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GCSE · Physics

P5c Pressure, Moments and Levers

AQA 8464 · Calculator allowed · about 110 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Levers, gears and pressure: making forces work harder

Original text written for Revision Library.

Whenever a spanner turns a stubborn nut, a crowbar levers up a heavy slab, or a cyclist changes gear on a hill, the same idea from physics is at work: the moment of a force, the turning effect produced when a force acts at a distance from a pivot. Engineers exploit the principle of moments to design levers and gears that let a small force, applied a long way from a pivot, balance or overcome a much larger force acting close to it. The same principle explains why a small child can balance a much heavier adult on a see-saw, simply by sitting further from the middle. Fluids exert forces too, in the form of pressure, calculated as force divided by the area it acts over. This is why a sharp knife cuts easily, while a wide ski keeps a skier from sinking into snow, and why a submarine's hull must withstand enormous pressure at great depth, where the weight of water pressing down from above is far greater than at the surface. Even the air around us has weight, producing an atmospheric pressure that gradually falls the higher you climb, until it becomes too thin to breathe without extra oxygen.

Figure 1 shows a spanner being used to tighten a nut. A force of 60 N is applied at right angles (perpendicular) to the spanner handle, at a distance of 0.20 m from the nut, which acts as the pivot.

(a)Define what is meant by the moment of a force.(1)

(b)State the equation linking the moment of a force, the force, and the perpendicular distance from the pivot to the line of action of the force, and state the unit of moment.(2)

(c)Calculate the moment of the 60 N force about the nut.(2)

(Total for Question 1 is 5 marks)

A mechanic, Priya, needs to loosen a stiff bolt. The moment needed to loosen the bolt is 18 Nm. She can apply a maximum force of 90 N with her hand. Use the equation moment = force x perpendicular distance.

(a)Calculate the minimum perpendicular distance from the bolt at which Priya must apply this force to loosen it.(2)

(b)Priya then fits an extension bar to her spanner, which doubles the perpendicular distance at which she can apply her force. Calculate the minimum force she now needs to apply to produce the same 18 Nm moment.(2)

(Total for Question 2 is 4 marks)

Figure 2 shows a playground see-saw, pivoted at its centre. Tariq, weight 500 N, sits on the left side, at a distance of 1.2 m from the pivot. His younger sister Amina, weight 300 N, sits on the right side. The see-saw is perfectly balanced (horizontal).

(a)State the principle of moments.(1)

(b)Calculate the moment of Tariq's weight about the pivot.(2)

(c)Use the principle of moments to calculate the distance, d, from the pivot at which Amina must sit for the see-saw to balance.(2)

(Total for Question 3 is 5 marks)

Figure 3 shows a uniform beam, pivoted at its centre, C. Two forces act downwards on the left of the pivot: 40 N at point A, 0.50 m from C, and 25 N at point B, 0.40 m from C. A single downward force, F, acts at point D, 0.60 m to the right of C. The beam is in equilibrium.

(a)Calculate the total moment produced by the two forces on the left of the pivot.(3)

(b)Use the principle of moments to calculate F.(2)

(c)State what is meant by the beam being 'in equilibrium'.(1)

(Total for Question 4 is 6 marks)

Figure 4 shows a crowbar used as a lever to lift a heavy paving slab. The crowbar pivots on a small stone that acts as a fulcrum. The slab pushes down on the crowbar with a force of 800 N, at a distance of 0.05 m from the fulcrum. Priya pushes down on the other end of the crowbar, at a distance of 0.75 m from the fulcrum, to just lift the slab.

(a)Calculate the minimum force that Priya must apply to just lift the slab.(3)

(b)Explain, in terms of the principle of moments, why the crowbar makes it easier for Priya to lift the slab compared with lifting it directly by hand.(2)

(Total for Question 5 is 5 marks)

Tools such as wheelbarrows, pliers and bottle openers all use the principle of moments to make everyday tasks easier, in the same way as the crowbar in Question 5.

