Simple Machines

165 soru

Soru 41Soru →

A mechanic is using a C-clamp with a threaded screw that has 1616 threads per inch (TPI). The clamp is currently open to a width of 5.05.0 inches. If the mechanic needs to tightly secure two steel plates that have a combined thickness of 1.51.5 inches, how many full revolutions of the screw are required to close the clamp to the exact thickness of the plates?

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Cevap: 56

Cevap

56
The screw must travel a linear distance of 3.53.5 inches to close the gap from 5.05.0 inches to 1.51.5 inches. With 1616 threads per inch, each inch of linear travel requires 1616 full revolutions. Multiplying the 3.53.5 inches of travel by 1616 revolutions per inch yields exactly 5656 revolutions.

Adım Adım Çözüm

1
Calculate the linear distance the clamp's jaw needs to travel.
5.0−1.5=3.55.0 - 1.5 = 3.5 inches
The screw only needs to cover the empty space between the current opening and the target thickness of the plates.
2
Calculate the total number of revolutions based on the travel distance and TPI.
3.5×16=563.5 \times 16 = 56 revolutions
TPI (threads per inch) indicates that one full revolution moves the screw 1/161/16 of an inch. Therefore, moving it 11 inch requires 1616 revolutions.

Anahtar Kavram

Linear Travel and Threads Per Inch (TPI)
Soru 42Soru →

A heavy-duty steel wedge is used to split a solid masonry block. The wedge has a penetration length of 24 cm24\text{ cm} and a maximum width (thickness) of 4 cm4\text{ cm} at its head. A pneumatic tool applies a constant downward driving effort force of 850 N850\text{ N} to the head of the wedge. Due to high surface resistance, the wedge operates with a mechanical efficiency of 60%60\%. What is the total lateral separating force exerted on the masonry block by the wedge? (Enter your answer as a number in Newtons)

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Cevap: 3060

Cevap

3060
The correct lateral separating force is found by determining the wedge's Ideal Mechanical Advantage (24 cm/4 cm=624\text{ cm} / 4\text{ cm} = 6), adjusting for energy lost to friction by applying the 60%60\% efficiency (6×0.60=3.6 AMA6 \times 0.60 = 3.6\text{ AMA}), and multiplying this actual advantage by the input force (850 N×3.6=3060 N850\text{ N} \times 3.6 = 3060\text{ N}).

Adım Adım Çözüm

1
Calculate the Ideal Mechanical Advantage (IMA) of the wedge.
IMA=6IMA = 6
The IMA of a wedge is defined as the ratio of its penetration length to its maximum thickness (24 cm/4 cm24\text{ cm} / 4\text{ cm}).
2
Determine the Actual Mechanical Advantage (AMA).
AMA=3.6AMA = 3.6
Because the wedge operates at 60%60\% efficiency due to friction, the actual mechanical advantage is the IMA multiplied by the efficiency (6×0.606 \times 0.60).
3
Calculate the total lateral separating force.
3060 N3060\text{ N}
The output force is the product of the input effort force and the Actual Mechanical Advantage (850 N×3.6850\text{ N} \times 3.6).

Anahtar Kavram

Mechanical Advantage and Efficiency of Wedges
Soru 43Soru →

A woodworker uses a steel splitting wedge to split a heavy log. The wedge has a length of 88 inches and a maximum thickness of 22 inches. If a lateral splitting force of 400400 lbs is needed to split the log, what ideal effort force must be applied to the top of the wedge? (Assume no friction)

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Cevap: 100100 lbs

Cevap

The ideal effort force is 100100 lbs.
The correct answer is found by first calculating the ideal mechanical advantage (IMA) of the wedge, which is its length divided by its thickness (8/2=48 / 2 = 4). Because the wedge multiplies the input force by a factor of 44, the required effort force is the splitting force divided by the IMA (400 lbs/4=100 lbs400 \text{ lbs} / 4 = 100 \text{ lbs}).

Adım Adım Çözüm

1
Calculate the ideal mechanical advantage (IMA) of the wedge.
IMA=LengthThickness=82=4IMA = \frac{\text{Length}}{\text{Thickness}} = \frac{8}{2} = 4
The mechanical advantage of a wedge depends on the ratio of its length to its maximum thickness.
2
Calculate the required ideal effort force using the IMA.
Effort Force=Load ForceIMA=4004=100 lbs\text{Effort Force} = \frac{\text{Load Force}}{IMA} = \frac{400}{4} = 100 \text{ lbs}
A wedge multiplies the effort force by its mechanical advantage to produce the required load force.

Anahtar Kavram

Wedges provide a mechanical advantage equal to their length divided by their maximum thickness, which reduces the effort force required to split an object.
Soru 44Soru →

A worker is using a wheelbarrow, which acts as a second-class lever, to transport materials. The wheel's axle serves as the fulcrum. The empty wheelbarrow has a downward weight of 200 N200 \text{ N}, with its center of gravity located 0.5 m0.5 \text{ m} horizontally from the axle. A load of bricks weighing 800 N800 \text{ N} is placed in the tray so that its center of gravity is 0.4 m0.4 \text{ m} horizontally from the axle. The worker lifts upward at the ends of the handles, which are located 1.2 m1.2 \text{ m} horizontally from the axle.

