Pulleys and Block-and-Tackle Systems: Lesson Test

Answer: A

Correct answer: 1. A pulley fastened to a beam does not move with the load, so it only changes the direction of the pull. Only one rope segment holds the backdrop, so the ideal effort equals the load and the IMA is 1.

• A (correct): A fixed pulley only redirects the rope, so the ideal effort equals the load and the advantage is 1.
• B: An advantage of 2 needs a pulley that rides with the load so two segments pull up on it; this wheel never moves.
• C: A value below 1 would mean the pulley makes the job harder, but an ideal fixed pulley neither helps nor hurts the force.
• D: Four would require four rope segments holding up a moving block, and here only one segment holds the backdrop.

Answer: D

Correct answer: 180 N. The movable pulley is held up by two rope segments, so the IMA is 2. Effort = 360

• A: Multiplying the load by 2 instead of dividing gives a force larger than the load, which a pulley that helps cannot require.
• B: This treats the pulley as fixed; a movable pulley has two segments sharing the load.
• C: Dividing by 3 assumes a third supporting segment that this simple system does not have.
• D (correct): Two segments pull up on the movable pulley, so each carries half of 360 N.

Answer: D

Correct answer: Reeving. The path a rope takes as it is threaded between the fixed and moving blocks is called reeving. The reeving decides how many segments support the moving block and so sets the mechanical advantage.

• A: An idler is a belt-drive pulley that guides or tensions a belt, unrelated to threading a tackle.
• B: Pitch describes the effective diameter where a belt rides in a pulley groove, not rope threading.
• C: Creep is the small speed loss as a friction belt stretches and relaxes, not a way of threading rope.
• D (correct): Reeving is the term for how the rope is passed through the sheaves of the blocks.

Answer: B

Correct answer: Clockwise. With an open belt on parallel shafts, the belt wraps both pulleys on the same side, so both turn in the same direction. The larger driven pulley turns clockwise, only more slowly.

• A: Pulley size changes speed, never direction, in an open belt drive.
• B (correct): In an open belt drive both pulleys turn the same way, so the driven pulley also turns clockwise.
• C: A steadily driven belt carries steady rotation; nothing in an open drive makes the pulley oscillate.
• D: Reversal happens only when the belt is crossed into a figure eight; an open belt does not reverse rotation.

Answer: A

Correct answer: Friction and moving parts use up part of the effort. Actual mechanical advantage uses the effort really applied. Friction in the sheaves and rope, plus the weight of moving parts, raise that effort, so the AMA is always below the IMA in a real hoist.

• A (correct): Some of the applied pull goes into overcoming bearing friction, rope bending and the weight of moving hardware, so the measured effort is larger than ideal.
• B: A lighter load does not explain why each newton of effort lifts less than the ideal amount.
• C: Both mechanical advantages are pure numbers without units, so units cannot explain the gap.
• D: The ideal value comes from the segment count; a lower measured ratio comes from losses, not a different count.

Answer: A

Correct answer: 500 rpm. With no slip,

• A (correct): Using D1N1 = D2N2, 1,500 times 100 / 300 = 500 rpm.
• B: Sharing a belt means the rim speeds match, not the rpm; the larger pulley turns fewer times.
• C: This divides by the diameter ratio squared (9), but speed changes by the simple diameter ratio of 3.
• D: This inverts the ratio; a larger driven pulley must turn slower, not faster.

Answer: C

Correct answer: 4. IMA equals effort distance divided by load distance: 3.0 m / 0.75 m = 4. Because both distances are in meters, the units cancel and the IMA is a pure number.

• A: Subtracting 0.75 from 3.0 does not give a ratio; IMA is found by dividing the distances.
• B: This divides the load distance by the rope distance, which inverts the ratio.
• C (correct): Dividing effort distance by load distance, 3.0 m / 0.75 m, gives a unitless IMA of 4.
• D: The meters cancel when one distance is divided by the other, so IMA has no unit.

Answer: A

Correct answer: Shut off the press and lock it out. Belt and pulley work must start with the machine turned off and locked out so it cannot restart unexpectedly. Only then is the guard removed, and it must be put back before the machine is restarted.

