Levers and Mechanical Advantage: Lesson Test

Answer: D

Correct answer: The fixed point the bar turns around. Every lever has a fulcrum, an effort and a load. The fulcrum is the fixed pivot point, usually drawn as a triangle or hinge, and the bar rotates around it.

• A: That is the effort, the force you supply, not the pivot.
• B: That is the load, the resistance the lever works against.
• C: A distance is not a part of the lever; the fulcrum is a physical point on which the bar turns.
• D (correct): The fulcrum is the pivot, the one point of the bar that stays put while the rest rotates about it.

Answer: B

Correct answer: Second class. Ask what is in the middle. Fulcrum in the middle means first class, load in the middle means second class, and effort in the middle means third class. Here the load is in the middle, so the lever is second class.

• A: A first-class lever has the fulcrum in the middle, not the load.
• B (correct): With the load in the middle (fulcrum, load, effort), the lever is second class.
• C: A third-class lever has the effort in the middle, not the load.
• D: A compound lever links two or more levers in stages; a single bar with the load in the middle is simply second class.

Answer: B

Correct answer: 0 N·m. Torque is force times the perpendicular distance from the pivot to the line of the force. A push aimed straight at the bolt has a line of action through the pivot, so the moment arm is 0 and the torque is

N·m, however hard you push.

• A: This multiplies the force by the handle length, which only works when the push is at a right angle to the handle.
• B (correct): The line of the push passes through the bolt, so the moment arm is zero and so is the torque.
• C: This treats the force itself as the torque and ignores the moment arm entirely.
• D: This divides the force by the length; torque is force times moment arm, and here that arm is zero.

Answer: A

Correct answer: 4. Actual mechanical advantage uses measured forces.

The newtons cancel, so the answer has no units.

• A (correct): Actual mechanical advantage is output force over input force: 720 divided by 180 is 4.
• B: This divides the effort by the load, which flips the ratio.
• C: This adds the forces; mechanical advantage compares them by division.
• D: This subtracts the forces; mechanical advantage is a ratio, not a difference.

Answer: D

Correct answer: 8. For an ideal lever, IMA equals the effort arm divided by the load arm:

The effort needed would be one eighth of the load.

• A: This multiplies the arm lengths; the ideal advantage is their ratio.
• B: This adds the two arm lengths instead of dividing them.
• C: This divides the load arm by the effort arm, which flips the ratio.
• D (correct): Ideal mechanical advantage is effort arm over load arm: 2.4 divided by 0.30 is 8.

Answer: D

Correct answer: Tweezers picking up a splinter. In a third-class lever the effort is applied between the fulcrum and the load. Tweezers are hinged at one end, pressed in the middle and grip at the tips, which is fulcrum, effort, load.

• A: The pivot sits between the two riders, so the fulcrum is in the middle: first class.
• B: The fulcrum sits between your hands and the crate, so a crowbar used this way is first class.
• C: The cargo sits between the wheel axle and your hands, so the load is in the middle: second class.
• D (correct): You squeeze tweezers between the hinge and the tips, so the effort is in the middle: third class.

Answer: B

Correct answer: The same amount of work. Work is force times distance. A lever with an advantage of 5 multiplies the force by 5 but the load moves only one fifth as far as your hands, so in the ideal case the work done on the load equals the work you supply.

• A: This assumes the lever multiplies energy; it multiplies force only, at the cost of distance.
• B (correct): In an ideal lever the force goes up five times while the load moves one fifth as far, so work out equals work in.
• C: This applies the distance reduction but forgets that the force grew by the same factor.
• D: This multiplies both force and distance gains; a lever cannot create work at all.

Answer: D

Correct answer: A first-class lever only. When the fulcrum is between the effort and the load, the bar acts like a seesaw: one end goes down while the other goes up. That is the first-class arrangement. Second- and third-class levers move the load in the same direction as the effort.

