Simple Machines Alıştırma Soruları — Armed Services Vocational Aptitude Battery (ASVAB)

Match the specific wedge geometries with their corresponding physical characteristics and force trade-offs.

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Cevap

High aspect ratio wedge matches maximum force multiplication with extensive driving distance. Low aspect ratio wedge matches rapid lateral displacement with high effort force. Asymmetrical wedge matches lateral force on one side with self-locking dependence. Symmetrical wedge matches output force divided into two opposing vectors.

The mechanical advantage of a wedge is determined by the ratio of its length to its thickness. A long, thin wedge (high aspect ratio) provides significant force multiplication but requires driving the wedge a longer distance and introduces high frictional losses. Conversely, a short, wide wedge (low aspect ratio) requires much more force to drive but splits the material quickly without binding. Single-incline (asymmetrical) wedges are used for lifting or holding against a flat surface, whereas double-incline (symmetrical) wedges split materials by pushing equally in both directions.

Adım Adım Çözüm

1

Analyze the mechanical advantage of wedge length-to-thickness ratios.

Long, thin wedges (high aspect ratio) have high IMA but suffer from high friction. Short, thick wedges (low aspect ratio) have low IMA but split materials rapidly.

Mechanical Advantage of a wedge is its length divided by its thickness. High IMA requires less effort force but greater effort distance.

2

Evaluate the directional forces of single versus double inclines.

Single-incline wedges exert force in one direction against a flat base. Double-incline wedges split force bidirectionally.

The geometry of the wedge's faces determines the direction of the normal forces applied to the material.

3

Map each wedge profile to its operational description.

Match the high aspect wedge to high force multiplication/friction, the low aspect wedge to rapid displacement/high effort, the asymmetrical wedge to one-sided force/friction dependence, and the symmetrical wedge to equal opposing vectors.

This aligns the theoretical mechanical properties with practical tool applications in the real world.

Anahtar Kavram

Wedge mechanical advantage, force vectors, and frictional trade-offs

Soru 2Soru →

A construction worker uses a straight steel bar to lift a $150\text{-pound}$ concrete slab. A metal pipe is placed underneath the bar to serve as a pivot point (fulcrum). If the slab rests on the bar $3\text{ feet}$ from the fulcrum, and the worker applies downward force on the opposite end at a distance of $9\text{ feet}$ from the fulcrum, how much force is required to perfectly balance the slab?

37.5\text{ lbs}

50\text{ lbs}

150\text{ lbs}

450\text{ lbs}

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Cevap: $50\text{ lbs}$

Cevap

The correct force required is

50\text{ lbs}

.

According to the principle of levers, a system is balanced when the input torque equals the output torque (

F_1 \times d_1 = F_2 \times d_2

). The concrete slab generates

450\text{ lb-ft}

of torque (

150\text{ lbs} \times 3\text{ ft}

). Dividing this torque by the effort arm distance of

9\text{ feet}

yields a required downward effort force of

50\text{ lbs}

.

Adım Adım Çözüm

1

Set up the torque balance equation for a first-class lever.

Force_{effort} \times Distance_{effort} = Force_{load} \times Distance_{load}

For a lever to be perfectly balanced, the torque applied by the worker must equal the torque exerted by the load.

2

Calculate the torque of the load.

150\text{ lbs} \times 3\text{ ft} = 450\text{ lb-ft}

Torque is calculated by multiplying the weight of the object by its distance to the fulcrum.

3

Solve for the required effort force.

450\text{ lb-ft} / 9\text{ ft} = 50\text{ lbs}

Dividing the total load torque by the length of the effort arm determines the exact force the worker needs to apply.

Anahtar Kavram

Torque balance in first-class levers (Law of Levers)

Soru 3Soru →

A hand-cranked emergency siren uses a compound gear train to amplify the rotational speed of the crank. The crank is directly attached to Gear A, which has $50$ teeth. Gear A meshes with Gear B, which has $10$ teeth. Gear B shares a solid shaft with Gear C, which has $40$ teeth. Finally, Gear C meshes with Gear D, which has $10$ teeth and is directly attached to the siren's rotor. If the operator turns the crank at a constant speed of $30\text{ rpm}$ , what is the rotational speed of the siren's rotor?