Assuming the wheelbarrow is in a horizontal position, what is the minimum upward force required to lift the rear supports off the ground?

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Cevap: 350

Cevap

The minimum upward force required to lift the wheelbarrow is 350 N.
The correct answer is derived by applying the principle of moments (torque). Because there are two separate masses acting on the lever, their torques must be calculated independently and summed. The wheelbarrow contributes 100 N⋅m100 \text{ N}\cdot\text{m} of torque and the bricks contribute 320 N⋅m320 \text{ N}\cdot\text{m}. Balancing the total 420 N⋅m420 \text{ N}\cdot\text{m} torque over a 1.2 m1.2 \text{ m} lever arm requires exactly 350 N350 \text{ N} of effort.

Adım Adım Çözüm

1
Calculate the torque from the empty wheelbarrow.
τcart=100 N⋅m\tau_{cart} = 100 \text{ N}\cdot\text{m}
The intrinsic weight of the lever system applies a downward force that must be accounted for.
2
Calculate the torque from the added load.
τload=320 N⋅m\tau_{load} = 320 \text{ N}\cdot\text{m}
The load applies an additional independent downward force at its specific distance from the fulcrum.
3
Sum the downward torques.
τtotal=420 N⋅m\tau_{total} = 420 \text{ N}\cdot\text{m}
In equilibrium, the total clockwise torque must be balanced by the total counterclockwise torque.
4
Divide the total torque by the effort distance to find the effort force.
Feffort=350 NF_{effort} = 350 \text{ N}
Torque equals force times distance. Rearranging 420=F×1.2420 = F \times 1.2 isolates the required lifting force.

Anahtar Kavram

Torque Equilibrium and Composite Loads in Second-Class Levers
Soru 45Soru →

A vintage printing press utilizes a compound gear train to feed paper. Shaft 1 is the input and is driven at 120120 RPM in a counter-clockwise direction. Gear A (4040 teeth) is mounted on Shaft 1 and meshes directly with Gear B (2020 teeth), which is mounted on Shaft 2. Also rigidly mounted on Shaft 2 is Gear C (5050 teeth). Gear C meshes directly with Gear D (2525 teeth) mounted on Shaft 3. Based on this configuration, what is the rotational speed and direction of Shaft 3?

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Cevap: 480480 RPM, counter-clockwise

Cevap

The output shaft rotates at 480480 RPM in a counter-clockwise direction.
In a compound gear train, mechanical advantage and speed ratios are multiplied across stages. In the first stage (Gears A to B), a 4040-tooth gear drives a 2020-tooth gear, doubling the speed to 240240 RPM and reversing the direction to clockwise. Both gears on Shaft 2 rotate together. In the second stage (Gears C to D), a 5050-tooth gear drives a 2525-tooth gear, doubling the speed again to 480480 RPM and reversing the direction back to counter-clockwise.

Adım Adım Çözüm

1
Calculate the rotational speed of Shaft 2.
Speed = 120×(40/20)=240120 \times (40 / 20) = 240 RPM.
The speed ratio between meshing gears is inversely proportional to their tooth count. A smaller driven gear rotates faster than the driving gear.
2
Determine the rotational direction of Shaft 2.
Clockwise.
When two external gears mesh, they rotate in opposite directions. Since Shaft 1 is counter-clockwise, Shaft 2 must be clockwise. Because Gears B and C are on the same shaft, Gear C also rotates at 240240 RPM clockwise.
3
Calculate the rotational speed of Shaft 3.
Speed = 240×(50/25)=480240 \times (50 / 25) = 480 RPM.
Applying the speed ratio principle again from Gear C (driving) to Gear D (driven).
4
Determine the rotational direction of Shaft 3.
Counter-clockwise.
Gear C (clockwise) meshes with Gear D, forcing Gear D to rotate in the opposite direction.

Anahtar Kavram

Compound Gear Trains and Speed Ratios
Soru 46Soru →

A manual boat winch is used to pull a small boat onto a trailer. The crank handle travels in a circle with a radius of 15 inches15 \text{ inches}, turning a central spool (the axle) with a radius of 3 inches3 \text{ inches}.

What is the Ideal Mechanical Advantage (IMA) of this winch?

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Cevap: 5

Cevap

The Ideal Mechanical Advantage is 5.
The Ideal Mechanical Advantage of a wheel and axle is calculated by dividing the radius of the wheel by the radius of the axle. Here, the wheel's radius is 15 inches15 \text{ inches} and the axle's radius is 3 inches3 \text{ inches}, which gives 15÷3=515 \div 3 = 5.