• A: In a third-class lever the effort and the load move the same way, so direction is kept.
• B: Second- and third-class levers keep the direction the same, so not every class reverses motion.
• C: In a second-class lever, lifting the effort end lifts the load as well, so direction is kept.
• D (correct): Only a lever with the fulcrum in the middle makes the two ends move in opposite directions.

Answer: C

Correct answer: 75%. Efficiency compares what you get with the ideal value.

• A: This divides IMA by AMA; real efficiency can never exceed 100 percent.
• B: This is the fraction lost to friction, not the efficiency.
• C (correct): Efficiency is AMA over IMA: 6 divided by 8 is 0.75, or 75 percent.
• D: This multiplies the two advantages instead of dividing them.

Answer: C

Correct answer: 20 N·m. Convert the distance to meters first: 40 cm = 0.40 m. Then

N·m.

• A: This uses 40 directly without converting centimeters to meters, making it 100 times too large.
• B: This converts 40 cm to 0.04 m; there are 100 cm in a meter, so 40 cm is 0.40 m.
• C (correct): Converting 40 cm to 0.40 m and multiplying by 50 N gives 20 N·m.
• D: This divides the force by the distance instead of multiplying them.

Answer: A

Correct answer: 1.2 m. Balance the moments about the pivot.

The heavier rider sits closer to the center, as expected.

• A (correct): Setting moments equal: 300 times 1.8 is 540 N·m, and 540 divided by 450 is 1.2 m.
• B: This flips the force ratio; the heavier person must sit closer to the pivot, not farther.
• C: Sitting at the same distance would only balance if the two weights were equal.
• D: This is just 300 divided by 450 and leaves out the child's 1.8 m distance.

Answer: A

Correct answer: 150 N. The axle is the fulcrum. Set the moments equal:

one quarter of the cargo weight.

• A (correct): The cargo moment is 600 times 0.40, or 240 N·m, and 240 divided by 1.6 is 150 N.
• B: This is the cargo's moment in N·m, not the force at the handles; it still needs to be divided by 1.6 m.
• C: This flips the arm ratio; with the hands farther from the axle the effort must be smaller than the load.
• D: This subtracts a quarter of the load from the load; the hand force itself is one quarter of the load.

Answer: B

Correct answer: 450 N. Set the moments equal and solve for the load.

• A: This flips the arms; a load close to the fulcrum can be much heavier than the effort, not lighter.
• B (correct): The effort moment is 90 times 1.5, or 135 N·m, and 135 divided by 0.30 is 450 N.
• C: This is the effort moment in N·m; it still has to be divided by the 0.30 m load arm.
• D: This multiplies the effort by the load arm instead of the effort arm.

Answer: A

Correct answer: 0.20 m. Every point on the bar turns through the same angle, so travel is proportional to distance from the fulcrum:

The force gain of 4 costs a travel loss of 4.

• A (correct): The load end moves 0.40/1.6, or one quarter, as far as the handle: 0.80 times 0.25 is 0.20 m.
• B: The load would match the handle travel only if the two arms were equal.
• C: This multiplies by the arm ratio the wrong way; the short load arm moves less than the handle, not more.
• D: This multiplies the handle travel by the load arm length rather than by the ratio of the arms.

Answer: A

Correct answer: 100 N. First convert mass to weight:

The load moment is

N·m. The effort is

• A (correct): The box weighs 250 N; its moment is 150 N·m, and 150 divided by 1.5 is 100 N.
• B: This flips the arm ratio; a longer effort arm should make the effort smaller than the load.
• C: This is the load's moment in N·m, which still has to be divided by the 1.5 m effort arm.
• D: This uses the 25 kg mass as if it were a force and skips converting it to a weight in newtons.

Answer: B

Correct answer: 40 N. In a compound lever the stage advantages multiply:

The ideal effort is the output force divided by the overall advantage:

• A: This uses only the first stage's advantage of 4 and ignores the second stage.
• B (correct): The overall advantage is 4 times 6, or 24, and 960 divided by 24 is 40 N.
• C: This adds the stage advantages to get 10; compound stages multiply.
• D: This uses only the second stage's advantage of 6 and ignores the first stage.

