All practice questions

Question 201Question →

A motorized winch is pulling a $1,500\text{ lb}$ boat and trailer assembly up a concrete boat ramp at a constant speed. The ramp has a total length of $40\text{ ft}$ and a vertical drop of $8\text{ ft}$ from the parking lot to the water level. If the mechanical efficiency of the ramp system is $80\%$ (due to friction in the wheel bearings and surface resistance), what is the actual effort force, in pounds, the winch must apply to pull the assembly up the ramp?

Show answer & explanation

Answer: 375

Answer

The winch must apply an actual effort force of 375 pounds.

To solve this problem, you must integrate the concepts of Ideal Mechanical Advantage (IMA) and mechanical efficiency. First, find the IMA by dividing the ramp's length by its vertical drop (

40 / 8 = 5

). In an ideal, frictionless system, the effort force required would be the load divided by the IMA (

1500 / 5 = 300\text{ lb}

). However, the problem states the system is only

80\%

efficient due to friction. Efficiency is the ratio of ideal effort to actual effort. To find the actual effort force, divide the ideal effort by the efficiency (

300 / 0.80 = 375\text{ lb}

).

Step-by-Step Solution

1

Calculate the Ideal Mechanical Advantage (IMA).

\text{IMA} = 5

The IMA of an inclined plane is found by dividing the length of the slope by the vertical height (

40 / 8 = 5

).

2

Calculate the theoretical effort force required under ideal, frictionless conditions.

\text{Theoretical Effort} = 300\text{ lb}

Theoretical effort force is the load divided by the IMA (

1500 / 5 = 300\text{ lb}

).

3

Calculate the actual effort force by accounting for the mechanical efficiency.

\text{Actual Effort} = 375\text{ lb}

Because the system is only

80\%

efficient, it requires more force than the ideal scenario. Divide the theoretical effort by the efficiency as a decimal (

300 / 0.80 = 375\text{ lb}

).

Key Concept

Actual Mechanical Advantage and Efficiency of Inclined Planes

Question 202Question →

A mechanic is using a C-clamp with a threaded screw that has $16$ threads per inch (TPI). The clamp is currently open to a width of $5.0$ inches. If the mechanic needs to tightly secure two steel plates that have a combined thickness of $1.5$ inches, how many full revolutions of the screw are required to close the clamp to the exact thickness of the plates?

Show answer & explanation

Answer: 56

Answer

56

The screw must travel a linear distance of

3.5

inches to close the gap from

5.0

inches to

1.5

inches. With

16

threads per inch, each inch of linear travel requires

16

full revolutions. Multiplying the

3.5

inches of travel by

16

revolutions per inch yields exactly

56

revolutions.

Step-by-Step Solution

1

Calculate the linear distance the clamp's jaw needs to travel.

5.0 - 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 \times 16 = 56

revolutions

TPI (threads per inch) indicates that one full revolution moves the screw

1/16

of an inch. Therefore, moving it

1

inch requires

16

revolutions.

Key Concept

Linear Travel and Threads Per Inch (TPI)

Question 203Question →

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

Show answer & explanation

Answer: 3060

Answer

3060

The correct lateral separating force is found by determining the wedge's Ideal Mechanical Advantage (

24\text{ cm} / 4\text{ cm} = 6

), adjusting for energy lost to friction by applying the

60\%

efficiency (

6 \times 0.60 = 3.6\text{ AMA}

), and multiplying this actual advantage by the input force (

850\text{ N} \times 3.6 = 3060\text{ N}

).

Step-by-Step Solution

1

Calculate the Ideal Mechanical Advantage (IMA) of the wedge.

IMA = 6

The IMA of a wedge is defined as the ratio of its penetration length to its maximum thickness (

24\text{ cm} / 4\text{ cm}

).

2

Determine the Actual Mechanical Advantage (AMA).

AMA = 3.6

Because the wedge operates at

60\%

efficiency due to friction, the actual mechanical advantage is the IMA multiplied by the efficiency (

6 \times 0.60

).

3

Calculate the total lateral separating force.