Adım Adım Çözüm

1
Identify the given dimensions of the wheel and the axle.
Rwheel=15 inchesR_{wheel} = 15 \text{ inches}, Raxle=3 inchesR_{axle} = 3 \text{ inches}
These are the required variables for the IMA equation.
2
Apply the Ideal Mechanical Advantage formula for a wheel and axle.
IMA=RwheelRaxleIMA = \frac{R_{wheel}}{R_{axle}}
IMA represents the ratio of the effort distance to the resistance distance.
3
Calculate the final value.
IMA=153=5IMA = \frac{15}{3} = 5
Dividing the wheel radius by the axle radius yields the mechanical advantage.

Anahtar Kavram

Ideal Mechanical Advantage (IMA) of a Wheel and Axle
Soru 47Soru →

A flight line technician must load a cargo pallet with a mass of 240 kg240\text{ kg} into a transport aircraft. The aircraft's cargo deck is 1.5 meters1.5\text{ meters} above the tarmac. To assist with loading, the technician deploys a straight, 6.0-meter6.0\text{-meter} long rigid ramp equipped with frictionless rollers. Assuming the acceleration due to gravity is 10 m/s210\text{ m/s}^2, which of the following identifies the ideal effort force required to push the pallet up the ramp, and correctly compares the total work performed to lifting the pallet vertically?

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Cevap: 600 N600\text{ N}; the total work performed is the same as lifting the pallet vertically.

Cevap

The effort force is 600 N600\text{ N}, and the total work performed is exactly the same as lifting the pallet vertically.
The correct answer accurately calculates both the required force and applies the principle of work conservation. First, weight is determined by multiplying mass by gravity (240 kg×10 m/s2=2400 N240\text{ kg} \times 10\text{ m/s}^2 = 2400\text{ N}). The ideal mechanical advantage (IMA) of the ramp is its length divided by its height (6.0 m/1.5 m=46.0\text{ m} / 1.5\text{ m} = 4). The effort force is the weight divided by the IMA (2400 N/4=600 N2400\text{ N} / 4 = 600\text{ N}). Finally, while the ramp reduces the effort force, it requires moving the object over a longer distance, meaning the total work performed (600 N×6.0 m=3600 Joules600\text{ N} \times 6.0\text{ m} = 3600\text{ Joules}) is identical to lifting it straight up (2400 N×1.5 m=3600 Joules2400\text{ N} \times 1.5\text{ m} = 3600\text{ Joules}).

Adım Adım Çözüm

1
Calculate the weight (load force) of the cargo pallet.
W=240 kg×10 m/s2=2400 NW = 240\text{ kg} \times 10\text{ m/s}^2 = 2400\text{ N}
Mass must be converted to weight (force) before mechanical advantage formulas can be properly applied.
2
Determine the Ideal Mechanical Advantage (IMA) of the inclined plane.
IMA=6.0 m1.5 m=4\text{IMA} = \frac{6.0\text{ m}}{1.5\text{ m}} = 4
IMA for an inclined plane is the ratio of the effort distance (ramp length) to the load distance (vertical height).
3
Calculate the ideal effort force.
Effort Force=2400 N4=600 N\text{Effort Force} = \frac{2400\text{ N}}{4} = 600\text{ N}
The effort force is equal to the total load force divided by the mechanical advantage of the machine.
4
Evaluate the total work performed in both scenarios.
Work via ramp: 600 N×6.0 m=3600 J600\text{ N} \times 6.0\text{ m} = 3600\text{ J}. Work vertical: 2400 N×1.5 m=3600 J2400\text{ N} \times 1.5\text{ m} = 3600\text{ J}. Work is equal.
According to the principle of conservation of energy, simple machines trade increased distance for reduced effort force, leaving the total work unchanged.

Anahtar Kavram

Mechanical advantage reduces effort force by increasing distance, but total work is conserved.
Soru 48Soru →

A construction worker needs to push a heavy wheelbarrow up to a raised foundation that is 2 m2\text{ m} above the ground. The worker has three different sturdy wooden planks available to use as a ramp. Arrange the following planks in order from the LEAST amount of effort force required to the MOST amount of effort force required.

Öğeleri doğru sıraya koymak için sürükleyin

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Cevap

The planks should be ordered from longest (10 m10\text{ m}) to shortest (4 m4\text{ m}). The 10 m10\text{ m} plank requires the least effort, the 8 m8\text{ m} plank requires a moderate amount, and the 4 m4\text{ m} plank requires the most effort.
The effort force needed to move an object up a ramp is inversely proportional to the ramp's length when the vertical height is fixed. The 10 m10\text{ m} plank provides the most gradual slope (highest mechanical advantage) and therefore requires the least effort force. As the plank length decreases to 8 m8\text{ m} and then 4 m4\text{ m}, the slope becomes steeper, reducing the mechanical advantage and requiring increasingly more effort force. Thus, ordering from least to most effort force is synonymous with ordering from longest to shortest plank.