(a)Give two examples of levers, other than a crowbar, that are used to make a task easier.(2)

(b)State the general effect that using a lever like this has on the size of force needed to perform a task.(1)

(Total for Question 6 is 3 marks)

Figure 5 shows two circular gears, A and B, with interlocking teeth. Gear A is turned by a motor and has a radius of 0.080 m. It drives gear B, which has a radius of 0.020 m.

(a)State how the direction of rotation of gear B compares with the direction of rotation of gear A.(1)

(b)The teeth of gear A push on the teeth of gear B with a contact force of 50 N. By Newton's Third Law, state the size and direction of the force that gear B's teeth exert on gear A's teeth.(1)

(c)Higher tier only. Calculate the moment that the 50 N contact force produces about the centre of gear A, and the moment it produces about the centre of gear B.(3)

(d)Higher tier only. Use your answers to part (c) to explain why gear B turns with a smaller turning force (moment) than gear A, even though the contact force between their teeth is the same size.(2)

(Total for Question 7 is 7 marks)

A student, Zara, carries out an experiment to investigate the principle of moments. She balances a metre ruler on a pivot at its 50 cm mark (its centre of mass), then hangs a fixed weight of 2.0 N from a point 40 cm to the left of the pivot. She then hangs a second weight, W, from various points to the right of the pivot, adjusting its distance, d, from the pivot until the ruler is exactly balanced (horizontal) each time. Figure 6 shows the equipment. Her results are given in the table: Weight on the right, W (N): 1.0, 2.0, 4.0, 5.0; Balancing distance from pivot, d (cm): 80.0, 40.0, 20.0, 16.0.

(a)State the independent variable and the dependent variable in this investigation.(2)

(b)State one variable that Zara should keep constant throughout the investigation, and explain why.(2)

(c)Explain why the ruler is pivoted at its centre of mass (the 50 cm mark), rather than at another point.(2)

(d)Calculate the moment of the fixed 2.0 N weight about the pivot, giving your answer in newton-centimetres (Ncm).(1)

(e)For each value of W in the table, calculate the moment produced by the weight on the right of the pivot, and use your results to state a conclusion about balanced moments.(3)

(f)Zara's measurements of d have an uncertainty of ± 0.5 cm each, because of the difficulty in judging exactly when the ruler is horizontal. Suggest one improvement to her method that would reduce this uncertainty.(2)

(g)State one hazard associated with this experiment, and one precaution to reduce the risk it presents.(2)

(Total for Question 8 is 14 marks)

A tractor's wheel exerts a force of 6000 N on the ground. The area of the wheel in contact with the ground is 0.50 m^2. Use the equation pressure = force / area (p = F / A).

(a)Define pressure.(1)

(b)State the equation linking pressure, force and area, and state the unit of pressure.(2)

(c)Calculate the pressure the wheel exerts on the ground.(2)

(Total for Question 9 is 5 marks)

Farida, weight 600 N, is standing still. When she wears trainers, the total area of both shoe soles in contact with the ground is 0.040 m^2. When she wears stiletto heels, the total contact area is only 0.00050 m^2. Use the equation p = F / A.

(a)Calculate the pressure Farida exerts on the ground when wearing trainers.(2)

(b)Calculate the pressure Farida exerts on the ground when wearing stiletto heels.(2)

(c)Explain, using your answers to parts (a) and (b), why stiletto heels are more likely to leave dents in a soft wooden floor than trainers, even though Farida's weight is the same in both cases.(2)

(d)State one other everyday example where a large contact area is used to reduce the pressure exerted on a surface.(1)

(Total for Question 10 is 7 marks)

A diver is fully submerged in a swimming pool. Pressure sensors are attached to her body, on her chest, her back and the top of her head, all at the same depth.