3060\text{ N}

The output force is the product of the input effort force and the Actual Mechanical Advantage (

850\text{ N} \times 3.6

).

Key Concept

Mechanical Advantage and Efficiency of Wedges

Question 204Question →

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 $200 \text{ N}$ , with its center of gravity located $0.5 \text{ m}$ horizontally from the axle. A load of bricks weighing $800 \text{ N}$ is placed in the tray so that its center of gravity is $0.4 \text{ m}$ horizontally from the axle. The worker lifts upward at the ends of the handles, which are located $1.2 \text{ m}$ 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?

Show answer & explanation

Answer: 350

Answer

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 \text{ N}\cdot\text{m}

of torque and the bricks contribute

320 \text{ N}\cdot\text{m}

. Balancing the total

420 \text{ N}\cdot\text{m}

torque over a

1.2 \text{ m}

lever arm requires exactly

350 \text{ N}

of effort.

Step-by-Step Solution

1

Calculate the torque from the empty wheelbarrow.

\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.

\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.

\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.

F_{effort} = 350 \text{ N}

Torque equals force times distance. Rearranging

420 = F \times 1.2

isolates the required lifting force.

Key Concept

Torque Equilibrium and Composite Loads in Second-Class Levers

Question 205Question →

A standard shipping crate is measured on Earth (where the acceleration due to gravity is $10 \text{ m/s}^2$ ) using a mechanical balance scale. The balance scale determines that the crate has a mass of $400 \text{ kg}$ . The crate is then transported to a planetary outpost where the local gravitational acceleration is $4 \text{ m/s}^2$ . At the outpost, the crate is hoisted using a heavy-duty spring scale. This spring scale was manufactured and calibrated on Earth to display mass in kilograms based on Earth's gravity. What numerical value will this spring scale display when weighing the crate at the outpost?

Show answer & explanation

Answer: 160

Answer

160

The true mass of the crate is 400 kg, which does not change based on location. On the outpost, the crate's weight (force) is 400 kg × 4 m/s² = 1600 N. The spring scale stretches based entirely on this 1600 N force. Because it was calibrated on Earth (where the acceleration due to gravity is 10 m/s²), its internal mechanism is designed to display '1 kg' for every 10 N of force it supports. Therefore, an applied force of 1600 N causes the scale to point to 1600 / 10 = 160.

Step-by-Step Solution

1

Determine the true mass of the crate.

400 kg

Mass is the amount of matter and is invariant. A balance scale compares masses and measures true mass accurately regardless of the local gravitational field.

2

Calculate the downward force (weight) exerted by the crate on the spring scale at the outpost.

1600 N

Weight depends on local gravity. It is calculated using the formula Weight = mass × local gravity, which is 400 kg × 4 m/s².

3

Determine the calibration factor of the Earth-calibrated spring scale.

10 N per 1 kg of display

A spring scale physically stretches in response to force. Because it was calibrated to display mass on Earth, it assumes that every 10 N of applied force corresponds to 1 kg of mass.

4

Calculate the value displayed by the spring scale.

160

The scale converts the 1600 N force it physically experiences into a kilogram reading using its Earth calibration: 1600 N / 10 m/s² = 160.

Key Concept

Mass is an invariant property of matter, while weight is the localized gravitational force acting on that mass. A spring scale measures force (weight) rather than mass, meaning a spring scale calibrated in kilograms on Earth will display incorrect mass readings in different gravitational fields.

Question 206Question →

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 $15 \text{ inches}$ , turning a central spool (the axle) with a radius of $3 \text{ inches}$ .

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

Show answer & explanation

Answer: 5

Answer

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 \text{ inches}

and the axle's radius is

3 \text{ inches}

, which gives

15 \div 3 = 5

.

Step-by-Step Solution

1

Identify the given dimensions of the wheel and the axle.

R_{wheel} = 15 \text{ inches}

,

R_{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 = \frac{R_{wheel}}{R_{axle}}

IMA represents the ratio of the effort distance to the resistance distance.

3

Calculate the final value.