Adım Adım Çözüm

1
Recall the formula for the Ideal Mechanical Advantage (IMA) of an inclined plane.
IMA=Length of the slopeHeight of the incline\text{IMA} = \frac{\text{Length of the slope}}{\text{Height of the incline}}
This formula helps us determine how much a ramp multiplies the input force.
2
Determine the relationship between length, IMA, and effort force.
Since the height (2 m2\text{ m}) is constant for all three planks, a longer plank results in a higher IMA. A higher IMA means less effort force is required to push the load.
Understanding this inverse relationship allows us to rank the effort force without doing complex calculations.
3
Rank the planks based on length to determine the required effort force.
Longest plank (10 m10\text{ m}) = Highest IMA = Least effort force. Shortest plank (4 m4\text{ m}) = Lowest IMA = Most effort force.
The question asks to order from LEAST effort force to MOST effort force, which corresponds to ordering from longest ramp to shortest ramp.

Anahtar Kavram

For an inclined plane of a given height, increasing the length of the plane increases its mechanical advantage, which proportionally decreases the effort force required to move a load up the plane.
Tahmini Süre:45s
Soru 49Soru →

Match each type of wedge or wedge application with its correct mechanical characteristic.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Long, thin wedge
Short, thick wedge
Doorstop

Eşleşmeler

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Cevap

The correct matches are: Long, thin wedge corresponds to high mechanical advantage and less effort; Short, thick wedge corresponds to lower mechanical advantage but rapid splitting; Doorstop corresponds to securing objects using friction.
The correct pairings are based on the principle that the ideal mechanical advantage (IMA) of a wedge is its length divided by its thickness. A long, thin wedge has a high mechanical advantage and therefore requires less effort force to penetrate a material. Conversely, a short, thick wedge has a lower mechanical advantage, meaning it takes more effort force, but it successfully splits the object wider in a shorter distance. A doorstop is a specific application that uses the wedge shape to generate a strong normal force and resulting friction between the door and the floor, holding the object stationary.

Adım Adım Çözüm

1
Analyze the mechanical advantage of a long, thin wedge.
Matches with 'Provides high mechanical advantage and requires less applied effort force.'
The mechanical advantage of a wedge is its length divided by its thickness. A larger ratio means less effort is needed to drive the wedge.
2
Analyze the mechanical advantage of a short, thick wedge.
Matches with 'Provides lower mechanical advantage but splits material apart rapidly over a short distance.'
A smaller length-to-thickness ratio results in a lower mechanical advantage, requiring more force, but the increased thickness spreads the material faster.
3
Analyze the practical application of a doorstop.
Matches with 'Secures objects in place primarily by utilizing friction against a surface.'
Instead of splitting material, a doorstop is driven under a door to increase normal force and create high friction, keeping it wedged in place.

Anahtar Kavram

The mechanical advantage of a wedge is determined by the ratio of its length to its thickness, governing the trade-off between the effort force required and the distance the material is separated.
Soru 50Soru →

A specialized industrial press uses a compound lever system composed of two uniform steel bars to compress a heavy coil spring.

- Lever A (Top handle): is 20 inches20\text{ inches} long, weighs 10 lbs10\text{ lbs}, and pivots on a fulcrum at its left end. A vertical rigid connecting link is attached to Lever A at a point 5 inches5\text{ inches} from its fulcrum.
- Lever B (Bottom pressing bar): is 30 inches30\text{ inches} long, weighs 30 lbs30\text{ lbs}, and pivots on a fulcrum at its left end. The connecting link from Lever A attaches to the far right end of Lever B (30 inches30\text{ inches} from its fulcrum). A heavy coil spring is placed under Lever B, located 6 inches6\text{ inches} from Lever B's fulcrum.

If a worker applies a 40 lb40\text{ lb} downward force at the far right end of Lever A (20 inches20\text{ inches} from its fulcrum), what is the total downward compressive force (in lbs) exerted on the spring?

*Assume both levers are horizontal, the connecting link is perfectly vertical, and the system is in static equilibrium.*

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Cevap: 975

Cevap

975 lbs
In static equilibrium, the sum of clockwise and counterclockwise torques about any pivot point must be exactly zero. By calculating the torques step-by-step through the compound system—incorporating both the applied forces and the center of gravity of each uniform bar—we find the final compressive force transmitted to the spring is 975 lbs.

Adım Adım Çözüm

1
Calculate total clockwise torque on Lever A
900 lb-in
The worker's force (40 lbs at 20 inches) and the lever's own uniform weight (10 lbs at its 10-inch center of gravity) both create clockwise rotation around the fulcrum.
2
Determine the force transmitted through the connecting link
180 lbs
The link must balance the 900 lb-in torque at a distance of 5 inches (900 / 5 = 180 lbs). This 180 lbs force is transmitted downward to Lever B.
3
Calculate total clockwise torque on Lever B
5850 lb-in
The downward force from the link (180 lbs at 30 inches) and Lever B's uniform weight (30 lbs at its 15-inch center of gravity) create clockwise rotation around Lever B's fulcrum.
4
Calculate the final compressive force applied to the spring
975 lbs
The spring must balance Lever B's total 5850 lb-in torque at a distance of 6 inches (5850 / 6 = 975 lbs).