(a)State how the readings on the three pressure sensors compare with each other.(1)

(b)Explain your answer to part (a), referring to the directions in which pressure due to a fluid acts.(2)

(Total for Question 11 is 3 marks)

Higher tier only. The diver from Question 11 is at a depth of 3.0 m below the surface of the pool. The density of the water is 1000 kg/m^3. Take the gravitational field strength as 9.8 N/kg. Use the equation pressure = height of column x density of liquid x gravitational field strength (p = h x ρ x g).

(a)Calculate the pressure due to the water alone at this depth.(2)

(b)Atmospheric pressure at the surface of the pool is 100000 Pa. Calculate the total pressure acting on the diver at this depth.(1)

(c)The diver then swims down to a depth of 6.0 m (double the original depth). State and explain what happens to the pressure due to the water alone at this new depth, compared with your answer to part (a).(2)

(d)State how the pressure due to the water at 3.0 m depth would change if the pool were filled with a liquid of greater density than water instead, with the depth unchanged.(1)

(Total for Question 12 is 6 marks)

A student, Ayesha, investigates how pressure in a liquid varies with depth. She takes a plastic bottle and makes three small holes in its side, at different heights above the base: hole X near the top, hole Y in the middle, and hole Z near the bottom. She fills the bottle completely with water and removes her finger from each hole in turn, measuring the horizontal distance each water jet travels before landing. Figure 7 shows the bottle and the jets of water. Her results are given in the table: Hole: X (near top), Y (middle), Z (near bottom); Depth below the water surface (cm): 5.0, 15.0, 25.0; Horizontal distance travelled by jet (cm): 12.0, 19.0, 21.0.

(a)State the hole (X, Y or Z) that produces the jet with the greatest horizontal distance, and use this observation to state how pressure varies with depth in a liquid.(2)

(b)Explain, in terms of the column of liquid above each hole, why the water at hole Z is pushed out with a greater force than the water at hole X.(2)

(c)Identify one variable, other than the depth of each hole, that should be controlled to make this a fair test between the three jets.(1)

(d)Suggest one improvement to Ayesha's method that would make her conclusion more reliable.(2)

(e)State one other application where understanding that pressure increases with depth in a liquid is important.(1)

(Total for Question 13 is 8 marks)

Atmospheric pressure is the pressure caused by the atmosphere, the layer of air surrounding the Earth.

(a)State what causes atmospheric pressure.(1)

(b)State how atmospheric pressure changes as altitude (height above sea level) increases.(1)

(Total for Question 14 is 2 marks)

A weather balloon is released from the ground and rises to a very high altitude. Explain why atmospheric pressure decreases as altitude increases, and why the rate of decrease is not constant (atmospheric pressure decreases more quickly at low altitudes than at high altitudes).

(Total for Question 15 is 6 marks)

Higher tier only. A steel ball bearing sinks when placed in water, but a large steel ship floats, even though the ship is far heavier than the ball bearing. Both the ball bearing and the ship experience an upthrust force from the water.

(a)State the condition, in terms of weight and upthrust, for an object to float in a fluid.(1)

(b)State how the upthrust on a submerged (or partly submerged) object is related to the weight of fluid it displaces.(1)

(c)Explain why the ball bearing sinks, while the ship floats, in terms of the volume of water each displaces.(2)

(Total for Question 16 is 4 marks)

Higher tier only. Daniel, a student, hangs a metal block from a newton meter. In air, the newton meter reads 6.0 N. When the block is fully lowered until it is fully submerged in water, the newton meter reads 4.0 N. The density of water is 1000 kg/m³ and the gravitational field strength, g, is 9.8 N/kg.