IMA = \frac{15}{3} = 5

Dividing the wheel radius by the axle radius yields the mechanical advantage.

Key Concept

Ideal Mechanical Advantage (IMA) of a Wheel and Axle

Question 207Question →

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

If a worker applies a $40\text{ lb}$ downward force at the far right end of Lever A ( $20\text{ inches}$ 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.*

Show answer & explanation

Answer: 975

Answer

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.

Step-by-Step Solution

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).

Key Concept

Torque Equilibrium in Compound and Uniform Levers

Question 208Question →

A dentist adjusts a patient's hydraulic examination chair. The system's input piston has a cross-sectional area of $8 \text{ in}^2$ , and the lifting piston beneath the chair has an area of $64 \text{ in}^2$ . If the dentist applies $35 \text{ lbs}$ of force to the input piston, how much upward force is generated by the lifting piston?

Show answer & explanation

Answer: 280

Answer

280

The hydraulic system provides a mechanical advantage equal to the ratio of the two piston areas (

64 / 8 = 8

). Multiplying the input force of

35 \text{ lbs}

by this mechanical advantage yields a total lifting force of

280 \text{ lbs}

.

Step-by-Step Solution

1

Identify the given variables.

Input area

A_1 = 8 \text{ in}^2

, output area

A_2 = 64 \text{ in}^2

, and input force

F_1 = 35 \text{ lbs}

.

These variables are required to apply Pascal's Principle.

2

Calculate the mechanical advantage (area ratio) of the hydraulic system.

\frac{A_2}{A_1} = \frac{64}{8} = 8

.

The ratio of the output area to the input area determines how much the input force will be multiplied.

3

Calculate the generated output force.

F_2 = F_1 \times \text{Area Ratio} = 35 \text{ lbs} \times 8 = 280 \text{ lbs}

.

Pascal's Principle states that pressure is transmitted equally throughout an enclosed fluid, meaning force increases proportionally with area.

Key Concept

According to Pascal's Principle, pressure applied to an enclosed fluid is transmitted equally in all directions. In a hydraulic system, this means the ratio of output force to input force is equal to the ratio of output area to input area (

F_1/A_1 = F_2/A_2

).

Estimated Time:45s

Question 209Question →

A motorized mixer uses two meshing gears to adjust the mixing speed. The driving gear, which is attached to the motor, has $12$ teeth and spins at $150$ RPM. It meshes directly with a driven gear that has $36$ teeth. What is the rotational speed of the driven gear in RPM?

Show answer & explanation

Answer: 50

Answer

50

The driven gear has three times as many teeth as the driving gear (

36

vs

12

). 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 \div 3 = 50

RPM).

Step-by-Step Solution

1

State the gear speed formula

T_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 \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 (

S_2

)

S_2 = \frac{1800}{36} = 50

To find the final RPM of the driven gear

Key Concept

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

Question 210Question →

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

Show answer & explanation

Answer: 5

Answer

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\text{ cm}

and a thickness of

3\text{ cm}

, the mechanical advantage is calculated as

15 / 3 = 5

.

Step-by-Step Solution

1

Identify the given dimensions of the wedge.

Length (

L

) =

15\text{ cm}

, Thickness (

t

) =

3\text{ cm}

.

These values are required to calculate the mechanical advantage.

2

Apply the formula for the ideal mechanical advantage of a wedge.

IMA = \frac{L}{t} = \frac{15}{3} = 5

.

The IMA is the ratio of the effort distance (length) to the resistance distance (thickness).

Key Concept

Ideal Mechanical Advantage of a Wedge

Question 211Question →

A newly designed lunar rover is being tested in a facility on Earth. The rover has a measured downward weight of $4500 \text{ N}$ . Assuming the acceleration due to gravity on Earth is $10 \text{ m/s}^2$ , what is the mass of the rover in kilograms?

Show answer & explanation

Answer: 450

Answer

The mass of the rover is 450 kg.