• A (correct): Shutting down and locking out keeps the machine from starting while hands are near the belt and pulleys.
• B: Reaching near a moving belt exposes the hands to the in-running nip point, so the machine must not be running.
• C: Overtightening overloads bearings and stretches the belt, and it does nothing to make the adjustment safe.
• D: Gloves can be caught by a nip point and drag the hand in; they do not make work on a powered drive safe.

Answer: C

Correct answer: Where the belt runs onto a pulley. An in-running nip point forms where the belt moves onto a pulley, creating a closing gap. Because the belt moves too fast to react, these points must be guarded and nobody should reach near a running belt.

• A: The middle of a free belt span has no pulley surface to pinch against.
• B: The hub turns with the shaft but does not form the narrowing belt-to-pulley gap.
• C (correct): The narrowing gap where the belt meets the pulley can draw in fingers, gloves, hair or sleeves in a fraction of a second.
• D: At the outgoing side the gap opens up, so it is not the in-running pinch the term describes.

Answer: A

Correct answer: 220 N. In the ideal model the rope tension is the same in every segment, so

• A (correct): Each of the six segments carries an equal share, so effort = 1,320 / 6 = 220 N.
• B: This is the effort for a single fixed pulley; six supporting segments share the load.
• C: This divides by 5, missing one of the six supporting segments.
• D: This multiplies the load by the segment count instead of dividing.

Answer: B

Correct answer: 200 N. Count the rope segments pulling up on the moving block. Starting from the dead end on the lower block, the rope makes four runs between the blocks, and the hauling end also leaves the lower block going up, for five supporting segments. Effort = 1,000

• A: Dividing by 4 counts only the segments between the blocks and forgets that the hauling end, pulled upward from the moving block, also holds it up.
• B (correct): Five segments pull up on the moving block (four between the blocks plus the hauling end), and 1,000 N / 5 = 200 N.
• C: Dividing by 2 counts only the sheaves on one block instead of the rope segments holding up the moving block.
• D: This assumes the rope gives no advantage, but the moving block is held up by several rope segments.

Answer: A

Correct answer: 3.2. Actual mechanical advantage uses the real effort: AMA = 640 N / 200 N = 3.2. It is smaller than the IMA of 4 because friction adds to the needed pull.

• A (correct): AMA is load divided by measured effort: 640 N / 200 N = 3.2.
• B: Four is the ideal value from the segment count; the actual value must use the measured pull.
• C: This is the efficiency, AMA divided by IMA, not the actual mechanical advantage itself.
• D: This divides the effort by the load, which inverts the ratio.

Answer: B

Correct answer: 1.35 m. Rope travel equals the number of supporting segments times the lift. Rope pulled = 3 x 0.45 m = 1.35 m.

• A: Dividing the lift by 3 gives a rope pull shorter than the lift, which is impossible for a force-reducing tackle.
• B (correct): Each of the three segments shortens by 0.45 m, so 3 x 0.45 m = 1.35 m of rope comes out.
• C: Equal rope and load travel happens only with an IMA of 1, not with three supporting segments.
• D: Multiplying by 2 uses the wrong segment count; this tackle has three supporting segments.

Answer: D

Correct answer: 200 N. The lower block and hook move with the compressor, so total lifted weight = 570 N + 30 N = 600 N. Ideal effort = 600

• A: This adds the block weight after dividing, but the block is shared by the three segments like the load.
• B: Dividing 600 N by 2 uses the wrong segment count for a three-segment tackle.
• C: This leaves out the 30 N block and hook, which the rope must also lift.
• D (correct): The rope lifts 570 N + 30 N = 600 N, and 600 N / 3 = 200 N.

Answer: C

Correct answer: 90%. Output work = 720 N x 1.2 m = 864 J. Input work = 400 N x 2.4 m = 960 J.

Efficiency = 864 J / 960 J = 0.90, or 90 percent.

