Cross-Disciplinary Integrated Question Practice for Year 9 AQA Engineering | Year 9 AQA 工程：跨学科综合题型训练

📚 Cross-Disciplinary Integrated Question Practice for Year 9 AQA Engineering | Year 9 AQA 工程：跨学科综合题型训练

Engineering is not a single subject; it is the art of blending science, mathematics, and design thinking to solve real-world problems. This article presents a series of integrated question practices specifically designed for Year 9 AQA Engineering students. Each section challenges you to connect knowledge from physics, materials, electronics, mechanics, and mathematical reasoning. By working through these problems, you will strengthen the cross-disciplinary thinking essential for any young engineer.

工程学并非一门单一学科，而是一门融合科学、数学与设计思维以解决现实问题的艺术。本文为 Year 9 AQA 工程学生量身打造了一系列综合题型训练。每个小节都要求你将物理、材料、电子、力学和数学推理的知识联系起来。通过练习这些问题，你将强化每一位年轻工程师所必备的跨学科思维能力。

1. Lever Systems and Mechanical Advantage | 杠杆系统与机械效益

Consider a first-class lever used in a lifting device. The effort arm measures 0.4 m and the load arm is 0.1 m. A worker applies an effort of 50 N downwards. Calculate the maximum load that can be lifted and determine the mechanical advantage of this system.

考虑一个用于提升装置的第一类杠杆。力臂长 0.4 m，负载臂长 0.1 m。一名工人施加 50 N 的向下的作用力。计算可以提升的最大负载，并确定该系统的机械效益。

kwise moment equals the anticlockwise moment. Here, effort × effort arm = load × load arm.

我们从力矩原理入手，当系统平衡时，顺时针力矩等于逆时针力矩。此处为作用力 × 力臂 = 负载 × 负载臂。

Rearranging: Load = (Effort × Effort Arm) / Load Arm = (50 N × 0.4 m) / 0.1 m.

整理得：负载 = (作用力 × 力臂) / 负载臂 = (50 N × 0.4 m) / 0.1 m。

Load = 200 N

负载 = 200 N

tiplies the input force by a factor of four. The problem links moments (physics) with ratio and proportion (mathematics), demonstrating cross-disciplinary application.

机械效益 (MA) 定义为负载与作用力之比，因此 MA = 200 N / 50 N = 4。这意味着杠杆将输入力放大了四倍。这道题目将力矩（物理）与比和比例（数学）联系起来，展现了跨学科的应用。

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2. Electronic Circuit Analysis: Ohm’s Law and Power | 电子电路分析：欧姆定律与功率

A simple circuit consists of a 12 V battery connected to a fixed resistor of 4 Ω. A student needs to find the current flowing and the power dissipated by the resistor. Later, a red LED with a forward voltage drop of 2.0 V is to be connected in series with the same voltage source, using a protective resistor. Determine the required resistance to limit the LED current to 20 mA.

一个简单电路由一个 12 V 电池和一个 4 Ω 的固定电阻组成。一名学生需要计算流过的电流以及电阻消耗的功率。随后，需要将一个正向电压降为 2.0 V 的红色 LED 与同一电压源串联，并使用保护电阻。为保证 LED 电流限制在 20 mA，求所需的阻值。

First part: apply Ohm’s Law V = I × R to find current I = V / R = 12 V / 4 Ω = 3 A.

第一部分：应用欧姆定律 V = I × R 求电流 I = V / R = 12 V / 4 Ω = 3 A。

Power P can be calculated using P = V × I or P = I² × R. Here, P = 12 V × 3 A = 36 W.

功率 P 可用 P = V × I 或 P = I² × R 计算。此处 P = 12 V × 3 A = 36 W。

Second part: the voltage across the resistor must be V_R = 12 V – 2.0 V = 10.0 V. The desired current is 20 mA = 0.020 A. Using Ohm’s Law again, R = V_R / I = 10.0 V / 0.020 A.

第二部分：电阻两端的电压必须为 V_R = 12 V – 2.0 V = 10.0 V。所需电流为 20 mA = 0.020 A。再次使用欧姆定律，R = V_R / I = 10.0 V / 0.020 A。

R = 500 Ω

R = 500 Ω

This exercise integrates basic electrical theory, semiconductor component knowledge, and unit conversion, demanding competence in both science and mathematics.

此练习综合了基础电学理论、半导体元器件知识以及单位换算，要求同时具备科学和数学能力。

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3. Material Selection: Density and Strength-to-Weight Ratio | 材料选择：密度与比强度

nium alloy (density 2700 kg/m³, yield strength 150 MPa). Both brackets will have the same volume of 2.0 × 10⁻⁴ m³. Calculate the mass of each bracket and compare the strength-to-weight ratio to recommend a material where weight saving is critical.

