Lesson Test: Inclined Planes, Wedges and Screws

Answer: A

Correct answer: Screw. A screw is an inclined plane wrapped around a cylinder. Its spiral thread acts as a ramp, so a long circular hand motion moves the load a short way along the axis.

• A (correct): A screw thread is a sloped path wrapped around a cylinder, so turning it moves the load along the axis.
• B: A wedge is a slope that moves into a gap; it is not wound around a shaft.
• C: A ramp is a fixed, flat sloped surface that the load travels up, with no spiral.
• D: A pulley is a grooved wheel with a rope, not a sloped surface at all.

Answer: A

Correct answer: 5. For a ramp, IMA = L / h.

• A (correct): IMA for a ramp is slope length divided by rise, and 10 divided by 2 is 5.
• B: Dividing the rise by the length flips the ratio and gives a value below 1.
• C: Subtracting the rise from the length is not how mechanical advantage is found.
• D: Multiplying the length by the rise does not give a ratio of distances.

Answer: B

Correct answer: 6. A wedge trades distance for force: the effort pushes it along its full length while the sides spread apart only by the thickness, so IMA = length ÷ thickness = 18 cm ÷ 3 cm = 6. The units cancel, which is a quick check that you formed a ratio. The common slip is multiplying the two dimensions, which gives 54, a size-like number rather than a comparison. A long, thin wedge such as a sharp axe blade has a high IMA.

• A: Adding the two dimensions does not produce a ratio of distances.
• B (correct): Wedge IMA is length divided by thickness, and 18 divided by 3 is 6.
• C: Subtracting the thickness from the length is not the wedge formula.
• D: Multiplying length by thickness gives an area-like number, not an advantage.

Answer: D

Correct answer: 0.0625 inch. Pitch is the spacing from one crest to the next. With 16 threads in each inch, each spacing is 1 divided by 16, or 0.0625 inch.

• A: This value is ten times too large; 1 divided by 16 is 0.0625, not 0.625.
• B: Moving the decimal point of 16 is not the same as taking 1 divided by 16.
• C: The thread count is not itself a distance between crests.
• D (correct): Pitch is 1 divided by the threads per inch, and 1 divided by 16 is 0.0625 inch.

Answer: A

Correct answer: Deliver more work out than is put in. A simple machine can change the size, direction, or distance of a force, but it cannot create energy. In the ideal case work out equals work in, and friction makes real output smaller.

• A (correct): Work output can never exceed work input; at best the two are equal in an ideal machine.
• B: Machines commonly lengthen the effort path in exchange for a smaller force.
• C: Lowering the needed force is the main purpose of these machines.
• D: A wedge turns a forward push into sideways pushes, so direction can change.

Answer: B

Correct answer: Set it on rated jack stands on firm, level ground. A jack can slip, and the vehicle can shift off its head. Before anyone reaches under a raised load, it must rest on rated jack stands or solid blocking on firm, level ground, with the wheels chocked.

• A: Extra lubrication can make the screw more likely to slip, and it does not make the jack a safe support.
• B (correct): A jack is made for lifting; stands or solid blocking on firm ground are needed to hold a load safely.
• C: A longer handle eases lifting but does nothing to keep the car from shifting or dropping.
• D: Lifting higher makes the car less stable and still leaves it resting only on the jack.

Answer: A

Correct answer: 75%. Efficiency = AMA / IMA × 100%. Here 3 divided by 4 is 0.75, so the ramp is 75% efficient.

• A (correct): Efficiency is AMA divided by IMA, and 3 divided by 4 is 0.75, or 75%.
• B: Multiplying the two advantages does not give efficiency.
• C: This is the fraction lost to friction, not the efficiency.
• D: Dividing IMA by AMA flips the ratio and gives an impossible efficiency above 100%.

Answer: B

Correct answer: The push acts over a longer distance. Work is force times distance. Because the push travels along the long slope instead of the short vertical rise, the same work can be done with a smaller force.

• A: Ideally the work equals that of a straight lift; a ramp never lowers it.
• B (correct): The same work is spread over the long slope, so the force needed at each step is smaller.
• C: The cart's weight is set by its mass and gravity, which a ramp does not change.
• D: Friction opposes the motion, so going uphill it adds to the effort.

Answer: D

Correct answer: 294 N. Weight is W = mg. Multiplying 30 kg by 9.8 m/s² gives 294 N.