Answer: A

Correct answer: 600 N. The IMA is

so the effort is

A first-class lever only gains force when the effort arm is longer than the load arm.

• A (correct): The effort arm is half the load arm, so the effort must be twice the load: 300 times 0.80 divided by 0.40 is 600 N.
• B: This assumes a first-class lever always multiplies force; here the effort arm is the shorter one, so the ratio is flipped.
• C: The effort would equal the load only if the two arms were the same length.
• D: This multiplies the load by the effort arm only, which gives a moment-like number rather than a balancing force.

Answer: C

Correct answer: 20 N·m. Torque with an angled force is

Convert 80 cm to 0.80 m, then

N·m.

• A: This uses 80 directly instead of 0.80 m, giving an answer 100 times too large.
• B: This treats the pull as perpendicular and ignores the 30-degree angle, which reduces the turning effect by half.
• C (correct): Converting 80 cm to 0.80 m and multiplying 50 by 0.80 by sin 30 degrees (0.5) gives 20 N·m.
• D: This uses cos 30 degrees instead of sin 30 degrees, but the angle is measured between the force and the bar, so the sine applies.

Answer: D

Correct answer: 600 N. A balanced bar must satisfy

as well as

Both children push down, so the pivot must push up with

• A: This averages the weights, but the force balance requires the pivot reaction to equal their sum.
• B: This subtracts the two weights, but both riders push down, so their weights add rather than cancel.
• C: This counts only the heavier rider; the pivot must also support the lighter child.
• D (correct): For the forces to cancel, the pivot pushes up with the total downward load, 240 N plus 360 N.

Answer: C

Correct answer: First class. Find the fulcrum first: the head resting on the board. The nail is on one side of it and the hand on the other, so the fulcrum is in the middle, which makes this a first-class lever.

• A: A second-class lever has the load in the middle, but here the nail is at one end of the arrangement.
• B: A compound lever links two or more levers in series, while the hammer here is a single rigid bar on one pivot.
• C (correct): The head touching the board is the fulcrum, and it sits between the nail (load) and the hand (effort).
• D: A third-class lever has the effort in the middle, but the hand pulls at the far end of the handle.

Answer: B

Correct answer: 4 N·m clockwise. Left moment:

N·m counterclockwise. Right moment:

N·m clockwise. They oppose, so subtract:

N·m in the clockwise direction of the larger moment.

• A: The larger moment comes from the right-hand weight, which turns a level beam clockwise, not counterclockwise.
• B (correct): The right weight gives 24 N·m clockwise and the left gives 20 N·m counterclockwise; the difference is 4 N·m clockwise.
• C: This adds the two moments, but they turn the beam in opposite directions and partly cancel.
• D: This adds opposing moments and also picks the direction of the smaller one.

Answer: D

Correct answer: 80%. Input work:

Output work:

• A: This is the fraction of input work lost, not the fraction delivered to the crate.
• B: This is the force ratio, 320 over 80, expressed as a percent; efficiency compares work, not force alone.
• C: This divides input work by output work; no real machine can exceed 100 percent.
• D (correct): Work out is 320 times 0.15, or 48 J; work in is 80 times 0.75, or 60 J; 48 divided by 60 is 80 percent.

Answer: A

Correct answer: 24 W. Power is force times speed:

An ideal lever multiplies force by 4 but cuts the load's speed to one quarter, so the load receives the same 24 W.

• A (correct): Input power is 60 times 0.40, or 24 W, and an ideal lever passes all of it to the load.
• B: This multiplies the power by the mechanical advantage, which would create energy.
• C: This divides the power by the mechanical advantage; the load gets more force but proportionally less speed, so power is unchanged.
• D: This divides the force by the speed; power is force times speed.

Answer: B

Correct answer: 62.5 N. Add the moments on the right:

N·m. The left force must supply the same moment:

• A: This is the total moment in N·m; it still has to be divided by the 0.80 m arm.
• B (correct): The right-side moments add to 20 + 30 = 50 N·m, and 50 divided by 0.80 is 62.5 N.
• C: This just adds the two loads and ignores their different distances from the pivot.
• D: This multiplies the total moment by the arm instead of dividing by it.