一个工程团队正在设计一个轻质支架。他们需要在低碳钢（密度 7800 kg/m³，屈服强度 250 MPa）和铝合金（密度 2700 kg/m³，屈服强度 150 MPa）之间选择。两个支架体积相同，均为 2.0 × 10⁻⁴ m³。计算每个支架的质量，并比较比强度，以推荐在重量要求苛刻的情况下的材料选择。

Mass is calculated from density (ρ) times volume (V): m_steel = 7800 kg/m³ × 2.0 × 10⁻⁴ m³ = 1.56 kg.

质量由密度 (ρ) 乘以体积 (V) 计算得出：m_steel = 7800 kg/m³ × 2.0 × 10⁻⁴ m³ = 1.56 kg。

m_aluminium = 2700 kg/m³ × 2.0 × 10⁻⁴ m³ = 0.54 kg. Aluminium is significantly lighter.

m_aluminium = 2700 kg/m³ × 2.0 × 10⁻⁴ m³ = 0.54 kg。铝合金明显更轻。

7800 kg/m³ ≈ 0.032 MPa·m³/kg. For aluminium alloy: 150 MPa / 2700 kg/m³ ≈ 0.056 MPa·m³/kg.

比强度可用屈服强度除以密度表示。钢材：250 MPa / 7800 kg/m³ ≈ 0.032 MPa·m³/kg。铝合金：150 MPa / 2700 kg/m³ ≈ 0.056 MPa·m³/kg。

Despite lower absolute strength, aluminium alloy offers a better strength-to-weight ratio. This typical materials selection problem blends science (material properties), mathematics (ratios and units), and sustainability thinking.

尽管绝对强度较低，铝合金却提供了更优的比强度。这道典型的材料选择问题融合了科学（材料性能）、数学（比值与单位）和可持续理念。

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4. Energy Efficiency of a Motorised Hoist | 电动葫芦的能效

A construction hoist lifts a 10 kg mass vertically by 2.0 m. The electric motor consumes 500 J of electrical energy from the supply during this lift. Ignore any energy stored in the system. Calculate the useful work done, the overall efficiency of the hoist, and explain where the remaining energy goes.

一台建筑用电动葫芦将 10 kg 的重物垂直提升 2.0 m。电机在此次提升过程中从电源消耗了 500 J 的电能。忽略系统中储存的任何能量。计算所做的有用功、葫芦的总效率，并解释剩余能量去了哪里。

against gravity: W = m × g × h. Using g = 9.8 m/s², W = 10 kg × 9.8 m/s² × 2.0 m = 196 J.

克服重力所做的有用功：W = m × g × h。取 g = 9.8 m/s²，W = 10 kg × 9.8 m/s² × 2.0 m = 196 J。

Efficiency η = (Useful Work Output / Total Energy Input) × 100% = (196 J / 500 J) × 100%.

效率 η =（有用功输出 / 总能量输入）× 100% =（196 J / 500 J）× 100%。

η = 39.2%

η = 39.2%

The remaining 304 J is mainly dissipated as thermal energy due to friction in bearings and gears, sound, and heat in the motor windings. This question links mechanics, energy transfers, and manufacturing tolerances, requiring evaluative skills.

剩下的 304 J 主要以热能形式耗散，原因包括轴承和齿轮的摩擦、噪音以及电机绕组发热。此题将力学、能量传递和制造公差联系起来，要求具备评价能力。

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5. Stress, Strain and Young’s Modulus in a Cable | 缆绳中的应力、应变与杨氏模量

A steel cable of original length 1.5 m and diameter 5.0 mm is used to hold a tensile load of 2000 N. The Young’s modulus for the steel is 200 GPa. Determine the tensile stress in the cable, the strain, and the extension under this load. Assume the cable behaves elastically.

一根原长 1.5 m、直径 5.0 mm 的钢缆用于承受 2000 N 的拉伸载荷。钢材的杨氏模量为 200 GPa。计算缆绳中的拉伸应力、应变以及在此载荷下的伸长量。假定缆绳处于弹性变形范围内。

Cross-sectional area A = π × (d/2)² = π × (2.5 × 10⁻³ m)² ≈ 1.9635 × 10⁻⁵ m².

横截面积 A = π × (d/2)² = π × (2.5 × 10⁻³ m)² ≈ 1.9635 × 10⁻⁵ m²。

Stress σ = Force / Area = 2000 N / 1.9635 × 10⁻⁵ m² ≈ 1.0186 × 10⁸ Pa or 101.9 MPa.