• A: Dividing the mass by g instead of multiplying gives a value far too small.
• B: Adding 9.8 to 30 is not the formula for weight.
• C: This treats the mass in kilograms as if it were already a force in newtons.
• D (correct): Weight is mass times g, and 30 times 9.8 is 294 N.

Answer: A

Correct answer: Chisel. A wedge is a moving inclined plane whose faces push material aside. A chisel's edge is driven into wood and splits or cuts it in exactly this way.

• A (correct): A chisel's sharp, sloped edge is driven forward and pushes the material apart sideways, which is wedge action.
• B: A wheelbarrow is a lever with the wheel as the pivot, not a moving slope.
• C: A pulley block uses wheels and rope to change force, not a sloped edge.
• D: A seesaw is a lever that pivots on a support in the middle.

Answer: D

Correct answer: 300 N. Find the IMA, then divide the load by it.

• A: Multiplying the load by the advantage makes the effort larger than the load.
• B: Subtracting one third of the load from the load does not use the advantage correctly.
• C: Dividing the load by the ramp length alone ignores the rise.
• D (correct): IMA is 4.5 divided by 1.5, or 3, and 900 N divided by 3 is 300 N.

Answer: A

Correct answer: 20 mm. Lead = pitch × starts = 1.25 mm × 2 = 2.5 mm per turn. In 8 turns the screw advances

• A (correct): Lead is 2 times 1.25, or 2.5 mm, and 8 turns times 2.5 mm is 20 mm.
• B: This doubles the travel a second time; the lead is 2.5 mm, not 5 mm.
• C: Using the pitch alone ignores the second start and gives half the true travel.
• D: Dividing the turns by the lead instead of multiplying gives the wrong result.

Answer: A

Correct answer: The length measured along the ramp surface. Work uses the distance moved in the direction of the force. The push is along the slope, so input work is push times sloped length; the vertical rise goes with the load for the output work.

• A (correct): The push acts along the slope, so the input distance is the sloped length.
• B: Adding the two sides does not match the path the push actually follows.
• C: The rise is the output distance used with the load's weight, not with the push.
• D: The run is not the direction of the push, so using it gives the wrong input work.

Answer: D

Correct answer: 320 N. On level ground the normal force equals the weight, 800 N. A moving crate feels kinetic friction,

so a 320 N push keeps it sliding steadily.

• A: Subtracting the two friction forces does not give the force needed for steady sliding.
• B: This is the crate's full weight, not the friction force.
• C: This is the static value needed to break the crate loose, not to keep it moving.
• D (correct): Once sliding, kinetic friction applies: 0.4 times 800 N is 320 N.

Answer: B

Correct answer: 440. Put both lengths in meters: 5 mm = 0.005 m.

• A: Dividing the radius by the lead leaves out the 2 pi needed for the full circle.
• B (correct): IMA is 2 pi times 0.35 m divided by 0.005 m, which is about 440.
• C: Using 0.35 m as a radius and then doubling again treats it as a diameter in the formula.
• D: Using pi times the radius counts only half of the handle's circular path.

Answer: C

Correct answer: About 1.67. First find the slope length with the Pythagorean theorem, then divide by the rise.

• A: Dividing rise by slope length flips the ratio, giving a value below 1.
• B: Dividing rise by run flips the ratio and uses the wrong side.
• C (correct): The slope length is the square root of 1.2 squared plus 1.6 squared, or 2.0 m, and 2.0 divided by 1.2 is about 1.67.
• D: This uses the horizontal run instead of the sloped length.

Answer: C

Correct answer: 80%. Input work = 300 N × 5 m = 1500 J. Output work = 1000 N × 1.2 m = 1200 J. Efficiency =

or 80%.

• A: Dividing input work by output work gives an impossible efficiency above 100%.
• B: Dividing the rise by the ramp length ignores the forces entirely.
• C (correct): Output work is 1000 times 1.2, or 1200 J; input is 300 times 5, or 1500 J; 1200 divided by 1500 is 80%.
• D: This is the share of input work lost to friction, not the efficiency.

Answer: A

Correct answer: 1500 N. Wedge IMA = 24 cm / 4 cm = 6. Ideal output force = 250

• A (correct): IMA is 24 divided by 4, or 6, and 250 N times 6 is 1500 N.
• B: Multiplying the push by the thickness alone does not give the wedge advantage.
• C: Dividing the push by the advantage makes the output smaller than the input.
• D: Multiplying the push by the length alone ignores the thickness.