(a)State the name of the upward force that causes the newton meter reading to decrease when the block is submerged.(1)

(b)Calculate the upthrust acting on the block when it is fully submerged.(2)

(c)The upthrust on a submerged object equals the weight of the fluid it displaces. Calculate the volume of water displaced by the block.(3)

(d)Because the block is fully submerged, the volume of water it displaces equals the volume of the block. Use this, and the block's weight in air, to calculate the density of the metal the block is made from.(3)

(Total for Question 17 is 9 marks)

A hydraulic excavator (digger) has wide caterpillar tracks instead of narrow wheels, and a long hydraulic arm, pivoted at a joint, to lift and move soil.

(a)Explain why the excavator is fitted with wide tracks rather than narrow wheels.(2)

(b)The hydraulic arm applies a force of 2000 N at a perpendicular distance of 0.15 m from a pivot joint, to lift a load. Calculate the moment produced about the pivot.(2)

(c)To lift a heavier load, an engineer suggests redesigning the arm so the same force acts further from the pivot, rather than increasing the force itself. State one advantage of increasing the distance from the pivot, rather than the force, to increase the moment.(1)

(d)State why the excavator's hydraulic arm can be described as a lever.(1)

(Total for Question 18 is 6 marks)

Mark scheme · P5c Pressure, Moments and Levers

Question 1

(a) B1 the turning effect of a force (about a pivot), oe
(a) Answer: The turning effect of a force about a pivot.
(b) B1 moment = force x perpendicular distance (from the pivot) oe, accept M = F x d
(b) B1 unit: newton-metre (Nm)
(b) Answer: Moment = force x perpendicular distance. Unit: newton-metre (Nm).
(c) M1 substitutes: moment = 60 x 0.20
(c) A1 12 Nm cao
(c) Answer: 12 Nm

Question 2

(a) M1 rearranges and substitutes: distance = 18 / 90
(a) A1 0.20 m cao
(a) Answer: 0.20 m
(b) M1 ft new distance = 2 x their (a) = 0.40 m; rearranges: force = 18 / 0.40
(b) A1 45 N cao ft
(b) Answer: 45 N

Question 3

(a) B1 for a system in equilibrium (balanced), the sum of the clockwise moments about a pivot equals the sum of the anticlockwise moments about the same pivot, oe
(a) Answer: For a balanced system, the sum of the clockwise moments equals the sum of the anticlockwise moments about the pivot.
(b) M1 substitutes: moment = 500 x 1.2
(b) A1 600 Nm cao
(b) Answer: 600 Nm
(c) M1 ft rearranges: d = their (b) / 300
(c) A1 2.0 m cao ft
(c) Answer: 2.0 m

Question 4

(a) M1 moment of 40 N force = 40 x 0.50 = 20 Nm
(a) M1 moment of 25 N force = 25 x 0.40 = 10 Nm
(a) A1 total = 30 Nm (20 + 10) cao
(a) Answer: 30 Nm
(b) M1 ft rearranges: F = their (a) / 0.60
(b) A1 50 N cao ft
(b) Answer: 50 N
(c) B1 the resultant force and the resultant (net) moment on the beam are both zero, so it does not move or turn, oe
(c) Answer: The resultant force and resultant moment on the beam are both zero.

Question 5

(a) M1 sets moments equal: F x 0.75 = 800 x 0.05
(a) M1 rearranges: F = (800 x 0.05) / 0.75
(a) A1 awrt 53.3 N cao
(a) Answer: awrt 53.3 N
(b) B1 Priya's force acts at a much greater perpendicular distance from the fulcrum than the slab's force (0.75 m compared with 0.05 m), oe
(b) B1 for the same moment, a much smaller force is needed at the larger distance than would be needed to lift the slab's full weight directly, oe
(b) Answer: Priya's force acts much further from the fulcrum than the slab's force, so a much smaller force produces the same moment needed to lift the slab.

Question 6

(a) B1 any one correct example, e.g. wheelbarrow, pliers, scissors, bottle opener, nutcracker, door handle
(a) B1 a second, different correct example from the list above
(a) Answer: Any two of: wheelbarrow, pliers, scissors, bottle opener, nutcracker, door handle.
(b) B1 it reduces (decreases) the force needed, oe
(b) Answer: It reduces the force needed.