Anahtar Kavram

Torque Equilibrium in Compound and Uniform Levers
Soru 51Soru →

A mechanic needs to lift a heavy engine using a screw jack. They can equip the jack with one of two interchangeable threaded rods: Rod X has a coarse thread with 44 threads per inch (TPI), and Rod Y has a fine thread with 88 TPI.

Assuming the jack's handle length is identical for both setups and friction is negligible, which of the following statements accurately describes the required effort force and total work done if the mechanic uses Rod Y instead of Rod X to lift the engine by exactly 1010 inches?

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Cevap: Rod Y requires half the effort force on the handle, but the total work done to lift the engine remains the same.

Cevap

Rod Y requires half the effort force on the handle, but the total work done to lift the engine remains the same.
A screw is essentially an inclined plane wrapped around a central cylinder. Increasing the Threads Per Inch from 44 to 88 halves the pitch of the thread, making the 'incline' less steep. This doubles the mechanical advantage of the jack, cutting the required effort force on the handle in half. However, the fundamental principle of all simple machines is that they do not reduce the total amount of work required to move a load; they merely trade increased input distance (more rotations of the handle) for decreased input force. Because the engine is still being lifted exactly 1010 inches, the total output work remains identical.

Adım Adım Çözüm

1
Analyze the relationship between Threads Per Inch (TPI) and pitch.
Rod Y (88 TPI) has half the pitch of Rod X (44 TPI), meaning it advances half the linear distance per full rotation.
Pitch is the mathematical inverse of TPI (1/TPI1/\text{TPI}). It determines the steepness of the screw's thread.
2
Determine the effect of a smaller pitch on Mechanical Advantage and effort force.
Because Rod Y advances half as far per rotation, its Ideal Mechanical Advantage is doubled. Thus, it requires half the effort force to lift the same load.
A screw is an inclined plane wrapped around a cylinder. A finer thread equates to a shallower incline, which trades distance for force.
3
Evaluate the total work done by the system in both scenarios.
The total work done remains identical regardless of the rod used.
Total output work is calculated as Load Force ×\times Lift Distance. Since the engine's weight and the 1010-inch lift distance do not change, the total work is conserved.

Anahtar Kavram

Mechanical Advantage of Screws and Conservation of Work
Tahmini Süre:1m 0s
Soru 52Soru →

A manual winch is used to lift a 240 lb240 \text{ lb} engine block exactly 2 feet2 \text{ feet} vertically. The winch consists of a steering-wheel style control with a diameter of 3 feet3 \text{ feet}, which is rigidly attached to an axle drum with a radius of 3 inches3 \text{ inches}. Assuming an ideal system with no friction, what continuous effort force must be applied to the outer edge of the wheel, and what total distance must the effort force move along the wheel's rotation?

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Cevap: 40 lbs40 \text{ lbs} of force; 12 feet12 \text{ feet} of distance

Cevap

The correct answer is 40 lbs40 \text{ lbs} of force and 12 feet12 \text{ feet} of distance.
The system's Ideal Mechanical Advantage is determined by the ratio of the effort radius (18 inches, half of the 3-foot diameter) to the load radius (3 inches), which yields an IMA of 6. This means the required effort force is divided by 6 (240 lbs/6=40 lbs240 \text{ lbs} / 6 = 40 \text{ lbs}), while the required effort distance is multiplied by 6 (2 feet×6=12 feet2 \text{ feet} \times 6 = 12 \text{ feet}) to conserve work.

Adım Adım Çözüm

1
Convert all measurements to the same unit (inches).
Wheel diameter = 3 feet=36 inches3 \text{ feet} = 36 \text{ inches}.
Units must be identical to calculate a valid, dimensionless mechanical advantage ratio.
2
Identify the effort radius and the load radius.
Wheel radius (RR) = 18 inches18 \text{ inches}. Axle drum radius (rr) = 3 inches3 \text{ inches}.
The lever arms in a wheel and axle system are the distances from the central axis (the radii), not the diameters.
3
Calculate the Ideal Mechanical Advantage (IMA).
IMA=Rr=183=6\text{IMA} = \frac{R}{r} = \frac{18}{3} = 6.
This establishes that the system multiplies the user's force by a factor of 6.
4
Calculate the required effort force.
Effort Force=LoadIMA=240 lbs6=40 lbs\text{Effort Force} = \frac{\text{Load}}{\text{IMA}} = \frac{240 \text{ lbs}}{6} = 40 \text{ lbs}.
Because the machine provides a mechanical advantage of 6, the required input force is one-sixth of the load.
5
Calculate the required effort distance.
Effort Distance=Load Distance×IMA=2 feet×6=12 feet\text{Effort Distance} = \text{Load Distance} \times \text{IMA} = 2 \text{ feet} \times 6 = 12 \text{ feet}.
According to the principle of work, whatever factor force is reduced by, distance must be multiplied by that same factor.