Answer: D

Correct answer: 360 N. First use moments about the axle:

Then balance forces: the wheel reaction plus 180 N must equal 540 N, so the wheel pushes up with 360 N.

• A: This is the lifting force at the handles, not the support force at the wheel.
• B: This ignores the share of the weight carried by the worker's hands.
• C: This adds the hand force to the load; the hands push up, so they reduce what the wheel must carry.
• D (correct): The hands lift 540 times 0.45 divided by 1.35, or 180 N, so the wheel supplies the rest: 540 minus 180 is 360 N.

Answer: A

Correct answer: 25 N. Treat the plank's weight as acting at its center, 1.2 m from the short end, which is

m beyond the sawhorse on the long side. Its moment is

N·m. The can sits 0.80 m from the sawhorse on the other side, so it must weigh

• A (correct): The plank's center is 1.2 m from each end, 0.40 m past the sawhorse; its 20 N·m moment divided by the can's 0.80 m arm is 25 N.
• B: This simply copies the plank's weight; the plank's weight acts only 0.40 m from the pivot, a shorter arm than the can's 0.80 m.
• C: This ignores the plank's own weight, which acts at its center and tips the plank toward the longer side.
• D: This flips the arms, multiplying by 0.80 and dividing by 0.40.

Answer: C

Correct answer: Deep in the jaws, close to the pivot. Scissors are paired first-class levers. Your hand moment is fixed, so the cutting force equals that moment divided by the load arm. The load arm is shortest right next to the pivot, which gives the greatest cutting force there.

• A: The cutting point changes the load arm, so the cutting force depends on where the material sits.
• B: Midway is better than the tips, but moving closer to the pivot shortens the load arm further.
• C (correct): Near the pivot the load arm is shortest, so the same hand moment produces the largest cutting force.
• D: At the tips the load arm is longest, so the cutting force is smallest there.

Answer: D

Correct answer: 85%. IMA =

and AMA =

Efficiency = AMA ÷ IMA =

or 85%.

• A: This divides the effort by the load, which inverts the actual mechanical advantage.
• B: This divides the ideal advantage by the actual advantage; a real lever can never exceed 100 percent efficiency.
• C: This writes the actual mechanical advantage of 4.25 as a percent without comparing it with the ideal value of 5.
• D (correct): The actual advantage of 4.25 divided by the ideal advantage of 5 gives 0.85, or 85 percent.

Answer: B

Correct answer: 200 N. The effort arm runs from the fulcrum to the hands, not the whole bar. The handle side is

and halfway along it is 0.525 m. Then

• A: This uses the full 1.05 m handle as the effort arm, but the worker grips only halfway along it.
• B (correct): The handle side is 1.05 m long, so the grip is 0.525 m from the fulcrum; 700 times 0.15 divided by 0.525 is 200 N.
• C: This uses the total 1.2 m bar length as the effort arm instead of the distance from the fulcrum to the hands.
• D: This takes half the load because the grip is at the halfway point, instead of using a moment balance.

Answer: A

Correct answer: 0.30 m. Combined position = total moment ÷ total weight.

• A (correct): The weighted average is (80 × 0.20 + 20 × 0.70) ÷ 100 = 30 ÷ 100 = 0.30 m, closer to the heavier load.
• B: This adds the two distances together instead of averaging them by weight.
• C: This pairs each weight with the other load's distance, which places the combined load near the lighter weight.
• D: This is the simple average of the two distances, which ignores that the 80 N load is four times heavier.

Answer: A

Correct answer: 240 N. The elbow is the fulcrum, the biceps is the effort and the dumbbell is the load, so the forearm is third class. The weight is

Balance moments (units cancel in the ratio):

• A (correct): The dumbbell weighs 30 N, and the muscle arm is one eighth of the load arm, so the muscle must pull 30 times 8, or 240 N.
• B: This is the dumbbell's weight; the muscle acts much closer to the elbow, so it must pull harder.
• C: This uses the 3 kg mass as if it were a force, skipping the conversion to 30 N.
• D: This flips the arm ratio; in a third-class lever the effort must exceed the load, not be smaller.