Question 7

(a) B1 gear B turns in the opposite direction to gear A, oe
(a) Answer: Gear B turns in the opposite direction to gear A.
(b) B1 50 N (equal in size), acting in the opposite direction
(b) Answer: 50 N, in the opposite direction.
(c) M1 moment about A = 50 x 0.080 = 4.0 Nm
(c) M1 moment about B = 50 x 0.020 = 1.0 Nm
(c) A1 both correct (4.0 Nm and 1.0 Nm) cao
(c) Answer: 4.0 Nm about A; 1.0 Nm about B
(d) B1 the contact force is the same size at the point where the teeth touch (a Newton's third law pair), oe
(d) B1 but gear B has a smaller radius (a shorter distance from its centre to the teeth), so the same force produces a smaller moment (moment = force x distance) about its centre than about gear A's centre, oe
(d) Answer: The contact force is the same size on both gears, but gear B's smaller radius means the same force produces a smaller moment about its centre.

Question 8

(a) B1 independent variable: the weight, W, hung on the right of the pivot
(a) B1 dependent variable: the balancing distance, d, from the pivot
(a) Answer: Independent: the weight W. Dependent: the balancing distance d.
(b) B1 the fixed weight (2.0 N) and its distance from the pivot (40 cm) on the left must stay the same throughout, oe
(b) B1 reason: so that the moment being balanced does not change, allowing a fair comparison between the different values of W, oe
(b) Answer: The fixed 2.0 N weight and its 40 cm distance, so the moment being balanced stays the same for a fair comparison.
(c) B1 at the centre of mass, the ruler's own weight acts through the pivot, so it produces no moment about the pivot, oe
(c) B1 so the weight of the ruler itself does not need to be included in the moment calculations, oe
(c) Answer: At its centre of mass, the ruler's weight acts through the pivot and produces no moment, so it can be ignored in the calculations.
(d) B1 80 Ncm (2.0 x 40)
(d) Answer: 80 Ncm
(e) M1 calculates at least two moments correctly, e.g. 1.0 x 80.0 = 80; 2.0 x 40.0 = 80
(e) A1 all four moments equal 80 Ncm (1.0x80.0, 2.0x40.0, 4.0x20.0, 5.0x16.0 all = 80)
(e) B1 conclusion: the moment on the right always equals the moment on the left (80 Ncm) when the ruler is balanced, confirming the principle of moments, oe
(e) Answer: All four moments equal 80 Ncm, confirming the principle of moments.
(f) B1 e.g. use a fixed pointer/fiducial marker and view it at eye level to judge when the ruler is exactly horizontal, or repeat each balance point several times and calculate a mean value of d
(f) B1 linked reason: reduces the effect of random error (e.g. parallax error) in judging the balance point, giving a more precise value of d, oe
(f) Answer: E.g. use a fixed pointer viewed at eye level, reducing parallax error and giving a more precise value of d.
(g) B1 hazard, e.g. the weights could fall and injure feet, or the ruler could fall off the pivot
(g) B1 precaution, e.g. work over a soft surface away from the desk edge, or keep feet clear of the weights, or use masses no larger than needed
(g) Answer: Hazard: falling weights could injure feet. Precaution: keep feet clear of the weights / work over a soft surface.