Anahtar Kavram

Ideal Mechanical Advantage (IMA) and Work in a Wheel and Axle
Tahmini Süre:1m 30s
Soru 53Soru →

Match each gear system configuration with its resulting mechanical properties.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Simple train: 15-tooth drive gear directly meshing with a 60-tooth driven gear.
Idler train: 60-tooth drive gear, 25-tooth idler gear, and 15-tooth driven gear.
Compound train: 20-tooth drive gear driving a 40-tooth gear on a countershaft; a 10-tooth gear on the same countershaft drives a 30-tooth final gear.
Rack and pinion: 12-tooth pinion gear meshed directly with a flat, toothed rack.

Eşleşmeler

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Cevap

The simple train multiplies torque by 4 and reverses direction. The idler train quadruples speed and keeps the same direction. The compound train provides a mechanical advantage of 6 and keeps the same direction. The rack and pinion converts rotational motion to linear motion.
Each specific gear arrangement produces distinct mechanical outcomes based on the ratio of teeth between the driving and driven gears, the number of intermediary gears, and the physical shape of the gears involved.

Adım Adım Çözüm

1
Determine the mechanical properties of the simple train.
Matches with: Torque is multiplied by 4; rotational direction is reversed.
A 15-tooth gear driving a 60-tooth gear has a ratio of 4:1. The driven gear is 4 times larger, multiplying torque by 4. Two meshing gears turn in opposite directions.
2
Determine the mechanical properties of the idler train.
Matches with: Output speed is 4 times the input speed; rotational direction is unchanged.
The 25-tooth idler gear does not change the ratio. The 60-tooth drive gear directly relates to the 15-tooth driven gear, meaning the output is 1/4 the size and spins 4 times as fast. An odd number of gears (3) preserves the rotational direction.
3
Determine the mechanical properties of the compound train.
Matches with: Overall mechanical advantage is 6; final gear rotates in the same direction as the input gear.
Calculate the ratio of each pair: 40/20 = 2, and 30/10 = 3. Multiply them: 2 * 3 = 6. This is the mechanical advantage. There are two gear meshes, causing the direction to reverse twice, returning to the original input direction.
4
Determine the mechanical properties of the rack and pinion.
Matches with: Rotational motion is converted into linear motion.
A round pinion turning against a straight rack is the standard mechanism for converting rotational motion into linear motion.

Anahtar Kavram

Calculating gear ratios and tracking rotational direction across simple, idler, compound, and rack-and-pinion trains.
Soru 54Soru →

A worker uses a uniform 6060-inch steel pry bar to lift a 150 lb150\text{ lb} concrete block. The worker places a pivot exactly 2020 inches from the load end of the bar, configuring it as a first-class lever. If the pry bar itself weighs 20 lbs20\text{ lbs}, how much downward vertical force must the worker apply at the opposite end of the bar to perfectly balance the concrete block?

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Cevap: 70 lbs70\text{ lbs}

Cevap

The correct force required is 70 lbs70\text{ lbs}.
To achieve equilibrium, the torques around the pivot must balance. The concrete block creates a counter-clockwise torque of 3000 in-lbs3000\text{ in-lbs} (150 lbs×20 inches150\text{ lbs} \times 20\text{ inches}). The pry bar's weight acts at its center (3030 inches from the end), which places it 1010 inches from the pivot on the effort side. This provides an assisting clockwise torque of 200 in-lbs200\text{ in-lbs} (20 lbs×10 inches20\text{ lbs} \times 10\text{ inches}). The effort force is applied at the very end, 4040 inches from the pivot. The equilibrium equation is 3000=200+(F×40)3000 = 200 + (F \times 40), resolving to 2800=F×402800 = F \times 40, which means the required force is 70 lbs70\text{ lbs}.

Adım Adım Çözüm

1
Calculate the counter-clockwise torque generated by the concrete block.
150 lbs×20 inches=3000 in-lbs150\text{ lbs} \times 20\text{ inches} = 3000\text{ in-lbs}.
The load is placed 2020 inches from the pivot, creating the primary rotational force to overcome.
2
Determine the location of the pry bar's center of gravity and its resulting torque.
The center of the 6060-inch bar is at 3030 inches. Since the pivot is at 2020 inches, the center of gravity is 1010 inches from the pivot on the effort side. Torque = 20 lbs×10 inches=200 in-lbs20\text{ lbs} \times 10\text{ inches} = 200\text{ in-lbs} (clockwise).
A uniform bar's weight acts at its exact center. Because this center is on the same side of the pivot as the worker's effort, it provides an assisting clockwise torque.
3
Set up the torque equilibrium equation to find the required effort force.
Counter-clockwise Torque = Clockwise Torque
3000=200+(F×40)3000 = 200 + (F \times 40)
For the lever to balance, the sum of opposing torques must be equal. The effort arm is the total length minus the load arm (60−20=4060 - 20 = 40 inches).
4
Solve the equation for the effort force (FF).
2800=F×402800 = F \times 40, which simplifies to F=70 lbsF = 70\text{ lbs}.
Isolating FF gives the exact supplemental force the worker needs to apply to achieve balance.