Answer: C

Correct answer: 200 N. The bar's center is 2.0 m from the left end, which is 1.0 m right of the pivot, giving

N·m. The 40 N weight is 3.0 m right of the pivot, giving 120 N·m. The left weight acts 1.0 m from the pivot, so it must be

• A: This balances only the 40 N weight at 3.0 m and ignores the moment from the bar's own weight.
• B: This measures the bar's center from the left end (2.0 m) instead of from the pivot (1.0 m).
• C (correct): Clockwise moments total 80 × 1.0 + 40 × 3.0 = 200 N·m, so a weight 1.0 m left of the pivot must be 200 N.
• D: This balances only the bar's own weight and leaves out the 40 N weight on the right end.

Answer: D

Correct answer: 150 N. Add the moments on the right:

N·m. The rope's moment is

Setting 0.6T = 90 gives T = 150 N.

• A: This uses cos 30 degrees, but the turning part of the pull depends on the sine of the angle between rope and bar.
• B: This places the total 150 N at the plain average distance of 0.65 m instead of adding the two separate moments.
• C: This assumes the rope pulls at a right angle and forgets that only half its force turns the bar at 30 degrees.
• D (correct): The loads give 36 + 54 = 90 N·m, and the rope supplies T × 1.2 × 0.5, so T = 90 ÷ 0.6 = 150 N.

Answer: C

Correct answer: 324 N. Input work =

Output work =

Load =

• A: This is the load for a perfect lever; it ignores the 10 percent of input work lost to friction.
• B: This flips the distance ratio, as though the load moved farther than the handle.
• C (correct): Input work is 54 J, 90 percent of that is 48.6 J, and 48.6 J ÷ 0.15 m = 324 N.
• D: This divides by the efficiency instead of multiplying, which would make the output work larger than the input work.

Answer: B

Correct answer: 150 N through 0.30 m. IMA

Effort

Handle travel

Check:

J in and

J out.

• A: The force is right, but the handle must move 6 times farther than the load, not the same distance.
• B (correct): The IMA is 6, so the effort is 900 divided by 6, and the handle must travel 6 times the 0.05 m rise.
• C: This divides the travel by 6; force gain always costs extra handle travel, so the handle moves more than the load.
• D: This multiplies the load by 6 instead of dividing; a long effort arm reduces the effort.

Answer: C

Correct answer: Arms 1.0 m and 0.25 m; a 50 N effort lifts 220 N. For each record, compare the ideal advantage (effort arm ÷ load arm) with the actual advantage (load ÷ effort). Actual can be at most equal to ideal. Only the 1.0 m and 0.25 m lever shows

above its ideal value of 4, so that record must be wrong.

• A: The ideal advantage is 0.5 and the actual advantage is 0.45; a force loss is normal for a lever that trades force for distance.
• B: The ideal advantage is 4 and the actual advantage is 3.6, a realistic result with some friction.
• C (correct): The ideal advantage is 4 but the record shows an actual advantage of 4.4, which no real lever can achieve.
• D: The ideal advantage is 3 and the actual advantage is 2.75, slightly below ideal as expected.

Answer: C

Correct answer: It increases by 50 N. Before: arms are 1.6 m and 0.40 m, so effort =

After: arms are 1.5 m and 0.50 m, so effort =

The effort rises by 50 N.

• A: This lengthens the load arm to 0.50 m but forgets that the effort arm shrinks to 1.5 m at the same time.
• B: The load's weight does not change, but the force needed depends on the arm lengths, which did change.
• C (correct): The effort goes from 600 × 0.40 ÷ 1.6 = 150 N to 600 × 0.50 ÷ 1.5 = 200 N, a 50 N increase.
• D: This gets the direction backward; moving the fulcrum toward the effort lengthens the load arm and shortens the effort arm.