Question 9

(a) B1 the force acting per unit area (normal to a surface), oe
(a) Answer: The force acting per unit area, at right angles to a surface.
(b) B1 pressure = force / area, accept p = F / A
(b) B1 unit: pascal (Pa), accept N/m²
(b) Answer: Pressure = force / area. Unit: pascal (Pa), equivalent to N/m^2.
(c) M1 substitutes: p = 6000 / 0.50
(c) A1 12000 Pa cao
(c) Answer: 12000 Pa

Question 10

(a) M1 substitutes: p = 600 / 0.040
(a) A1 15000 Pa cao
(a) Answer: 15000 Pa
(b) M1 substitutes: p = 600 / 0.00050
(b) A1 1200000 Pa cao (accept 1.2 x 10⁶ Pa)
(b) Answer: 1200000 Pa (1.2 x 10⁶ Pa)
(c) B1 the stiletto heels have a much smaller contact area than the trainers, oe
(c) B1 for the same force (weight), a smaller area produces a much greater pressure, which is more likely to damage the floor, oe
(c) Answer: The stiletto heels have a much smaller area, so for the same weight they produce a much greater pressure, which is more likely to dent the floor.
(d) B1 any sensible example, e.g. snowshoes, skis, wide tractor tyres, wide rucksack straps, a wide foundation for a building
(d) Answer: E.g. snowshoes, skis, or wide tractor tyres.

Question 11

(a) B1 they are all equal (the same), oe
(a) Answer: They are all equal.
(b) B1 pressure at a point in a fluid acts equally in all directions, oe
(b) B1 because all three sensors are at the same depth in the same fluid, they experience the same pressure, oe
(b) Answer: Pressure in a fluid acts equally in all directions, so at the same depth in the same fluid, all three sensors read the same.

Question 12

(a) M1 substitutes: p = 3.0 x 1000 x 9.8
(a) A1 29400 Pa cao
(a) Answer: 29400 Pa
(b) B1 ft 129400 Pa (their (a) + 100000)
(b) Answer: 129400 Pa
(c) B1 the pressure due to the water doubles, to 58800 Pa, oe
(c) B1 because pressure due to a column of liquid is directly proportional to depth (h), for a constant density and g, oe
(c) Answer: The pressure doubles to 58800 Pa, because pressure due to a liquid column is directly proportional to depth.
(d) B1 the pressure due to the liquid would be greater (increase), oe
(d) Answer: The pressure would be greater.

Question 13

(a) B1 hole Z (near the base) produces the jet that travels furthest
(a) B1 this shows that pressure in a liquid increases with depth, oe
(a) Answer: Hole Z; this shows that pressure in a liquid increases with depth.
(b) B1 there is a greater height (depth) of water above hole Z than above hole X, oe
(b) B1 a greater height of liquid produces a greater pressure (p = h x ρ x g), so a greater force pushes the water out of hole Z, oe
(b) Answer: There is a greater height of water above hole Z, producing a greater pressure and a greater force pushing the water out.
(c) B1 e.g. the size (diameter) of each hole must be the same, or the bottle must be completely full at the start each time
(c) Answer: E.g. the size (diameter) of each hole should be the same.
(d) B1 e.g. repeat each measurement, refilling the bottle to the same level each time, and calculate a mean distance for each hole
(d) B1 linked reason: reduces the effect of random error (e.g. from timing when the finger is removed, or how the landing distance is measured), giving more reliable results, oe
(d) Answer: E.g. repeat each measurement and take a mean, reducing random error and giving a more reliable result.
(e) B1 any sensible example, e.g. submarine and diving suit design must withstand higher pressure at greater depth; dam walls are built thicker at the bottom
(e) Answer: E.g. submarine hulls and diving suits must withstand greater pressure at greater depth.

Question 14

(a) B1 collisions between air molecules in the atmosphere and a surface, oe (accept: the weight of the column of air above a surface)
(a) Answer: Collisions between air molecules and a surface (caused by the weight of the air above).
(b) B1 atmospheric pressure decreases as altitude increases, oe
(b) Answer: Atmospheric pressure decreases as altitude increases.