• A: This uses only the distance ratio 1.2 m / 2.4 m and ignores the forces.
• B: This divides effort by load, which is neither a work ratio nor an efficiency.
• C (correct): Output work 720 N x 1.2 m = 864 J divided by input work 400 N x 2.4 m = 960 J gives 90 percent.
• D: This divides input work by output work; efficiency can never be above 100 percent.

Answer: B

Correct answer: 75%. Six supporting segments give an IMA of 6. Efficiency = AMA / IMA =

or 75 percent.

• A: This divides IMA by AMA, inverting the ratio and giving an impossible efficiency above 100 percent.
• B (correct): With six segments the IMA is 6, and efficiency = 4.5 / 6 = 0.75, or 75 percent.
• C: This is the fraction lost, (6 - 4.5) / 6, not the fraction of work delivered.
• D: Subtracting 4.5 from 6 gives a difference, not an efficiency; the values must be divided.

Answer: D

Correct answer: 309 N. Total upward force =

x

Two segments share it, so each carries 618

• A: This is the total upward force, not the share carried by each of the two segments.
• B: This is the tension for holding the load still; it leaves out the extra force needed to accelerate it.
• C: This is only the extra accelerating force per segment and ignores the weight of the load.
• D (correct): The total force is 60 x (9.8 + 0.5) = 618 N, shared by two segments, so each carries 309 N.

Answer: A

Correct answer: 520 N. The rope tension is 250 N on both sides, and both segments pull down on the fixed pulley. Hook force = 250 N + 250 N + 20 N = 520 N.

• A (correct): Both rope segments pull down 250 N each, and the pulley adds 20 N: 250 + 250 + 20 = 520 N.
• B: This assumes the hook holds only the load, but the worker's downward pull also acts on the pulley.
• C: This counts both rope segments but forgets the weight of the fixed pulley itself.
• D: This counts only the load-side segment and the pulley, leaving out the downward pull of the effort segment.

Answer: D

Correct answer: 7.54 m/s.

• A: This uses pi D N but leaves the answer per minute, forgetting to divide by 60.
• B: This uses 2 pi times the diameter, doubling the distance per revolution.
• C: This divides by 60 but leaves out pi, so it is not the circumference per turn.
• D (correct): Belt speed is pi times 0.08 times 1,800 / 60, about 7.54 m/s.

Answer: D

Correct answer: 1,800 rpm.

The smaller driven pulley turns faster, as expected.

• A: The two pulleys share a rim speed, not an rpm, so the smaller one turns faster.
• B: This mixes the 50 mm radius of the driven pulley with the 250 mm diameter of the driver.
• C: This inverts the ratio; a smaller driven pulley must turn faster than the driver.
• D (correct): N2 = 720 times 250 / 100 = 1,800 rpm, faster because the driven pulley is smaller.

Answer: C

Correct answer: 16 N·m. The torque ratio is the diameter ratio, driven over driver.

The output turns at one quarter of the motor speed, so power is unchanged.

• A: This divides the torque by the ratio, which happens only when the driven pulley is smaller and speeds up.
• B: This multiplies by the square of the diameter ratio, but torque scales with the simple ratio of 4.
• C (correct): The larger pulley turns 300 / 75 = 4 times slower, so in an ideal drive the torque rises by the same factor: 4 times 4 = 16 N·m.
• D: A belt shares rim speed between pulleys, but torque changes inversely with rpm, so it does not stay the same.

Answer: B

Correct answer: 450 rpm counterclockwise. Speed:

The belt crosses between the pulleys, so the driven pulley turns counterclockwise, opposite to the driver.

• A: The speed is right, but a crossed belt reverses rotation.
• B (correct): 1,200 times 150 / 400 = 450 rpm, and the crossed belt makes the driven pulley turn opposite to the driver.
• C: This inverts the ratio and also ignores the reversal caused by the crossed belt.
• D: The direction is right, but the ratio is inverted; the larger pulley must turn slower.

Answer: C

Correct answer: 400 rpm. Belt speed is the same everywhere, so the idler cancels out.

• A: This inverts the driver-to-fan ratio, making the larger pulley turn faster.
• B: This is the speed of the small idler, not of the fan pulley.
• C (correct): The idler does not change the ratio, so fan speed is 1,200 times 100 / 300 = 400 rpm.
• D: The fan pulley is three times the driver's diameter, so it cannot match the motor's rpm.