Answer: D

Correct answer: 150 N·m. A weight acts straight down, so its moment arm is the horizontal distance from the pivot: 1.5 m × cos 60° = 0.75 m. The moment is

N·m.

• A: This uses sin 60 degrees, which gives the vertical height of the point rather than its horizontal distance from the pivot.
• B: This uses 150 cm without converting to 1.5 m, giving a value 100 times too large.
• C: This uses the full distance along the bar, but for a hanging weight the moment arm is the horizontal distance to the pivot.
• D (correct): The horizontal moment arm is 1.5 m × cos 60 degrees = 0.75 m, and 200 N × 0.75 m = 150 N·m.

Answer: B

Correct answer: 187.5 N. IMA

so the ideal effort is

The actual mechanical advantage is

so the real effort is

• A: This divides the load itself by 0.80 and forgets the lever's mechanical advantage of 4.
• B (correct): The ideal effort is 600 divided by 4, or 150 N; with 80 percent efficiency the real effort is 150 divided by 0.80, or 187.5 N.
• C: This is the ideal effort for a frictionless lever and leaves out the 80 percent efficiency.
• D: This multiplies the ideal effort by 0.80; friction makes the real effort larger, not smaller.

Answer: C

Correct answer: 1.5 m from the pivot. Counterclockwise moment:

N·m. Clockwise moments must also total 600 N·m:

• A: This ignores the 300 N child already on the right, making the 200 N child balance all 600 N·m alone.
• B: This subtracts the weights first (400 minus 300) instead of subtracting the moments.
• C (correct): The left moment is 600 N·m; the 300 N child supplies 300 N·m, so the 200 N child must supply 300 N·m at 1.5 m.
• D: This adds the 300 N·m on the right to the left side's moment instead of subtracting it.

Answer: A

Correct answer: Third class; it gives up force to gain speed and reach at the blade. Read the order of the three points from the pivot: top hand (fulcrum), lower hand (effort), blade (load). With the effort in the middle, this is a third-class lever.

Its effort arm is shorter than its load arm, so the ideal mechanical advantage is below 1: the landscaper pushes harder than the load but gains speed and reach at the blade.

• A (correct): The effort sits between the fulcrum and the load, so the effort arm is shorter and the lever trades force for motion.
• B: The fulcrum is at the top end, not between the effort and the load, so this is not a first-class arrangement.
• C: In a second-class lever the load is in the middle; here the effort hand is in the middle.
• D: The class is right, but a third-class lever always has an effort arm shorter than the load arm, so it reduces force.

Answer: C

Correct answer: 100 N. In an ideal lever, work in equals work out: effort × effort distance = load × load distance.

The handle moves 12 times as far as the block, so the effort is one-twelfth of the load.

• A: This multiplies the load by the distance ratio instead of dividing; a long handle travel must mean a small effort.
• B: This assumes the lever gives no force gain, but the handle moves 12 times farther than the load.
• C (correct): Equal work in and out: 1,200 × 4 ÷ 48 = 100 N.
• D: This divides the load by 48 alone and ignores the 4 cm the block rises.

Answer: D

Correct answer: 1. Ideal mechanical advantage is effort arm ÷ load arm. With the pivot centered, the two arms are equal, so the ratio is 1. The balance gives no force gain and no distance gain; it only reverses the direction of motion.

• A: A ratio of two equal, nonzero lengths cannot be zero; the balance still passes force from one side to the other.
• B: This is the value for an effort arm half as long as the load arm, not for equal arms.
• C: This would need an effort arm twice as long as the load arm; centered arms are equal.
• D (correct): Equal arms make the ratio of effort arm to load arm exactly 1.

Answer: D

Correct answer: The load between the fulcrum and the effort. When the load sits between the fulcrum and the effort, the effort is always the farthest point from the fulcrum. Its arm is therefore always longer than the load arm, so the ideal mechanical advantage is always above 1. That is a second-class lever.