Answer: D

Correct answer: 3.98 N·m. Set input work per turn equal to output work:

• A: Multiplying by 2 pi instead of dividing makes the torque far too large.
• B: Leaving the lead in millimeters instead of meters inflates the answer a thousandfold.
• C: This is the work per turn in joules; it still needs to be divided by 2 pi.
• D (correct): Work per turn is 5000 N times 0.005 m, or 25 J, and dividing by 2 pi gives about 3.98 N·m.

Answer: D

Correct answer: It is cut in half. Screw IMA is $2\pi r/\ell$. Halving $\ell$ doubles the IMA, so the ideal effort is cut in half. The tradeoff is that twice as many turns are needed for the same lift.

• A: A smaller lead gives more advantage, so the effort goes down, not up.
• B: The IMA changes in direct proportion to the lead, so the effort halves rather than quarters.
• C: The lead directly affects how far the load rises per turn, so it changes the IMA.
• D (correct): Lead is in the denominator of the screw IMA, so halving the lead doubles the IMA and halves the ideal effort.

Answer: B

Correct answer: 30 N. Screw IMA is proportional to the handle radius. Going from 0.10 m to 0.30 m triples the IMA, so the ideal effort falls from 90 N to

• A: The handle radius sets the hand's circular path, so changing it does change the effort.
• B (correct): Tripling the radius triples the IMA, so the effort becomes 90 divided by 3, or 30 N.
• C: Dividing by 9 treats the effect as squared, but the IMA grows only in proportion to the radius.
• D: A longer handle lowers the effort; multiplying by 3 moves it the wrong way.

Answer: C

Correct answer: Same ideal advantage, opposite turning direction. A right-hand thread tightens clockwise and a left-hand thread tightens counterclockwise. Since the pitch and handle radius are equal, the IMA is the same; handedness changes direction, not advantage.

• A: The two threads tighten in opposite directions, and their advantages are the same.
• B: Thread direction does not change the lead or the handle path, so the IMA does not double.
• C (correct): Handedness only decides which way the screw must turn to advance; the pitch and handle set the advantage.
• D: Thread direction has no effect on the distances that set the IMA.

Answer: D

Correct answer: 63%. In series, each stage passes on only part of its input, so

or 63%.

• A: Adding the efficiencies gives an impossible value above 100%.
• B: The combination is less efficient than its weakest stage, not equal to it.
• C: Averaging the two efficiencies ignores that each stage loses energy in turn.
• D (correct): Efficiencies in series multiply: 0.70 times 0.90 is 0.63, or 63%.

Answer: D

Correct answer: Down the slope, along the ramp surface. Friction acts along the contact surface and opposes the direction of motion. Since the crate slides up the ramp, friction on it points down the slope.

• A: The force at right angles to the surface is the normal force, not friction.
• B: Straight down is the direction of the weight, not friction.
• C: Pointing friction uphill would mean it helps the push, but friction opposes the motion.
• D (correct): Friction opposes the motion, and the crate moves up the slope, so friction acts down along the surface.

Answer: C

Correct answer: High thread friction makes the screw self-locking. Many screws, especially those below roughly 50% efficiency, are self-locking: thread friction is too high for the load to turn them backward. This is useful, but it is not a guarantee, so loads still need proper support.

• A: A highly efficient screw would be more likely to run backward under load.
• B: Once released, the handle applies no force to the screw.
• C (correct): Low-efficiency screws have enough thread friction that the load cannot spin them backward.
• D: A lead larger than pitch means a multi-start thread, which tends to make reversing easier, not harder.

Answer: D

Correct answer: Both need 500 J. Ideal work equals weight times height: 500 N × 1 m = 500 J. On the 2 m ramp the push is 250 N, and on the 4 m ramp it is 125 N, but both give 500 J.

• A: A longer ramp lowers the force, not the work.
• B: The longer ramp halves the force while doubling the distance, so the work stays the same.
• C: These values multiply the full weight by the ramp length instead of by the rise.
• D (correct): Lifting 500 N through 1 m takes 500 J no matter how long the frictionless ramp is.

Answer: B

Correct answer: 30 cm long and 3 cm thick. Wedge IMA is length divided by thickness:

The largest IMA, 10, needs the smallest effort.