Answer: C

Correct answer: 3%.

• A: This treats the 27 rpm shortfall as if it were a percentage.
• B: This is the fraction of ideal speed that is delivered, not the fraction lost.
• C (correct): The shortfall is 900 - 873 = 27 rpm, and 27 / 900 = 0.03, or 3 percent.
• D: This slips a decimal place when converting 27 / 900 = 0.03 into a percentage.

Answer: D

Correct answer: 300 N. Only the vertical part of each tension holds the sign: 2T cos 60 degrees = 300 N.

• A: This uses sin 60 degrees, but the angle is measured from vertical, so the vertical part uses cosine.
• B: This divides the whole 300 N by cos 60 degrees for one segment, forgetting that two segments share the load.
• C: This is the tension only if the segments hung straight down; tilting them requires more tension.
• D (correct): Each segment must supply 150 N vertically, and T = 300 / (2 x cos 60 degrees) = 300 / (2 x 0.5) = 300 N.

Answer: B

Correct answer: 220 N. Draw a line just above the lower block: it cuts the dead-end segment and the two segments around the lower sheave, so n = 3. The downward hauling end comes off the upper block and is not counted.

• A: This also counts the hauling end, but it leaves the upper block and is pulled downward, so it does not support the load.
• B (correct): Three segments pull up on the lower block: the dead end and both sides of the lower sheave, so effort = 660 / 3 = 220 N.
• C: This counts only the two segments around the lower sheave and misses the dead end tied to the lower block.
• D: This treats the system as a single fixed pulley, ignoring the segments shared by the moving block.

Answer: B

Correct answer: 0.2 m/s. Rope and load move for the same time, so their speeds have the same ratio as their distances. Load speed = 0.8

• A: Dividing by 2 uses the wrong number of supporting segments.
• B (correct): Load speed equals rope speed divided by the segment count: 0.8 / 4 = 0.2 m/s.
• C: Multiplying by 4 makes the load move faster than the rope, but a force-multiplying tackle slows the load.
• D: Equal speeds would mean an IMA of 1, not a four-segment tackle.

Answer: A

Correct answer: 120 N. The overall IMA of machines in series is the product: 5 x 3 = 15. Ideal effort = 1,800

• A (correct): Machines in series multiply, so the total IMA is 5 x 3 = 15, and 1,800 N / 15 = 120 N.
• B: This uses only the tackle advantage of 3 and ignores the winch.
• C: This uses only the winch advantage of 5 and ignores the tackle.
• D: This adds the advantages (5 + 3 = 8) instead of multiplying them.

Answer: D

Correct answer: A toothed timing belt. Friction belts lose a little speed to creep and can slip under load. A timing belt has teeth that engage toothed pulleys, so the shafts keep an exact speed ratio, which is why engines use it to time the camshaft.

• A: Crossing the belt reverses rotation and adds wear, but it is still a friction drive that can slip.
• B: A flat belt grips by friction only, so creep and slip prevent an exact speed ratio.
• C: A V-belt grips well by wedging into its groove, but it still relies on friction and can creep or slip.
• D (correct): Its molded teeth mesh with grooved pulleys like gears, so it cannot slip in normal use and keeps an exact ratio.

Answer: B

Correct answer: 600 rpm. Speed calculations use the pitch diameter, where the belt actually rides.

• A: This uses the outside diameters, but a V-belt rides lower in the groove at the pitch diameter.
• B (correct): Using pitch diameters, 1,500 x 120 / 300 = 600 rpm.
• C: This mixes the driver's outside diameter with the driven pitch diameter instead of using pitch diameters for both.
• D: This inverts the ratio; the larger driven pulley must turn slower than the driver, not faster.

Answer: C

Correct answer: 8 N·m. The smaller driven pulley turns

times faster. In a lossless drive, torque changes opposite to speed, so arbor torque = 24 N·

N·m.