Anahtar Kavram

Torque Equilibrium Involving Lever Mass
Soru 55Soru →

Arrange the following material handling tasks in order from the LEAST amount of effort force required to the GREATEST amount of effort force required. (Assume frictionless ramps)

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Cevap

The correct sequence from least to greatest effort force is: Task A (30 lbs30\text{ lbs}), Task D (40 lbs40\text{ lbs}), Task C (50 lbs50\text{ lbs}), Task B (60 lbs60\text{ lbs}).
Task A requires 30 lbs30\text{ lbs} of force, Task D requires 40 lbs40\text{ lbs}, Task C requires 50 lbs50\text{ lbs}, and Task B requires 60 lbs60\text{ lbs}. Ordering these mathematically from lowest to highest yields Task A, Task D, Task C, and Task B.

Adım Adım Çözüm

1
Calculate the Ideal Mechanical Advantage (IMA) for each ramp.
Task A: 12/3=412/3 = 4. Task B: 6/2=36/2 = 3. Task C: 8/4=28/4 = 2. Task D: 5/1=55/1 = 5.
The IMA determines how much the ramp multiplies the applied force, calculated as the length of the ramp divided by its height.
2
Calculate the required effort force for each task.
Task A: 120/4=30 lbs120/4 = 30\text{ lbs}. Task B: 180/3=60 lbs180/3 = 60\text{ lbs}. Task C: 100/2=50 lbs100/2 = 50\text{ lbs}. Task D: 200/5=40 lbs200/5 = 40\text{ lbs}.
Effort force equals the load's total weight divided by the IMA of the ramp.
3
Order the tasks from least effort to greatest effort.
Task A (30 lbs30\text{ lbs}) < Task D (40 lbs40\text{ lbs}) < Task C (50 lbs50\text{ lbs}) < Task B (60 lbs60\text{ lbs}).
The question specifically asks for the sequence starting with the lowest force and ending with the highest.

Anahtar Kavram

Ideal Mechanical Advantage and Effort Force on an Inclined Plane
Soru 56Soru →

A homeowner needs to move a 300 lb300\text{ lb} refrigerator onto a porch that is 2 feet2\text{ feet} above ground level. They place a straight, frictionless wooden ramp that is 6 feet6\text{ feet} long from the ground to the porch edge. How much effort force is required to push the refrigerator up the ramp?

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Cevap: 100 lbs100\text{ lbs}

Cevap

100 lbs100\text{ lbs}
To find the force required to push an object up a frictionless ramp, first find the ramp's Ideal Mechanical Advantage (IMA). The IMA is the ramp's length divided by its vertical height (6 ft/2 ft=36\text{ ft} / 2\text{ ft} = 3). The effort force needed is the object's weight divided by this mechanical advantage (300 lbs/3=100 lbs300\text{ lbs} / 3 = 100\text{ lbs}).

Adım Adım Çözüm

1
Calculate the Ideal Mechanical Advantage (IMA) of the ramp.
IMA=LengthHeight=6 ft2 ft=3\text{IMA} = \frac{\text{Length}}{\text{Height}} = \frac{6\text{ ft}}{2\text{ ft}} = 3
The IMA tells us how many times easier the ramp makes the task compared to lifting the object straight up.
2
Calculate the required effort force.
Effort Force=WeightIMA=300 lbs3=100 lbs\text{Effort Force} = \frac{\text{Weight}}{\text{IMA}} = \frac{300\text{ lbs}}{3} = 100\text{ lbs}
An inclined plane reduces the necessary force by a factor equal to its mechanical advantage.

Anahtar Kavram

Ideal Mechanical Advantage and Effort Force of an Inclined Plane
Tahmini Süre:1m 0s
Soru 57Soru →

A worker uses a steel wedge to split a large block of ice. If the worker switches to a longer wedge that has the exact same maximum thickness, which of the following statements correctly describes the result?

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Cevap: It decreases the effort force required to split the ice.

Cevap

It decreases the effort force required to split the ice.
The ideal mechanical advantage of a wedge is its length divided by its thickness. By increasing the length while keeping the thickness the same, the mechanical advantage increases, which reduces the effort force needed to do the job.