Mass represents the amount of matter in an object and is found by dividing its weight (the force of gravity acting upon it) by the local gravitational acceleration. Using the formula

m = \frac{W}{g}

, substituting

4500 \text{ N}

for

W

and

10 \text{ m/s}^2

for

g

yields exactly

450 \text{ kg}

.

Step-by-Step Solution

1

Identify the given variables for weight and gravity.

W = 4500 \text{ N}

,

g = 10 \text{ m/s}^2

These are the necessary components to determine the object's mass.

2

State the relationship between weight and mass.

W = m \times g

This is Newton's Second Law applied to the force of gravity.

3

Rearrange the formula to isolate mass (

m

).

m = \frac{W}{g}

We need to solve for the unknown variable, mass.

4

Substitute the values and divide.

m = \frac{4500}{10} = 450

Performing the arithmetic gives the amount of matter inside the rover.

Key Concept

Weight is the gravitational force exerted on an object, calculated as the product of its mass and local gravitational acceleration (

W = mg

). Mass is an invariant property that can be found by dividing weight by gravity.

Estimated Time:45s

Question 212Question →

A hydraulic rescue spreader operates using a closed fluid system. The tool's spreading arms are driven by an output piston with a cross-sectional area of $20 \text{ in}^2$ . During an operation, the output piston must exert a constant force of $6000 \text{ lbs}$ and extend exactly $4 \text{ inches}$ . The hand-operated pump uses an input piston with a cross-sectional area of $2 \text{ in}^2$ .

To achieve this full $4 \text{ inch}$ extension of the output piston, what is the total distance, in inches, that the input piston must travel?

Show answer & explanation

Answer: 40

Answer

The input piston must travel a total distance of 40 inches.

In a closed hydraulic system, the volume of fluid displaced by the input piston must equal the volume received by the output piston (

A_1 \times d_1 = A_2 \times d_2

). Since the output piston requires

80 \text{ in}^3

of fluid (

20 \text{ in}^2 \times 4 \text{ in}

), the input piston, with an area of

2 \text{ in}^2

, must travel

40 \text{ inches}

(

80 / 2

) to displace that same volume. The force value (

6000 \text{ lbs}

) is not needed to find the distance.

Step-by-Step Solution

1

State the principle of conservation of volume for a closed hydraulic system.

V_{in} = V_{out}

or

A_1 \times d_1 = A_2 \times d_2

Because liquids are essentially incompressible, the volume of fluid pushed by the input piston equals the volume that moves the output piston.

2

Calculate the volume of fluid displaced at the output cylinder.

V_{out} = 20 \text{ in}^2 \times 4 \text{ inches} = 80 \text{ in}^3

This is the total volume of fluid required to extend the spreading arms by the specified distance.

3

Calculate the required travel distance for the input piston.

2 \text{ in}^2 \times d_1 = 80 \text{ in}^3 \rightarrow d_1 = 40 \text{ inches}

Dividing the total required volume by the input piston's area gives the total distance the input piston must be pushed.

Key Concept

Conservation of fluid volume and mechanical advantage in hydraulic systems.

Question 213Question →

A portable hand-cranked emergency generator utilizes a compound gear train to step up the rotational speed of its internal rotor. The hand crank is directly attached to the driving gear (Gear A), which has $60$ teeth. Gear A meshes with Gear B, which has $15$ teeth. Gear B is fixed to the same solid shaft as Gear C, which has $45$ teeth. Finally, Gear C meshes with Gear D, which has $10$ teeth and is attached to the generator's rotor.

If a person turns the hand crank at a steady speed of $40\text{ RPM}$ , what is the resulting speed of the generator's rotor in $\text{RPM}$ ?

Show answer & explanation

Answer: 720

Answer

720

In a compound gear train, the overall mechanical advantage (or speed ratio) is found by evaluating each meshing pair separately. The first pair (A to B) increases the speed by a factor of

4

(since

60/15 = 4

). The input speed is

40\text{ RPM}

, so Gear B spins at

160\text{ RPM}

. Because Gear B and Gear C are fixed to the same shaft, Gear C also spins at

160\text{ RPM}

. The second meshing pair (C to D) increases the speed by a factor of

4.5

(since

45/10 = 4.5

). Multiplying Gear C's speed by this ratio gives

160 \times 4.5 = 720\text{ RPM}

.

