thread handedness determines linear direction relative to rotation.

Question 1637Question →

A logistics crew needs to load an $800\text{ lb}$ supply crate onto a transport plane whose cargo deck is $6\text{ ft}$ above the tarmac. The crew can apply a maximum combined effort force of $200\text{ lb}$ . Assuming friction is negligible, what is the minimum ramp length required, and what is the total work done to lift the crate?

Minimum ramp length of

24\text{ ft}

; total work of

4,800\text{ ft-lb}

Minimum ramp length of

24\text{ ft}

; total work of

1,200\text{ ft-lb}

Minimum ramp length of

4\text{ ft}

; total work of

4,800\text{ ft-lb}

Minimum ramp length of

24\text{ ft}

; total work of

19,200\text{ ft-lb}

Show answer & explanation

Answer: Minimum ramp length of $24\text{ ft}$ ; total work of $4,800\text{ ft-lb}$

Answer

Minimum ramp length of

24\text{ ft}

; total work of

4,800\text{ ft-lb}

The Ideal Mechanical Advantage required is the load divided by the effort (800 / 200 = 4). Since IMA also equals length divided by height, the length must be 4 times the height (4 * 6 = 24 ft). Total work is the force times the distance it moves in the direction of the force, which can be calculated as Load * Height (800 * 6 = 4,800 ft-lb) or Effort * Length (200 * 24 = 4,800 ft-lb).

Step-by-Step Solution

1

Calculate the required Ideal Mechanical Advantage (IMA).

\text{IMA} = \frac{\text{Load}}{\text{Effort}} = \frac{800\text{ lb}}{200\text{ lb}} = 4

The IMA tells us how much the machine (ramp) multiplies the applied force.

2

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

\text{Length} = \text{IMA} \times \text{Height} = 4 \times 6\text{ ft} = 24\text{ ft}

For an inclined plane,

\text{IMA} = \frac{\text{Length}}{\text{Height}}

. Rearranging gives

\text{Length} = \text{IMA} \times \text{Height}

.

3

Calculate the total work done.

\text{Work} = \text{Load} \times \text{Height} = 800\text{ lb} \times 6\text{ ft} = 4,800\text{ ft-lb}

Work can be calculated as the vertical work required to lift the load. Alternatively,

\text{Work} = \text{Effort} \times \text{Length} = 200\text{ lb} \times 24\text{ ft} = 4,800\text{ ft-lb}

.

Key Concept

The Ideal Mechanical Advantage of an inclined plane dictates its proportions (length vs. height) to reduce effort force, while the total work done remains constant (conservation of energy).

Question 1638Question →

When a simply supported horizontal beam bends downward due to a heavy load applied at its center, the top surface of the beam experiences tensile stress, while the bottom surface experiences compressive stress.

True

False

Show answer & explanation

Answer: False

Answer

False

When a simply supported horizontal beam bends downward, it forces the material on the top surface to push together and the material on the bottom surface to stretch apart. Therefore, the top experiences compressive stress and the bottom experiences tensile stress, making the statement false.

Step-by-Step Solution

1

Visualize the deformation of the beam under the central load.

The beam bows downward in the center, creating a U-like curvature.

Understanding the physical deformation is the first step to identifying the internal stresses.

2

Identify the effect on the top surface of the bent beam.

The top surface curves inward, causing its overall length to effectively shorten.

When a material's fibers are pushed together or shortened, it is under compressive stress.

3

Identify the effect on the bottom surface of the bent beam.

The bottom surface curves outward, causing its overall length to stretch.

When a material's fibers are pulled apart or elongated, it is under tensile stress.

Key Concept

Bending Stress in Structural Members

Estimated Time:30s

Question 1639Question →

A specialized rover is designed to operate on both Earth (where gravitational acceleration is $10\text{ m/s}^2$ ) and an exoplanet (where gravitational acceleration is $15\text{ m/s}^2$ ). The rover carries a scientific payload that is measured to have a weight of $600\text{ N}$ on Earth. When the rover arrives on the exoplanet, it suspends the payload from a spring scale to verify its properties, and later pushes the payload horizontally across a frictionless ice field to accelerate it at $4\text{ m/s}^2$ . What will the spring scale read, and what horizontal force must the rover apply?