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Cevap: 600

Cevap

600 rpm

The system consists of two gear pairs that both increase speed. The first pair (

50

teeth driving

10

teeth) increases the speed by a factor of

5

. The second pair (

40

teeth driving

10

teeth) increases the speed by a factor of

4

. The total speed multiplier is

5 \times 4 = 20

. Multiplying the input speed of

30\text{ rpm}

by

20

yields the final output speed of

600\text{ rpm}

.

Adım Adım Çözüm

1

Determine the speed multiplier for the first gear pair.

Ratio =

50 / 10 = 5

Gear A drives Gear B. A larger gear driving a smaller gear increases rotational speed proportionally to their tooth count.

2

Determine the speed multiplier for the second gear pair.

Ratio =

40 / 10 = 4

Gear C drives Gear D. Both pairs are independent in tooth count, but Gear C rotates at the same speed as Gear B because they share a shaft.

3

Calculate the total mechanical advantage for speed.

Total multiplier =

5 \times 4 = 20

In a compound gear train, the total speed ratio is the product of the individual gear pair ratios.

4

Calculate the final output speed.

30 \times 20 = 600\text{ rpm}

Multiply the input speed by the total speed multiplier to find the output speed.

Anahtar Kavram

Compound Gear Trains and Speed Ratios

Soru 4Soru →

Match each mechanical tool or scenario with its correct lever classification and mechanical advantage (MA) characteristic.

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Cevap

Pry bar matches to Class 1 (MA > 1); Hand truck matches to Class 2 (MA > 1); Tweezers match to Class 3 (MA < 1); Symmetrical seesaw matches to Class 1 (MA = 1).

Each tool utilizes a specific lever class based on the relative arrangement of the Fulcrum, Load, and Effort. By identifying which component is positioned in the middle, we determine the correct class. The ratio of the effort arm to the load arm then dictates whether the mechanical advantage is greater than, equal to, or less than 1.

Adım Adım Çözüm

1

Analyze the pry bar scenario to locate the relative positions of the fulcrum, effort, and load.

The fulcrum (wooden block) is between the effort (hands) and the load (engine), but closer to the load.

This configuration defines a Class 1 lever that is optimized for force multiplication.

2

Analyze the hand truck to identify its lever class.

The wheels act as the fulcrum at one end, the effort is applied at the opposite end (handles), and the load (crates) rests in the middle.

Having the load in the middle defines a Class 2 lever, which always provides a mechanical advantage greater than 1.

3

Examine how tweezers apply force.

The fulcrum is the joined end, the effort is applied by fingers pinching the middle, and the load is grasped at the opposite open end.

Having the effort in the middle defines a Class 3 lever, which trades force for precision and distance (MA < 1).

4

Evaluate the symmetrical seesaw scenario.

The fulcrum is directly in the center with equal distances to both applied forces.

This is a Class 1 lever where neither distance nor force is multiplied, meaning MA equals 1.

Anahtar Kavram

Lever Classifications and Mechanical Advantage

Soru 5Soru →

A traditional water well is operated by a hand crank that raises a bucket of water. The crank handle has a turning radius of $12 \text{ inches}$ , and it drives a wooden cylinder (the axle) that has a diameter of $8 \text{ inches}$ . If the total weight of the water and bucket is $45 \text{ pounds}$ , what is the ideal effort force required on the handle to lift the bucket at a constant speed?

15 \text{ pounds}

22.5 \text{ pounds}

30 \text{ pounds}

135 \text{ pounds}

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Cevap: $15 \text{ pounds}$

Cevap

The ideal effort force required is

15 \text{ pounds}

.

The system operates as a force multiplier because the effort is applied to the larger radius (the crank handle). First, find the axle's radius by dividing its

8 \text{ inch}

diameter by 2, yielding

4 \text{ inches}

. The mechanical advantage is the ratio of the effort radius to the load radius:

12 \text{ inches} / 4 \text{ inches} = 3

. This means the crank multiplies the applied force by 3. To lift a

45\text{-pound}

load, divide the load by the mechanical advantage (

45 / 3

), which results in a required effort force of

15 \text{ pounds}

.

Adım Adım Çözüm

1

Determine the radius of the axle.

The radius of the axle (

r

) is

4 \text{ inches}

.