Step-by-Step Solution

1

Determine the rotational speed of Gear B.

160\text{ RPM}

Gear A (

60

teeth) drives Gear B (

15

teeth). The speed multiplier is

60 / 15 = 4

. Therefore, Gear B spins at

40 \times 4 = 160\text{ RPM}

.

2

Determine the rotational speed of Gear C.

160\text{ RPM}

Gears mounted on the exact same solid shaft rotate at the same speed, regardless of their size.

3

Determine the rotational speed of Gear D (the rotor).

720\text{ RPM}

Gear C (

45

teeth) drives Gear D (

10

teeth). The speed multiplier is

45 / 10 = 4.5

. Therefore, Gear D spins at

160 \times 4.5 = 720\text{ RPM}

.

Key Concept

Compound Gear Train Speed Ratios

Question 214Question →

A depot worker must slide a $1200\text{ lb}$ steel safe up a straight ramp to a storage platform that is $3\text{ feet}$ above the main floor. The worker uses a mechanical winch that can pull with a maximum force of $240\text{ lbs}$ . Assuming a frictionless surface, what is the minimum ramp length required to move the safe?

Show answer & explanation

Answer: 15

Answer

15

The mechanical advantage required to move a

1200\text{ lb}

load using only

240\text{ lbs}

of force is

1200 / 240 = 5

. Because the Ideal Mechanical Advantage of an inclined plane is its length divided by its height, a ramp with an IMA of

5

lifting a load

3\text{ feet}

vertically must be

5 \times 3 = 15\text{ feet}

long.

Step-by-Step Solution

1

Calculate the required Ideal Mechanical Advantage (IMA).

IMA = 5

The mechanical advantage is the ratio of the load force to the maximum effort force (

1200 / 240 = 5

).

2

Set up the equation for the mechanical advantage of an inclined plane.

5 = Length / 3

For an inclined plane, IMA is equal to the length of the slope divided by the vertical height.

3

Solve for the ramp length.

Length = 15 feet

Multiplying the IMA by the height (

5 \times 3

) gives the minimum required ramp length to achieve that mechanical advantage.

Key Concept

Ideal Mechanical Advantage (IMA) of an Inclined Plane

Question 215Question →

During a training exercise, a signaling flare is launched vertically upward from ground level with an initial velocity of $30\text{ m/s}$ . Assuming air resistance is negligible and using the approximation $g = 10\text{ m/s}^2$ , how many seconds will it take for the flare to reach its maximum height?

Show answer & explanation

Answer: 3

Answer

It will take 3 seconds for the flare to reach its maximum height.

At its maximum height, any vertically launched object momentarily stops before falling back down, meaning its instantaneous velocity is

0\text{ m/s}

. Given an initial upward velocity of

30\text{ m/s}

, gravity decreases this velocity by

10\text{ m/s}

for every passing second. Therefore, it takes exactly 3 seconds for gravity to completely counteract the initial upward speed (

30 / 10 = 3

).

Step-by-Step Solution

1

Identify known variables and the final goal

v_i = 30\text{ m/s}

,

v_f = 0\text{ m/s}

(velocity is zero at the apex), and

a = -10\text{ m/s}^2

. The goal is to find time (

t

).

Establishing the initial and final states is required to choose the correct kinematic equation for free fall.

2

Select the appropriate kinematic formula

Use the equation

v_f = v_i + at

.

This formula connects initial velocity, final velocity, acceleration, and time, without needing to know the distance traveled.

3

Substitute the values and solve for

t

0 = 30 - 10t \Rightarrow 10t = 30 \Rightarrow t = 3

.

Solving the linear algebraic equation yields the exact time it takes for gravity to decelerate the flare to a stop.

Key Concept

Calculating the time to reach the apex of vertical projectile motion under uniform gravitational acceleration.