• A: Its IMA is only 5, so it needs twice the effort of the best wedge.
• B (correct): Its IMA is 30 divided by 3, or 10, the largest of the four, so it needs the least effort.
• C: Its IMA is just 4, the smallest of the group, so it needs the most effort.
• D: Its IMA is 8, which is good but still less than 10.

Answer: B

Correct answer: 1.5 mm. In the label M16 x 1.5, 16 mm is the diameter and 1.5 mm is the pitch. A single-start thread advances one pitch per turn, so the bolt moves 1.5 mm.

• A: Dividing the diameter by the pitch has no physical meaning for travel per turn.
• B (correct): For a single-start thread, lead equals pitch, and the second number in the label is the 1.5 mm pitch.
• C: Multiplying diameter by pitch does not give the travel per turn.
• D: 16 mm is the bolt's diameter, which does not set how far it moves per turn.

Answer: C

Correct answer: 180 W. Work = 150 N × 6 m = 900 J. Power = work / time = 900 J / 5 s = 180 W.

• A: Dividing the force by the time skips the distance needed to find work.
• B: 900 is the work in joules; it must still be divided by the time.
• C (correct): Work is 150 N times 6 m, or 900 J, and 900 J divided by 5 s is 180 W.
• D: Multiplying work by time instead of dividing gives a value far too large.

Answer: C

Correct answer: 4. Put both lengths in the same unit: 60 cm = 0.60 m. Then IMA = 2.4 m / 0.60 m = 4.

• A: Dividing the rise by the length flips the ratio below 1.
• B: Mixing meters and centimeters, 2.4 divided by 60, gives a meaningless tiny ratio.
• C (correct): Converting 60 cm to 0.60 m gives 2.4 divided by 0.60, which is 4.
• D: Dividing 2.4 by 0.06 slips the conversion by a factor of ten.

Answer: B

Correct answer: 227 N. Downhill component:

Friction:

pointing uphill because the crate moves down. The helper supplies the rest: 400 - 173 ≈ 227 N.

• A: Adding friction to the downhill pull treats friction as if the crate were moving up.
• B (correct): Gravity pulls 400 N downhill and friction pushes about 173 N uphill, leaving about 227 N for the helper.
• C: This ignores friction, which helps hold the crate back as it slides down.
• D: This uses the full 800 N weight for the normal force instead of the smaller W cos 30°.

Answer: C

Correct answer: 380 N. Slope component:

Normal force:

so friction

N downhill. Push needed: 250 + 130 ≈ 380 N.

• A: This bases friction on the full 500 N weight instead of the normal force W cos 30°.
• B: Subtracting friction treats it as helping, but friction acts downhill when the crate moves up.
• C (correct): Slope component is 250 N and friction is 0.3 times 433 N, about 130 N; together they need about 380 N.
• D: This is the frictionless value and leaves out the friction that opposes the uphill motion.

Answer: D

Correct answer: 59.7 N. Lead = 2 × 2.5 mm = 5 mm = 0.005 m.

Actual effort = 17.9 / 0.30 ≈ 59.7 N.

• A: This is the ideal effort; a 30% efficient screw needs more force, not the ideal amount.
• B: Multiplying the ideal effort by the efficiency lowers it, but friction always raises real effort.
• C: This uses the 2.5 mm pitch as the lead, ignoring the second start and doubling the IMA.
• D (correct): Lead is 5 mm, so IMA is about 503 and ideal effort about 17.9 N; dividing by 0.30 gives about 59.7 N.

Answer: B

Correct answer: 80%. Slope length:

so IMA = 1.5 / 0.9 ≈ 1.67. AMA = 1200 / 900 ≈ 1.33. Efficiency =

or 80%.

• A: This uses the 1.2 m run instead of the 1.5 m slope length when finding the IMA.
• B (correct): Slope length is 1.5 m, so IMA is 1.5/0.9, about 1.67; AMA is 1200/900, about 1.33; 1.33 divided by 1.67 is 80%.
• C: Dividing the effort by the load gives the inverse of AMA, not the efficiency.
• D: Dividing IMA by AMA flips the ratio and gives an impossible value above 100%.