The diameter is given as

8 \text{ inches}

, and radius is half of the diameter (

8 / 2 = 4

).

2

Calculate the Ideal Mechanical Advantage (IMA) of the wheel and axle system.

The IMA is

3

.

The IMA for a wheel and axle where the effort is applied to the wheel is the radius of the wheel divided by the radius of the axle (

IMA = R / r = 12 / 4 = 3

).

3

Calculate the required ideal effort force.

The effort force is

15 \text{ pounds}

.

Effort force equals the load divided by the IMA (

45 / 3 = 15

). This means the crank multiplies the user's force by 3.

Anahtar Kavram

Ideal Mechanical Advantage in Wheel and Axle Systems

Soru 6Soru →

A carpenter uses a claw hammer to extract a stubborn nail from a wooden board. The hammer functions as a first-class lever, pivoting on the curve of its head. The perpendicular distance from the pivot point to the nail is $6 \text{ cm}$ , and the carpenter applies an effort force on the handle at a perpendicular distance of $24 \text{ cm}$ from the pivot point. If the nail requires a resistance force of $480 \text{ N}$ to be pulled out, what minimum effort force must the carpenter apply?

96 \text{ N}

160 \text{ N}

120 \text{ N}

384 \text{ N}

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Cevap: $120 \text{ N}$

Cevap

The minimum effort force required is

120 \text{ N}

.

In a first-class lever system in equilibrium, the effort torque must equal the resistance torque (Effort Force × Effort Arm = Resistance Force × Resistance Arm). Plugging in the given values yields: Effort Force × 24 cm = 480 N × 6 cm. Solving for the Effort Force gives 120 N.

Adım Adım Çözüm

1

Identify the type of lever and the location of the fulcrum.

The claw hammer is acting as a first-class lever. The pivot point (fulcrum) is located between the applied effort (the carpenter's hand) and the load (the nail).

Correctly identifying the lever setup is necessary to assign the proper lengths to the effort arm and resistance arm.

2

Determine the length of the effort arm and the resistance arm.

The effort arm is the distance from the pivot to the hand (

24 \text{ cm}

). The resistance arm is the distance from the pivot to the nail (

6 \text{ cm}

).

Torque is calculated using the distance from the pivot point to where the force is applied.

3

Calculate the mechanical advantage (MA) of the lever.

MA = \frac{\text{Effort Arm}}{\text{Resistance Arm}} = \frac{24}{6} = 4

.

The mechanical advantage indicates how much the lever multiplies the applied force.

4

Calculate the required effort force to overcome the resistance.

\text{Effort Force} = \frac{\text{Resistance Force}}{MA} = \frac{480}{4} = 120 \text{ N}

.

The effort force multiplied by the mechanical advantage must equal the resistance force to balance the torques.

Anahtar Kavram

Torque Balance and Mechanical Advantage in First-Class Levers

Soru 7Soru →

Match each lifting scenario with the simplest ideal pulley configuration that provides the necessary mechanical advantage to complete the task.

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Cevap

The mechanic requires an IMA of 4 (Two-fold purchase). The warehouse worker requires an IMA of 1 (Single fixed pulley). The deckhand requires an IMA of 3 (Luff tackle). The builder requires an IMA of 2 (Single movable pulley).

The required Ideal Mechanical Advantage (IMA) for any lifting task is calculated by dividing the load's weight by the applied effort force (

\text{IMA} = \text{Load} / \text{Effort}

). Once the IMA is determined, it directly corresponds to the number of rope strands that must physically pull upward on the movable blocks in an ideal pulley system.

Adım Adım Çözüm

1

Calculate the required Ideal Mechanical Advantage (IMA) for each lifting scenario.

IMA is calculated as the ratio of Load to Effort (

\text{IMA} = \frac{\text{Load}}{\text{Effort}}

).

Identifying the required mechanical multiplier is necessary to select the correct pulley configuration.

2

Evaluate the engine lift scenario.

Load is

400 \text{ lb}

, effort is

100 \text{ lb}

. Required

\text{IMA} = \frac{400}{100} = 4

. This matches the two-fold purchase.

The system must multiply the effort by four.

3

Evaluate the crate lift scenario.