Question 216Question →

A stonemason uses a steel wedge to fracture a granite boulder. The wedge has a penetration length of $160\text{ mm}$ and tapers to a maximum width (thickness) of $20\text{ mm}$ . Assuming an ideal mechanical system with zero friction, if the mason applies a downward effort force of $450\text{ N}$ to the head of the wedge, what is the maximum lateral outward force, in Newtons, that the wedge exerts on the stone?

Show answer & explanation

Answer: 3600

Answer

3600

The correct lateral outward force is 3600 N. This is found by first determining the ideal mechanical advantage (IMA) of the wedge, which is its length divided by its thickness (160 / 20 = 8). Multiplying this IMA by the input effort force (450 N) gives the total ideal output force exerted on the stone.

Step-by-Step Solution

1

Calculate the Ideal Mechanical Advantage (IMA) of the wedge.

IMA = \frac{160\text{ mm}}{20\text{ mm}} = 8

The mechanical advantage of a wedge is determined by the ratio of its penetration length to its maximum thickness.

2

Calculate the output resistance force.

Output\ Force = 450\text{ N} \times 8 = 3600\text{ N}

In an ideal machine, the output force is equal to the input effort force multiplied by the mechanical advantage.

Key Concept

Mechanical Advantage of a Wedge

Question 217Question →

A twin-engine military cargo aircraft with a total mass of $80,000 \text{ kg}$ is accelerating down a runway for takeoff. Each of the two engines generates $140,000 \text{ N}$ of forward thrust. If the combined forces of rolling friction and aerodynamic drag opposing the aircraft's motion amount to $80,000 \text{ N}$ , what is the forward acceleration of the aircraft in $\text{m/s}^2$ ?

Show answer & explanation

Answer: 2.5

Answer

The forward acceleration of the aircraft is 2.5 m/s².

To accurately determine the acceleration, one must first find the net force acting on the aircraft. The total forward thrust is the sum of both engines (

140,000 \text{ N} \times 2 = 280,000 \text{ N}

). Subtracting the opposing forces of friction and drag (

80,000 \text{ N}

) yields a net forward force of

200,000 \text{ N}

. Dividing this net force by the aircraft's mass of

80,000 \text{ kg}

gives a forward acceleration of

2.5 \text{ m/s}^2

.

Step-by-Step Solution

1

Calculate the total forward thrust applied to the aircraft.

2 \times 140,000 \text{ N} = 280,000 \text{ N}

The aircraft has two engines, and the problem provides the thrust generated by each individual engine.

2

Calculate the net force (

F_{net}

) acting on the aircraft.

F_{net} = 280,000 \text{ N} - 80,000 \text{ N} = 200,000 \text{ N}

Net force is the vector sum of all forces. The opposing forces (drag and friction) must be subtracted from the forward thrust.

3

Apply Newton's Second Law to calculate the acceleration.

a = \frac{200,000 \text{ N}}{80,000 \text{ kg}} = 2.5 \text{ m/s}^2

According to Newton's Second Law, acceleration equals the net force divided by the mass (

a = \frac{F_{net}}{m}

).

Key Concept

Newton's Second Law of Motion (

F_{net} = ma

)

Question 218Question →

A sheet metal fabricator uses a custom compound lever tool to punch holes in aluminum panels. The tool consists of two connected levers:

1. The Handle (Lever A): A second-class lever with its fulcrum at the very top. The handle is $25\text{ inches}$ long overall. A vertical connecting rod is attached to this handle $5\text{ inches}$ away from the fulcrum.
2. The Punch Arm (Lever B): A first-class lever positioned horizontally. The connecting rod pulls down on one end, which is $8\text{ inches}$ away from its fulcrum. The punching die is located on the opposite side of the fulcrum, $2\text{ inches}$ away.

If the fabricator applies a perpendicular force of $24\text{ lbs}$ to the extreme end of the handle, what is the total downward force, in pounds, exerted by the punch die onto the aluminum panel?

Show answer & explanation

Answer: 480

Answer

The correct answer is

480\text{ lbs}

.