• A: This multiplies the torque by 3, but the arbor speeds up, so its torque must go down.
• B: Torque stays the same only when the pulleys are equal; here the speed triples, so torque changes.
• C (correct): The arbor turns 240 / 80 = 3 times faster, so in a lossless drive its torque is 24 / 3 = 8 N·m.
• D: This mistakes the speed ratio of 3 for the output torque itself.

Answer: A

Correct answer: 300 N. Efficiency = AMA / IMA, so AMA = 0.80 x 4 = 3.2. Actual pull = 960

larger than the ideal 240 N because of friction.

• A (correct): The AMA is 0.8 x 4 = 3.2, so the actual pull is 960 / 3.2 = 300 N.
• B: This is the ideal effort, 960 / 4; friction makes the real pull larger.
• C: This divides the load by the efficiency but ignores the four supporting segments.
• D: This multiplies the ideal effort by 0.8, making the real pull smaller than the ideal, which friction cannot do.

Answer: B

Correct answer: 420 N. The movable pulley and load rise together, so the lifted weight is 800 N + 40 N = 840 N. The anchored end and the segment running up to the fixed pulley both pull up on the movable pulley, giving 2 segments; the downward pull from the fixed pulley is only the effort line.

Effort = 840

• A: This assumes the fixed pulley cancels any advantage, but the movable pulley is still held up by two segments.
• B (correct): Two segments hold up the movable pulley, which carries 800 + 40 = 840 N, so the effort is 840 / 2 = 420 N.
• C: This counts the downward effort line from the fixed pulley as a third supporting segment, but it does not touch the moving pulley.
• D: This divides only the 800 N load by 2 and leaves out the weight of the movable pulley that rises with it.

Answer: B

Correct answer: 150 N and 1.2 m. Three segments pull up on the lower block: the dead end tied to it, the segment between the sheaves, and the hauling end pulled upward from it.

Effort = 450

and rope travel = 3 x 0.4 m = 1.2 m.

• A: This uses 4 segments, more than the three rope segments that actually reach the lower block.
• B (correct): The dead end, the middle segment and the upward hauling end all pull on the lower block, so n = 3: 450 / 3 = 150 N and 3 x 0.4 = 1.2 m.
• C: This uses 2 segments, the count for a gun tackle with the hauling end coming off the upper block.
• D: The force is right, but dividing the lift by 3 makes the rope travel shorter than the lift, which cannot happen.

Answer: C

Correct answer: 195 N. Moving mass:

Total upward force:

Each of the four segments carries

• A: This includes the acceleration but forgets the 5 kg lower block that rises with the engine.
• B: This holds the load steady but leaves out the extra force needed to accelerate it.
• C (correct): The total moving mass is 75 kg, so total force is 75 times (9.8 + 0.6) = 780 N, shared by four segments: 195 N each.
• D: This is the total upward force on the moving block, not the tension in one segment.

Answer: A

Correct answer: 125 N. Ideal total:

so ideal effort is

Overall efficiency is

so the real effort is

• A (correct): Total IMA is 8 times 3 = 24, giving an ideal 90 N; overall efficiency is 0.9 times 0.8 = 0.72, so effort = 90 / 0.72 = 125 N.
• B: This is the ideal effort for the total IMA of 24; it ignores both efficiencies.
• C: This allows for the tackle's 80 percent efficiency but leaves out the winch's losses.
• D: This multiplies the ideal effort by the efficiency, which would make the real effort smaller than ideal.

Answer: C

Correct answer: 1,885 W. Belt speed:

Power = (400 N - 160 N) x 7.85 m/s, about 1,885 W.

• A: This leaves pi out of the belt speed, using D x N / 60 instead of the circumference.
• B: This uses only the tight-side tension and ignores the slack side pulling back.
• C (correct): Belt speed is pi x 0.12 x 1,250 / 60, about 7.85 m/s, and (400 - 160) x 7.85 is about 1,885 W.
• D: This adds the two tensions, but the turning force is their difference.

Answer: C

Correct answer: 240 rpm and 36 N·m. Stage 1:

Stage 2:

The overall reduction is 6, so ideal torque is

N·m.