Load is

150 \text{ lb}

, effort is

150 \text{ lb}

. Required

\text{IMA} = \frac{150}{150} = 1

. This matches the single fixed pulley.

No mechanical advantage is needed; only a change in force direction.

4

Evaluate the deckhand hoist scenario.

Load is

300 \text{ lb}

, effort is

100 \text{ lb}

. Required

\text{IMA} = \frac{300}{100} = 3

. This matches the luff tackle.

The system must multiply the effort by three.

5

Evaluate the beam lift scenario.

Load is

200 \text{ lb}

, effort is

100 \text{ lb}

. Required

\text{IMA} = \frac{200}{100} = 2

. This matches the single movable pulley.

The system must multiply the effort by two.

Anahtar Kavram

The required mechanical advantage is the ratio of load weight to effort force, and this dictates the number of rope strands that must directly support the movable block in a pulley system.

Soru 8Soru →

A worker needs to load a heavy crate onto the back of a truck. Instead of lifting it straight up from the ground, the worker uses a long wooden ramp. Assuming friction is negligible, what is the primary mechanical benefit of using this inclined plane?

It decreases the amount of effort force required to move the crate.

It decreases the total amount of work required to load the crate.

It decreases the actual weight of the crate while it is moving up the slope.

It reduces the surface area of the crate in contact with the ground, decreasing friction.

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Cevap: It decreases the amount of effort force required to move the crate.

Cevap

The primary benefit of using an inclined plane is that it decreases the amount of effort force required to move the crate.

An inclined plane allows the same amount of work to be completed over a longer distance. Because work equals force multiplied by distance, increasing the distance proportionally reduces the amount of effort force needed to move the object.

Adım Adım Çözüm

1

Identify the purpose of using simple machines like an inclined plane.

Simple machines are used to provide a mechanical advantage.

Understanding the fundamental function of simple machines is key to solving the problem.

2

Evaluate the relationship between work, force, and distance for an inclined plane.

The formula for work is

W = F \times d

. The total work to lift the crate to the truck bed remains exactly the same.

Conservation of energy dictates that the total work done against gravity is independent of the path taken.

3

Determine how the inclined plane changes the variables in the work equation.

Because the ramp increases the distance (

d

) over which the work is performed, the required effort force (

F

) decreases proportionally.

This inverse relationship between force and distance is the defining characteristic of an inclined plane's mechanical advantage.

Anahtar Kavram

Mechanical Advantage of Inclined Planes

Tahmini Süre:45s

Soru 9Soru →

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

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

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Screw Pitch and Linear Travel

Soru 10Soru →

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.

Cevabı ve açıklamayı göster

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

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Actual Mechanical Advantage and Efficiency

Soru 11Soru →

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.

Cevabı ve açıklamayı göster

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

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Screws and Threads

Tahmini Süre:1m 0s

Soru 12Soru →

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.

Cevabı ve açıklamayı göster

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

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Work and Force Relationship on Inclined Planes

Tahmini Süre:1m 0s

Soru 13Soru →

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.

Cevabı ve açıklamayı göster

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

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

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

Tahmini Süre:2m 0s

Soru 14Soru →

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.

Cevabı ve açıklamayı göster

Cevap: 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.

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

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.

Soru 15Soru →

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}

Cevabı ve açıklamayı göster

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

Cevap

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).

Adım Adım Çözüm

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}

.

Anahtar Kavram

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).

Soru 16Soru →

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).

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Cevabı ve açıklamayı göster

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

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.

Tahmini Süre:1m 0s

Soru 17Soru →

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

Cevabı ve açıklamayı göster

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

Cevap

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}

).

Adım Adım Çözüm

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.

Anahtar Kavram

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

Tahmini Süre:2m 0s

Soru 18Soru →

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

Cevabı ve açıklamayı göster

Cevap: 105 lbs

Cevap

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}

.

Adım Adım Çözüm

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.

Anahtar Kavram

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.

Soru 19Soru →

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.

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Cevabı ve açıklamayı göster

Cevap

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).

Adım Adım Çözüm

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.

Anahtar Kavram

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.

Tahmini Süre:45s

Soru 20Soru →

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.

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Cevabı ve açıklamayı göster

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Mechanical Advantage of Screws Based on TPI

Simple Machines

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