应力 σ = 力 / 面积 = 2000 N / 1.9635 × 10⁻⁵ m² ≈ 1.0186 × 10⁸ Pa，即 101.9 MPa。

ress / Young’s Modulus = σ / E = 1.0186 × 10⁸ Pa / 200 × 10⁹ Pa ≈ 5.093 × 10⁻⁴ (no units).

应变 ε = 应力 / 杨氏模量 = σ / E = 1.0186 × 10⁸ Pa / 200 × 10⁹ Pa ≈ 5.093 × 10⁻⁴（无量纲）。

material science, structural mechanics, and geometry are combined to predict real-world deformation.

伸长量 ΔL = ε × L₀ = 5.093 × 10⁻⁴ × 1.5 m = 7.64 × 10⁻⁴ m，约 0.76 mm。这里结合了材料科学、结构力学和几何学，以预测真实的变形量。

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6. Hydraulic Jack and Pascal’s Principle | 液压千斤顶与帕斯卡原理

A hydraulic jack has a small piston of area 2.0 cm² and a large piston of area 50 cm². A force of 100 N is applied to the small piston. Using Pascal’s principle, calculate the force exerted by the large piston and the distance moved by the large piston if the small piston moves down by 10 cm. Assume no leakage and incompressible fluid.

一台液压千斤顶的小活塞面积为 2.0 cm²，大活塞面积为 50 cm²。向小活塞施加 100 N 的力。利用帕斯卡原理，计算大活塞产生的力，以及如果小活塞下移 10 cm，大活塞将移动的距离。假定无泄漏且液体不可压缩。

= F₁/A₁ = F₂/A₂. So, F₂ = (A₂/A₁) × F₁ = (50 cm² / 2.0 cm²) × 100 N = 25 × 100 N = 2500 N.

帕斯卡原理指出压强等值传递：p = F₁/A₁ = F₂/A₂。因此，F₂ = (A₂/A₁) × F₁ = (50 cm² / 2.0 cm²) × 100 N = 25 × 100 N = 2500 N。

t: V = A₁ × d₁ = A₂ × d₂. Hence d₂ = (A₁ / A₂) × d₁ = (2.0 cm² / 50 cm²) × 10 cm = 0.4 cm.

液体排量恒定：V = A₁ × d₁ = A₂ × d₂。因此 d₂ = (A₁ / A₂) × d₁ = (2.0 cm² / 50 cm²) × 10 cm = 0.4 cm。

The system trades displacement for force, a principle used in car braking systems and jacks. The problem links fluid physics, mechanics, and proportional reasoning, all core to engineering systems.

该系统以位移换取力，这一原理应用在汽车制动系统和千斤顶中。此题将流体物理、力学和比例推理联系起来，这些均为工程系统的核心。

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7. Control Systems: Flowchart for a Conveyor Belt | 控制系统：传送带流程图

Design a control flowchart for a simple conveyor belt system. The system must start when a green push button is pressed, run the motor, and stop when a red button is pressed OR when an optical sensor detects that a box has reached the end of the belt. Additionally, an emergency stop should override both conditions and stop the motor immediately. Represent the logic using standard flowchart symbols.

为一个简易传送带系统设计控制流程图。按下绿色按钮时系统启动，电机运转；当按下红色按钮或光学传感器检测到箱子到达传送带末端时，系统停止。此外，一个急停按钮应能覆盖上述所有条件，立即停止电机。使用标准流程图符号呈现该逻辑。

The solution involves starting with a 'Start' oval, then a decision diamond for green button press. If yes, the motor turns on. The process then continuously checks two shutdown conditions in parallel: red button OR sensor activated. These are combined in a decision diamond. If either condition is true, the motor turns off; if emergency stop is activated at any moment, the motor stops regardless of the state. The flowchart contains an infinite loop until shutdown.

解决方案包括一个“开始”椭圆形，然后是一个判断绿色按钮按下的菱形框。如果是，电机启动。程序随后持续并行检查两个停机条件：红色按钮或传感器被触发。这两者在一个决策菱形中组合。如果任一条件为真，电机停止；如果急停在任意时刻被激活，无论状态如何均立即停止电机。该流程图包含一个无限循环，直到关机。

This exercise bridges systems engineering, programming logic, and electrical safety, reinforcing the importance of logic diagrams in automated systems.