• A: The torque is right, but the speed ratio is inverted; larger driven pulleys slow the output.
• B: These are the values for the intermediate shaft; the second belt stage still halves speed and doubles torque.
• C (correct): Speed: 1,440 times 80/240 = 480 rpm, then 480 times 100/200 = 240 rpm; the speed drops by 6, so torque rises to 6 times 6 = 36 N·m.
• D: The speed is right, but torque rises as speed falls, so dividing the torque is backwards.

Answer: D

Correct answer: 3,200 J. Useful work:

Since efficiency is output over input, input work is

As a check, the worker pulls 8 m of rope with

• A: This is the useful output work done on the crate; the losses add to the input.
• B: This multiplies the output work by the efficiency, giving less input than output.
• C: This multiplies the full load by the 8 m of rope pulled, as if each segment needed the whole load.
• D (correct): Output work is 1,200 times 2 = 2,400 J; input work = 2,400 / 0.75 = 3,200 J.

Answer: D

Correct answer: 400 rpm counterclockwise. Stage 1 (open belt): 1,800 x

still clockwise. The 100 mm pulley shares the countershaft, so it also turns 600 rpm clockwise.

Stage 2 (crossed belt): 600 x

and the crossed belt reverses the direction to counterclockwise.

• A: This stops at the countershaft speed and skips the second belt stage.
• B: This inverts the second-stage ratio; the larger 150 mm pulley must turn slower than 600 rpm.
• C: The speed is right, but the crossed belt in the second stage reverses the rotation.
• D (correct): Stage 1 gives 1,800 x 90 / 270 = 600 rpm clockwise, and stage 2 gives 600 x 100 / 150 = 400 rpm, reversed by the crossed belt.

Answer: B

Correct answer: An idler. A pulley that only presses on or guides the belt, without powering or being powered by a machine, is an idler. It turns at whatever speed its own diameter dictates and leaves the speed ratio between driver and driven unchanged.

• A: The driver is the pulley connected to the power source, such as the motor, and it supplies the motion.
• B (correct): An idler guides or tensions the belt without driving anything, so it does not change the driver-to-driven speed ratio.
• C: The driven pulley receives power and turns a machine, which this pulley does not do.
• D: A movable block belongs to a rope hoist and rides up with the load; it is not part of a belt drive.

Answer: C

Correct answer: 150 N. In an ideal machine, work in equals work out:

The distance ratio

also shows the IMA is 3, and

• A: This multiplies the load by the distance ratio instead of dividing, which would make the effort larger than the load.
• B: This is the full load; it ignores that pulling three times as far as the load rises cuts the force to one third.
• C (correct): Work in equals work out: the effort times 2.4 m must equal 450 N times 0.8 m, which is 360 J, so the effort is 150 N.
• D: 360 is the output work in joules, not a force; it still has to be divided by the 2.4 m of rope pulled.

Answer: B

Correct answer: 900 rpm. With no slip, rim speeds match, so

Sense check: the pump pulley is the bigger one, so it should turn slower than 1,500 rpm.

• A: This inverts the diameter ratio; a larger driven pulley must turn slower than the driver, not faster.
• B (correct): Using $N_2=N_1{D}_1/D_2$: 1,500 times 120 divided by 200 gives 900 rpm, slower because the driven pulley is larger.
• C: This assumes both pulleys share the same rpm, which is only true for pulleys on the same shaft, not pulleys linked by a belt.
• D: This squares the diameter ratio as if it were an area ratio; belt speed ratios use the plain diameter ratio.

Answer: C

Correct answer: 400 N. Add the moving-block weight:

Ideal effort is 1,500 divided by 5 segments, or 300 N.

The actual effort must always come out larger than the ideal effort.

• A: This applies the efficiency correctly but leaves out the 120 N lower block and hook, which rise with the load.
• B: This is the ideal effort; a real tackle that is only 75 percent efficient needs a larger pull than the ideal one.
• C (correct): Total lifted weight is 1,380 + 120 = 1,500 N; the ideal effort is 1,500/5 = 300 N, and dividing by 0.75 for losses gives 400 N.
• D: This multiplies the ideal effort by 0.75; friction makes the actual effort larger, so you must divide by the efficiency.