• A: A first-class lever gains force only if the effort side is longer; with the fulcrum near the effort, it needs more effort than the load.
• B: This third-class order always makes the effort arm shorter, so the effort is always larger than the load.
• C: Equal arms give a mechanical advantage of exactly 1, so the effort equals the load rather than being less.
• D (correct): This second-class order always gives a longer effort arm, so the ideal effort is always less than the load.

Answer: B

Correct answer: 100 N. Take moments about the pivot. The bag makes

N·m on the left. The plank's center is at 2.5 m, which is 1.0 m right of the pivot, so its weight makes

N·m on the right and already helps.

The right end is

m from the pivot:

so F = 100 N.

• A: This ignores the plank's weight; its center lies on the right side of the pivot and helps balance the bag.
• B (correct): The plank's own weight supplies 100 N·m on the right, leaving 350 N·m for the end force: 350 ÷ 3.5 = 100 N.
• C: This adds the plank's moment to the bag's side, but the plank's center is right of the pivot, so it works against the bag.
• D: This divides by the full 5.0 m plank length instead of the 3.5 m distance from the pivot to the right end.

Answer: D

Correct answer: 30 cm. Every point on the rigid handle turns through the same angle, so each point moves in proportion to its distance from the pivot.

Bristle travel = 10 cm ×

This third-class lever trades force for a larger, faster motion at the far end.

• A: This flips the ratio; the bristles are farther from the pivot, so they move more than the hand, not less.
• B: Points at different distances from the pivot move different amounts; only the angle of rotation is the same.
• C: This measures the bristle distance from the lower hand (0.80 m) instead of from the pivot, giving a ratio of 2.
• D (correct): The bristles are three times as far from the pivot as the lower hand, so they move three times as far.

Answer: B

Correct answer: 0 J. Work is force times the distance moved along the force. Torque only measures a twisting tendency about a pivot. Because the nut is not turning, nothing moves through any distance, so the work done is 0 J even though 110 N·m of torque is applied.

• A: This treats torque in N·m as if it were energy in joules; the units look alike, but the quantities differ.
• B (correct): No part of the nut moves, so no work is done, no matter how large the torque.
• C: This halves the torque, but with no motion, any nonzero amount of work is impossible.
• D: Doubling the torque has no meaning here; work requires movement, and the nut is not turning.

Answer: C

Correct answer: 75 N, moving the handle 48 cm. Load =

IMA =

so the ideal effort is

Friction raises the effort:

Handle travel is set by geometry, not friction: 6 cm × 8 = 48 cm.

• A: This is the ideal effort; a 75 percent efficient lever needs more effort than the ideal value.
• B: The effort is right, but efficiency does not change the travel; the arm ratio alone sets it at 48 cm.
• C (correct): Ideal effort 56.25 N divided by 0.75 gives 75 N, and the 8:1 arm ratio makes the handle travel 8 × 6 = 48 cm.
• D: This multiplies by the efficiency instead of dividing and also shrinks the travel; losses raise the effort.

Answer: C

Correct answer: 130 N. Put both arms in the same unit: 25 cm = 0.25 m. The IMA is

Ideal effort =

• A: This uses 25 for the load arm and 1.5 for the effort arm without converting, so the units do not match.
• B: This converts the wrong way, treating 1.5 m as 150 while using 0.25 m; both lengths must be in one unit.
• C (correct): With both arms in meters, the IMA is 6, and 780 ÷ 6 = 130 N.
• D: This multiplies the load by the IMA instead of dividing; a longer effort arm means less effort.

Answer: A

Correct answer: It becomes half as large. The load's moment stays the same, so effort × effort arm must stay the same. Doubling the effort arm therefore cuts the needed effort in half; the hands simply move twice as far.

• A (correct): The same moment supplied at twice the distance needs half the force.
• B: This reverses the relationship; effort is inversely proportional to the effort arm.
• C: Moving the hands farther from the fulcrum increases leverage, so the needed effort drops.
• D: Effort varies directly with one over the arm length, not with its square, so doubling the arm halves it.