Adım Adım Çözüm

1
Recall the formula for the ideal mechanical advantage (IMA) of a wedge.
IMA = Length / Thickness
This relationship determines how much a wedge multiplies the applied force.
2
Analyze the change in the wedge's dimensions described in the scenario.
The length increases, while the thickness remains exactly the same.
Understanding the variables helps determine the new mechanical advantage.
3
Determine the effect on the effort force.
Since the numerator (length) is larger and the denominator (thickness) is constant, the IMA increases. A higher IMA means less effort force is needed.
An increased mechanical advantage trades a longer distance for a reduced effort force requirement.

Anahtar Kavram

Mechanical Advantage of Wedges
Soru 58Soru →

A technician is analyzing the thread specifications of four different fasteners to determine their linear travel. Assuming all fasteners have single-start threads, arrange them in order from the LEAST to the GREATEST linear distance they will travel into a fixed base for each full 360∘360^\circ rotation.

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Cevap

The correct sequence from least to greatest linear travel per rotation is: the machine screw (20 TPI), the wood screw (1/16 inch pitch), the lag bolt (10 TPI), and finally the power screw (0.125 inch lead).
The sequence correctly calculates and compares the linear travel per rotation for each fastener. By converting all specifications to decimal inches (0.050.05 in, 0.06250.0625 in, 0.100.10 in, and 0.1250.125 in), they can be accurately ordered from least to greatest.

Adım Adım Çözüm

1
Define the relationship between threads per inch (TPI), pitch, lead, and linear travel.
For single-start threads, the linear travel per rotation equals the lead, which is equal to the pitch. Pitch can be calculated as 1÷TPI1 \div \text{TPI}.
Establishing the mathematical formulas is necessary to compare different types of thread specifications on a single scale.
2
Calculate the decimal linear travel for the fasteners specified in TPI.
The machine screw travels 1÷20=0.051 \div 20 = 0.05 inches per rotation. The lag bolt travels 1÷10=0.101 \div 10 = 0.10 inches per rotation.
Converting TPI into decimal pitch allows for direct numerical comparison.
3
Convert the fractional pitch of the wood screw to a decimal.
The wood screw travels 1÷16=0.06251 \div 16 = 0.0625 inches per rotation. The power screw is already given as 0.1250.125 inches.
Decimals provide a consistent format for sequential ordering.
4
Compare all decimal values and arrange them from least to greatest.
0.050.05 (machine screw) <0.0625< 0.0625 (wood screw) <0.10< 0.10 (lag bolt) <0.125< 0.125 (power screw).
Finalizes the sequence according to the explicit instructions in the stem.

Anahtar Kavram

The linear distance a screw travels in one full rotation is determined by its lead. For single-start threads, the lead equals the pitch, and the pitch is the reciprocal of the threads per inch (P=1/TPIP = 1 / \text{TPI}).
Soru 59Soru →

A motorized mixer uses two meshing gears to adjust the mixing speed. The driving gear, which is attached to the motor, has 1212 teeth and spins at 150150 RPM. It meshes directly with a driven gear that has 3636 teeth. What is the rotational speed of the driven gear in RPM?

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Cevap: 50

Cevap

50
The driven gear has three times as many teeth as the driving gear (3636 vs 1212). Because gears have an inverse relationship between size and speed, the larger driven gear will rotate at one-third the speed of the driving gear (150÷3=50150 \div 3 = 50 RPM).

Adım Adım Çözüm

1
State the gear speed formula
T1×S1=T2×S2T_1 \times S_1 = T_2 \times S_2
To establish the inverse relationship between the number of gear teeth and rotational speed
2
Substitute the given values into the formula
12×150=36×S212 \times 150 = 36 \times S_2
To set up the equation with the known tooth counts and driving speed
3
Solve for the unknown driven speed (S2S_2)
S2=180036=50S_2 = \frac{1800}{36} = 50
To find the final RPM of the driven gear

Anahtar Kavram

The rotational speed of meshing gears is inversely proportional to their number of teeth.
Soru 60Soru →

A carpenter uses a wooden wedge as a doorstop to hold a heavy door open. The wedge has a length of 15 cm15\text{ cm} and a maximum thickness of 3 cm3\text{ cm}. What is the ideal mechanical advantage (IMA) of this wedge?

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Cevap: 5

Cevap

The ideal mechanical advantage is 5.
The ideal mechanical advantage of a wedge is found by dividing its length (the distance the wedge penetrates) by its maximum thickness (the distance the object is separated). For a wedge with a length of 15 cm15\text{ cm} and a thickness of 3 cm3\text{ cm}, the mechanical advantage is calculated as 15/3=515 / 3 = 5.

Adım Adım Çözüm

1
Identify the given dimensions of the wedge.
Length (LL) = 15 cm15\text{ cm}, Thickness (tt) = 3 cm3\text{ cm}.
These values are required to calculate the mechanical advantage.
2
Apply the formula for the ideal mechanical advantage of a wedge.
IMA=Lt=153=5IMA = \frac{L}{t} = \frac{15}{3} = 5.
The IMA is the ratio of the effort distance (length) to the resistance distance (thickness).

Anahtar Kavram

Ideal Mechanical Advantage of a Wedge
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