Answer: B

Correct answer: 600 rpm. Pulleys keyed to the same shaft rotate as one piece, so each makes the same number of turns per minute: 600 rpm. Diameter only matters when a belt links two pulleys on different shafts; here it changes the rim speed, not the rpm.

• A: This applies the belt-drive ratio 50/200, but that ratio only applies to pulleys connected by a belt, not to a shared shaft.
• B (correct): Pulleys fixed to one shaft must turn together, so both turn at the shaft speed regardless of their diameters.
• C: This applies the inverted ratio 200/50; diameter has no effect on the rpm of pulleys mounted on the same shaft.
• D: There is no reason to halve the speed; a shared shaft gives every pulley on it exactly the same rpm.

Answer: A

Correct answer: 300 N. Only the vertical part of each tension holds the sign up, and with the angle measured from vertical that part is $T\cos \theta$.

Tilted segments always need more tension than the 240 N they would carry hanging straight.

• A (correct): Each segment's vertical share is $T\cos 37^{\circ }$, so $2\times 0.80T=480$ N and T = 300 N.
• B: This is the tension if the segments hung straight down; tilting them means each needs more tension than half the load.
• C: This multiplies the straight-hanging share by cos 37 degrees; you must divide by the cosine, which raises the tension.
• D: This uses sin 37 degrees, but the angle is measured from vertical, so the vertical part of the tension uses the cosine.

Answer: A

Correct answer: 80%. First find the AMA from measured forces, then compare it with the IMA from the segment count.

The sheave count does not matter; only the three supporting segments set the IMA.

• A (correct): AMA is 840/350 = 2.4 and IMA is the 3 supporting segments, so efficiency is 2.4/3 = 0.80, or 80 percent.
• B: This divides the AMA by 4, the number of sheaves; the IMA comes from supporting rope segments, not wheels.
• C: This divides the IMA by the AMA; efficiency is AMA over IMA and can never exceed 100 percent.
• D: This is the effort divided by the load, 350/840, which is the inverse of the AMA, not the efficiency.

Answer: D

Correct answer: 160 rpm. Work stage by stage. Stage one:

and the 75 mm pulley on that shaft also turns at 480 rpm.

Stage two:

The overall reduction is 3 times 3 = 9, and

• A: This is only the intermediate shaft speed; the second belt stage reduces the speed by another factor of three.
• B: This adds the two stage ratios (3 + 3 = 6); in a compound drive the stage ratios multiply to give 9.
• C: This inverts both ratios, making a drive with larger driven pulleys speed up instead of slow down.
• D (correct): Stage one gives 1,440 times 100/300 = 480 rpm, which the 75 mm pulley shares; stage two gives 480 times 75/225 = 160 rpm.

Answer: A

Correct answer: 1,300 N. Ideal tension is

N throughout the rope. The upper block is pulled down by the four supporting segments and the hauling line, five segments in all, plus its own weight.

Check from outside: the load (1,000 N), the worker's pull (250 N) and the block (50 N) also total 1,300 N.

• A (correct): Each segment carries 1,000/4 = 250 N; five segments pull down on the upper block (four supporting plus the hauling line), giving 1,250 N, plus the 50 N block.
• B: This counts only the load; the worker's downward pull on the hauling line and the block's own weight also hang on the hook.
• C: This adds the block weight to the load but leaves out the 250 N downward pull of the hauling line.
• D: This correctly counts five segments at 250 N but forgets the 50 N weight of the upper block, which the hook also carries.

Answer: D

Correct answer: 400 rpm. Each turn moves the belt one circumference,

The belt covers 6.28 m each second, which is about 6.67 turns per second.

• A: This leaves out the factor of 60, so it is the number of turns per second, not per minute.
• B: This leaves out $\pi$; each turn moves the belt one circumference, $\pi D$, not one diameter.
• C: This uses $2\pi \times 0.30$ for the circumference, treating the diameter as a radius.
• D (correct): Rearranging $v=\pi DN/60$ gives N = 60 times 6.28 divided by (3.14 times 0.30), about 400 rpm.