Question 15

Level 1 (1-2): One or two isolated, correct statements are made, for example that atmospheric pressure decreases with height, with little or no reference to air molecules, weight of air, or density.
Level 2 (3-4): Identifies that there is less air (fewer air molecules, and less weight of air) above a point as altitude increases, and links this to fewer collisions and a decrease in pressure, but does not explain why the rate of decrease changes with altitude.
Level 3 (5-6): A clear, logically sequenced explanation: as altitude increases, there is less air (fewer molecules, less weight of air) above a given point, so fewer/weaker collisions occur per second and pressure decreases; and explains that because the atmosphere is denser near the ground, a small increase in height near sea level removes a relatively large amount of air, so pressure falls quickly at low altitude, whereas at high altitude the air is already thin, so climbing further removes much less air, so pressure falls more slowly.
Indicative content:
- atmospheric pressure at a point is caused by the weight of the column of air above that point
- as altitude (height) increases, there is less air (fewer air molecules) above a given point
- this means the weight of air above the point decreases as altitude increases
- fewer air molecules also means fewer collisions between air molecules and a surface per second
- pressure is force per unit area, so fewer or weaker collisions per second means lower pressure
- the atmosphere becomes less dense with increasing height (the air itself gets thinner)
- because the air is much denser near sea level, a small increase in height near the ground removes a relatively large amount of air (and weight) from above the point, so pressure decreases quickly at low altitude
- at high altitude the air is already thin, so the same increase in height removes much less mass of air, so pressure decreases more slowly
- therefore atmospheric pressure decreases with height, but not at a constant rate; it decreases increasingly slowly as altitude increases

Question 16

(a) B1 an object floats when the upthrust acting on it is equal to its weight, oe
(a) Answer: An object floats when the upthrust on it equals its weight.
(b) B1 the upthrust is equal to the weight of fluid displaced by the object, oe
(b) Answer: The upthrust is equal to the weight of fluid displaced.
(c) B1 the ball bearing has a small volume, so it displaces only a small weight of water, giving a small upthrust that is less than its weight, so it sinks, oe
(c) B1 the ship's hull has a very large volume (much of it filled with air), so it displaces a very large weight of water, giving an upthrust large enough to equal its (much greater) weight, so it floats, oe
(c) Answer: The ball bearing displaces little water, so its upthrust is less than its weight and it sinks. The ship's large hull displaces enough water to give an upthrust equal to its weight, so it floats.

Question 17

(a) B1 upthrust
(a) Answer: Upthrust.
(b) M1 upthrust = 6.0 - 4.0
(b) A1 2.0 N cao
(b) Answer: 2.0 N
(c) M1 ft mass of displaced water = their (b) / 9.8 = 2.0 / 9.8
(c) M1 ft volume = their mass / 1000
(c) A1 awrt 2.0 x 10^-4 m³ cao ft (accept 204 cm³ or 2.04 x 10^-4 m³)
(c) Answer: awrt 2.0 x 10^-4 m³
(d) M1 mass of block = weight in air / g = 6.0 / 9.8
(d) M1 ft density = their mass / their (c) volume
(d) A1 3000 kg/m³ cao ft
(d) Answer: 3000 kg/m³

Question 18

(a) B1 wide tracks give a much larger contact area with the ground than narrow wheels, for the same weight of machine, oe
(a) B1 since pressure = force / area, a larger area produces a smaller pressure on the ground, reducing the chance of the excavator sinking into soft ground, oe
(a) Answer: Wide tracks give a larger contact area, which produces a smaller pressure on the ground for the same weight, reducing the chance of sinking.
(b) M1 substitutes: moment = 2000 x 0.15
(b) A1 300 Nm cao
(b) Answer: 300 Nm
(c) B1 a smaller force is needed to produce the same (or a greater) moment, reducing the strain on the hydraulic system, oe
(c) Answer: A smaller force can produce the same moment, reducing the strain on the hydraulic system.
(d) B1 it is a rigid bar that turns about a pivot, using the principle of moments to make lifting a load easier, oe
(d) Answer: It is a rigid bar that turns about a pivot, using the principle of moments to make lifting easier.

Pressure, Moments and Levers - Worksheets, Questions and Revision