Answer: C

Correct answer: The data contain an error, since work out exceeds work in. Check work: input = 200 N × 3 m = 600 J, output = 700 N × 1 m = 700 J. Output can never exceed input, so a measurement or calculation must be wrong; the ideal push would be at least 700/3 ≈ 233 N.

• A: An ideal ramp would give equal work, but 600 J and 700 J are not equal.
• B: Friction turns work into heat; it never adds useful work to the load.
• C (correct): Input is 200 times 3, or 600 J, but output is 700 times 1, or 700 J, which no machine can deliver.
• D: An efficiency above 100% would mean the ramp created energy, which is impossible.

Answer: B

Correct answer: 141 N. On an incline the normal force is

Kinetic friction is

• A: This is the downhill pull of gravity, W sin 20°, not the friction force.
• B (correct): Normal force is 600 cos 20°, about 564 N, and 0.25 times 564 N is about 141 N.
• C: This multiplies the coefficient by the downhill component W sin 20° instead of the normal force.
• D: This uses the full 600 N weight as the normal force, which overstates friction on a slope.

Answer: C

Correct answer: 17.2 N·m. Lead = 3 × 2 mm = 6 mm = 0.006 m.

Actual torque = 6.88 / 0.40 ≈ 17.2 N·m.

• A: This is the ideal torque; a 40% efficient screw needs more.
• B: This uses the 2 mm pitch as the lead, ignoring the three starts.
• C (correct): Lead is 6 mm; ideal torque is 7200 times 0.006 divided by 2 pi, about 6.88 N·m, and dividing by 0.40 gives about 17.2 N·m.
• D: Multiplying the ideal torque by the efficiency lowers it, but friction raises the needed torque.

Answer: C

Correct answer: 100 N. In series, advantages multiply: IMA =

Efficiencies multiply:

AMA =

so effort =

• A: Multiplying the ideal effort by the efficiency lowers it, but losses raise the real effort.
• B: Adding the two IMAs instead of multiplying them understates the combined advantage.
• C (correct): Total IMA is 60 times 5, or 300; total efficiency is 0.5 times 0.8, or 0.40; AMA is 120, so effort is 12,000 divided by 120, or 100 N.
• D: This is the ideal effort and ignores the losses in both stages.

Answer: A

Correct answer: 0 J. Work is force times distance moved,

With

the work on the crate is 0 J. The push still matters for balancing forces, but it does no work on a load that stays still.

• A (correct): Work needs movement in the direction of the force; the crate moves zero distance, so no work is done on it.
• B: This assumes the crate moved 1 m, but it did not move at all.
• C: Dividing force by time does not produce work in joules.
• D: Multiplying force by time does not give work; work uses distance moved.

Answer: A

Correct answer: 1200 N, with the sides moved 1.5 cm apart. IMA =

Spreading force = 150

and the spread = 12 cm / 8 = 1.5 cm. Check: 150 N × 0.12 m = 18 J in and 1200 N × 0.015 m = 18 J out.

• A (correct): IMA is 8, so force is 150 times 8, or 1200 N, and the spread is 12 cm divided by 8, or 1.5 cm.
• B: Dividing the force by the IMA reverses its effect; a wedge multiplies the push.
• C: The output cannot move as far as the input when the force is multiplied; that would create work.
• D: Multiplying the input distance by the IMA would make output work far exceed input work.

Answer: B

Correct answer: 98 N. Convert mass to weight:

The ramp's IMA is

• A: This divides the mass, 40 kg, by the IMA; mass must first be converted to weight in newtons.
• B (correct): The weight is 40 times 9.8 = 392 N, the IMA is 3.0/0.75 = 4, and 392/4 = 98 N.
• C: This is the generator's full weight; the ramp's IMA of 4 cuts the needed effort to one quarter of it.
• D: This multiplies the weight by the IMA; a helping machine divides the load by its mechanical advantage.

Answer: A

Correct answer: The vertical height the load rises. Output work on a ramp is the work of lifting: weight times vertical rise, $Wh$. The slope length belongs to the input side, where the effort acts along the surface.

• A (correct): The job is lifting against gravity, so output work is weight times vertical rise.
• B: Slope length is the input distance, the path of the push, not the output distance.
• C: Gravity acts vertically, so the horizontal run does no lifting work on the load.
• D: Adding the two distances has no physical meaning; output work uses only the rise.