Spring scale:

900\text{ N}

; Horizontal force:

240\text{ N}

Spring scale:

900\text{ N}

; Horizontal force:

3600\text{ N}

Spring scale:

9000\text{ N}

; Horizontal force:

2400\text{ N}

Spring scale:

900\text{ N}

; Horizontal force:

15\text{ N}

Show answer & explanation

Answer: Spring scale: $900\text{ N}$ ; Horizontal force: $240\text{ N}$

Answer

Spring scale:

900\text{ N}

; Horizontal force:

240\text{ N}

The payload's mass is an invariant property, calculated from its Earth weight (

m = \frac{W}{g} = \frac{600}{10} = 60\text{ kg}

). On the exoplanet, the spring scale measures the local gravitational force (weight), which is mass times local gravity (

60 \times 15 = 900\text{ N}

). The horizontal force depends only on the invariant mass and the desired acceleration (

F = ma = 60 \times 4 = 240\text{ N}

).

Step-by-Step Solution

1

Determine the invariant mass of the payload using its Earth weight.

m = \frac{W_{\text{Earth}}}{g_{\text{Earth}}} = \frac{600\text{ N}}{10\text{ m/s}^2} = 60\text{ kg}

Mass is the amount of matter and remains constant regardless of location. It is required to determine both the new weight and the horizontal force.

2

Calculate the new weight on the exoplanet to find the spring scale reading.

W_{\text{exo}} = m \times g_{\text{exo}} = 60\text{ kg} \times 15\text{ m/s}^2 = 900\text{ N}

A spring scale measures the local force of gravity (weight), which depends on the local gravitational acceleration.

3

Calculate the required horizontal force using Newton's Second Law.

F = m \times a = 60\text{ kg} \times 4\text{ m/s}^2 = 240\text{ N}

Horizontal acceleration on a frictionless surface depends only on the object's mass (inertia) and the applied force, not on its weight.

Key Concept

Mass is an invariant property that determines inertia and relates to force (

F=ma

), while weight is a location-dependent gravitational force (

W=mg

) measured by spring scales.

Question 1640Question →

A logistics team is evaluating different ramp configurations to push heavy equipment into a warehouse. Calculate the ideal mechanical advantage (IMA) for each ramp. Order the following ramps from the GREATEST mechanical advantage (requires the least effort force) to the LEAST mechanical advantage (requires the most effort force).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from greatest to least mechanical advantage is: the 5m long / 1m high ramp, the 8m long / 2m high ramp, the 9m long / 3m high ramp, and finally the 4m long / 2m high ramp.

The correct order requires calculating the Ideal Mechanical Advantage (IMA = Length / Height) for each ramp and arranging them in descending order. The 5m/1m ramp has an IMA of 5. The 8m/2m ramp has an IMA of 4. The 9m/3m ramp has an IMA of 3. The 4m/2m ramp has an IMA of 2.

Step-by-Step Solution

1

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

IMA = Length of the incline ÷ Vertical height (IMA = L / h).

This formula determines how much the ramp multiplies the applied force.

2

Calculate the IMA for the ramp with a length of 5 meters and a height of 1 meter.

IMA = 5 ÷ 1 = 5.

Direct application of the IMA formula.

3

Calculate the IMA for the ramp with a length of 8 meters and a height of 2 meters.

IMA = 8 ÷ 2 = 4.

Direct application of the IMA formula.

4

Calculate the IMA for the ramp with a length of 9 meters and a height of 3 meters.

IMA = 9 ÷ 3 = 3.

Direct application of the IMA formula.

5

Calculate the IMA for the ramp with a length of 4 meters and a height of 2 meters.

IMA = 4 ÷ 2 = 2.

Direct application of the IMA formula.

6

Order the calculated IMA values from greatest to least.

5, then 4, then 3, then 2.

The question asks for the order from greatest mechanical advantage to least.

Key Concept

The ideal mechanical advantage of an inclined plane is calculated by dividing the length of the incline by the vertical height (IMA = L / h). A greater IMA means less effort force is required to move an object up the ramp.

Estimated Time:1m 0s

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