In a compound lever system, the mechanical advantage of each lever multiplies sequentially to produce the final force. The handle (Lever A) provides a mechanical advantage of

25/5 = 5

. Therefore, the

24\text{ lbs}

input is multiplied by

5

, becoming

120\text{ lbs}

of force on the connecting rod. The punch arm (Lever B) provides a mechanical advantage of

8/2 = 4

. The

120\text{ lbs}

input to Lever B is then multiplied by

4

, resulting in a total output force of

480\text{ lbs}

at the punching die.

Step-by-Step Solution

1

Calculate the input torque generated by the fabricator on Lever A.

24\text{ lbs} \times 25\text{ inches} = 600\text{ in-lbs}

Torque is determined by multiplying the applied force by the perpendicular distance from the fulcrum.

2

Determine the force exerted on the connecting rod by Lever A.

600\text{ in-lbs} / 5\text{ inches} = 120\text{ lbs}

In equilibrium, the input torque equals the output torque. The rod is attached

5\text{ inches}

from the fulcrum.

3

Calculate the input torque on Lever B caused by the connecting rod.

120\text{ lbs} \times 8\text{ inches} = 960\text{ in-lbs}

The connecting rod applies

120\text{ lbs}

of force at a distance of

8\text{ inches}

from Lever B's fulcrum.

4

Calculate the final output force at the punching die.

960\text{ in-lbs} / 2\text{ inches} = 480\text{ lbs}

The punching die is

2\text{ inches}

from the fulcrum, balancing the

960\text{ in-lbs}

of torque on the opposite side.

Key Concept

Compound Lever Systems and Torque Balance

Estimated Time:1m 30s

Question 219Question →

A small water pump uses a direct gear drive system. The driving gear, attached to an electric motor, has $20$ teeth and spins at $300\text{ rpm}$ . It meshes directly with a driven gear on the pump shaft that has $60$ teeth. What is the rotational speed of the driven gear in rpm?

Show answer & explanation

Answer: 100

Answer

100 rpm

Because the driven gear has three times as many teeth as the driving gear (

60

compared to

20

), it requires three times the effort to turn. Therefore, it will rotate at exactly one-third the speed of the driving gear. Dividing the driving speed of

300\text{ rpm}

by

3

yields

100\text{ rpm}

.

Step-by-Step Solution

1

Recall the fundamental gear relationship formula.

S_1 \times T_1 = S_2 \times T_2

The product of the speed and number of teeth of the driving gear equals the product of the speed and number of teeth of the driven gear.

2

Substitute the given values into the formula.

300 \times 20 = S_2 \times 60

We are given the driving gear speed (

300\text{ rpm}

) and teeth (

20

), along with the driven gear teeth (

60

).

3

Solve for the driven gear speed (

S_2

).

S_2 = \frac{6000}{60} = 100\text{ rpm}

Isolating

S_2

provides the final rotational speed of the driven gear.

Key Concept

The speed of a driven gear is inversely proportional to its number of teeth compared to the driving gear.

Question 220Question →

A small metal anchor weighs $120 \text{ N}$ in the air. When the anchor is completely submerged in water, the tension on its lifting cable is measured to be $85 \text{ N}$ . What is the magnitude of the buoyant force acting on the submerged anchor?

Show answer & explanation

Answer: 35

Answer

35

According to Archimedes' principle, the upward buoyant force exerted on a submerged object reduces its apparent weight. The buoyant force is equal to the difference between its true weight in the air and its apparent weight in the fluid. Subtracting the apparent weight (85 N) from the true weight (120 N) gives a buoyant force of 35 N.

Step-by-Step Solution

1

Identify the actual weight and the apparent weight.

Actual weight = 120 N, Apparent weight = 85 N

These values represent the downward force of gravity and the net downward force when submerged, respectively.

2

Calculate the buoyant force.

120 N - 85 N = 35 N

The buoyant force is the upward force that accounts for the reduction in weight. It is found by subtracting the apparent weight from the actual weight.

Key Concept

Apparent Weight and Buoyant Force

Answer

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