Answer: D

Correct answer: 17.5 cm. Find the IMA from the forces, then use

• A: Seven is the IMA, a pure number; it still must be multiplied by the 2.5 cm thickness to give a length.
• B: This adds the IMA to the thickness; IMA is length divided by thickness, so you multiply to get the length.
• C: This divides the thickness by the IMA, giving a wedge thinner than it is thick, which cannot have IMA 7.
• D (correct): The IMA is 1,260/180 = 7, and length equals IMA times thickness, so 7 times 2.5 cm = 17.5 cm.

Answer: A

Correct answer: 8 turns. Lead equals pitch times starts:

per turn.

• A (correct): The lead is 3 times 1.5 = 4.5 mm per turn, and 36/4.5 = 8 turns.
• B: This uses the pitch as the travel per turn; a triple-start screw moves three pitches each turn.
• C: This divides the distance by the number of starts only, ignoring the 1.5 mm pitch.
• D: This multiplies the distance by the pitch instead of dividing the distance by the lead.

Answer: C

Correct answer: Friction on its faces resists sliding. Friction lowers a wedge's efficiency, but it is also what keeps a wedge in place. Without friction on its faces, the door's squeeze would push a smooth wedge straight back out.

• A: A small rubber wedge is very light; its weight alone could not resist the door's push.
• B: Mechanical advantage multiplies an applied force; it does not create a force pulling the wedge in.
• C (correct): Friction between the wedge faces and the floor and door resists the squeeze that would push it back out.
• D: Air pressure acts on all sides of the wedge and gives no net force holding it in.

Answer: D

Correct answer: 38.2 N. In one turn the hand travels $2\pi r$ while the load moves one lead.

• A: This uses the handle radius instead of the handle circle, $2\pi r$, as the effort distance per turn.
• B: This uses $\pi r$, half the circle, so the effort distance per turn is too short and the effort too large.
• C: This converts 4 mm to 0.04 m instead of 0.004 m, making the lead ten times too large.
• D (correct): The IMA is $2\pi (0.20)/0.004\approx 314$, and 12,000/314 is about 38.2 N.

Answer: B

Correct answer: About 70%. Slope component:

Normal force:

so friction is about 260 N and the push is about 860 N.

AMA =

The IMA is

/sin 30 degrees = 2.

• A: This bases friction on the full 1,200 N weight; on a slope the normal force is only $W\cos \theta$, about 1,039 N.
• B (correct): The push is 600 + 0.25 times 1,039 = about 860 N, so AMA is 1,200/860 = 1.40; IMA is 1/sin 30 degrees = 2, giving about 70 percent.
• C: This uses $W\sin \theta$ for the normal force; the force pressing into the surface is $W\cos \theta$.
• D: This divides IMA by AMA; efficiency is AMA over IMA and can never exceed 100 percent.

Answer: B

Correct answer: 240. The screw multiplies the hand force by 40, and the block multiplies that output again by 6.

• A: This adds the two advantages; in series each stage multiplies the output of the one before it.
• B (correct): Ideal advantages of machines in series multiply, so 40 times 6 = 240.
• C: This subtracts the advantages, which has no basis; series stages multiply.
• D: This divides 40 by 6; the block multiplies the screw's output force again rather than reducing it.

Answer: A

Correct answer: About 3,140 N. Find the IMA, scale it by the efficiency to get the AMA, then multiply by the effort.

Output force

• A (correct): IMA is $2\pi (0.15)/0.003\approx 314$, AMA is 25 percent of that, about 78.5, and 40 times 78.5 is about 3,140 N.
• B: This is the ideal output, 40 times 314; it ignores that only 25 percent of the ideal advantage is realized.
• C: This divides by the efficiency instead of multiplying, giving more than the ideal output, which is impossible.
• D: This leaves $2\pi$ out of the effort distance, using the radius instead of the circumference of the handle circle.

Answer: C

Correct answer: 320 N. The weight splits into $W\sin \theta$ along the slope and $W\cos \theta$ into the surface. With no friction, the holding force must balance the slope component.

• A: This is $W\cos \theta$, the part pressing into the ramp surface, not the part pulling down the slope.
• B: This is the full weight; on a 30 degree ramp only part of the weight acts along the slope.
• C (correct): The part of the weight pulling down along the slope is $W\sin \theta =640\times 0.50=320$ N, so that much force holds it.
• D: This uses the tangent of 30 degrees; the component of weight along the slope uses the sine.