该练习架起了系统工程、编程逻辑和电气安全之间的桥梁，强化了逻辑图在自动化系统中的重要性。

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8. Heat Sink Design and Surface Area Calculation | 散热器设计与表面积计算

spaced fins, each 50 mm high and 2 mm thick. The base thickness is 5 mm. Calculate the total exposed surface area available for convection cooling. Ignore the areas where the base and fins join. Use mm² for area and give the answer in mm².

一个平板铝散热器由一个 50 mm × 50 mm 的底座和五个等间距的散热片组成，每片高 50 mm、厚 2 mm。底座厚度为 5 mm。计算可用于对流冷却的总暴露表面积。忽略底座与散热片连接处的面积。使用 mm² 作为面积单位并给出答案。

es = 2 × (50 × 5) + 2 × (50 × 5) = 500 + 500 = 1000 mm². So total base exposed = 2500 + 2500 + 1000 = 6000 mm².

底座顶部表面积（扣除鳍片接触区域）：area_base_top = 50 × 50 = 2500 mm²。底座底部和侧面面积：底部 = 50 × 50 = 2500 mm²，四条边缘 = 2 × (50 × 5) + 2 × (50 × 5) = 500 + 500 = 1000 mm²。因此底座总暴露面积 = 2500 + 2500 + 1000 = 6000 mm²。

fin = 2 × (50 × 50) + (50 × 2) = 5000 + 100 = 5100 mm². Five fins = 5 × 5100 = 25500 mm².

每个散热片有两个侧面（50 mm × 50 mm）和一个顶部边缘（50 mm × 2 mm）。单片面积 = 2 × (50 × 50) + (50 × 2) = 5000 + 100 = 5100 mm²。五片 = 5 × 5100 = 25500 mm²。

00 mm². Connecting geometry with thermodynamics, students learn how surface area impacts heat dissipation in electronic devices.

总表面积 = 6000 + 25500 = 31500 mm²。将几何与热力学联系起来，学生可以了解表面积如何影响电子设备中的散热。

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9. Reading and Interpreting Engineering Drawings | 阅读与解读工程图纸

Study the following description of an engineering component: a rectangular block 80 mm long, 40 mm wide, and 20 mm high has a cylindrical through-hole of diameter 16 mm drilled centrally through its width (the 40 mm dimension). Interpret this to sketch the front, top, and right-side views in third-angle projection. Determine the minimum distance from the hole’s circumference to the nearest edge of the block.

阅读以下工程零部件的描述：一个长 80 mm、宽 40 mm、高 20 mm 的长方体块，在其宽度方向（40 mm 尺寸）中心钻有一个直径 16 mm 的通孔。据此在第三角投影中绘制前视图、俯视图和右视图。计算孔圆周到块体最近边缘的最小距离。

the centre to the nearest edge (front or back) is 20 mm. The radius of the hole is 8 mm, so the minimum distance from the hole circumference to the front edge is 20 mm – 8 mm = 12 mm.

在俯视图中，孔以一个圆呈现。孔的中心位于宽度的中点：距前边缘 40 mm / 2 = 20 mm。中心到最近边缘（前或后）的距离为 20 mm。孔的半径为 8 mm，因此孔圆周到前边缘的最小距离为 20 mm – 8 mm = 12 mm。

Dimensioning and projection are essential for communicating design intent. This question merges spatial geometry, measurement, and technical drawing conventions.

尺寸标注与投影对于传达设计意图至关重要。这道题目融合了空间几何、测量和技术制图规范。

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10. Integrated Problem: Mini Wind Turbine Blade Design | 综合问题：小型风力涡轮叶片设计

wood (density 160 kg/m³, Young’s modulus 3.5 GPa). The blade length is 0.3 m and its cross-section is roughly an aerofoil with chord 0.04 m. Estimate the mass of one blade, the kinetic energy available in the wind per second passing through the rotor swept area, and discuss how lift and drag forces are generated to turn the rotor. Use air density 1.2 kg/m³. Swept area for a single blade is approximated as 0.3 m length × 0.04 m chord.

你需要为一台小型教育用风力涡轮设计叶片。涡轮需在 5 m/s 的风速下运行。可用材料为轻木（密度 160 kg/m³，杨氏模量 3.5 GPa）。叶片长度为 0.3 m，其横截面近似为翼型，弦长 0.04 m。估算单个叶片的质量，计算每秒通过转子扫掠面积的风中可用动能，并讨论升力和阻力如何产生以驱动转子转动。取空气密度 1.2 kg/m³。单支叶片的扫掠面积近似为 0.3 m 长 × 0.04 m 弦长。

= 4.8 × 10⁻⁵ m³. Mass m = ρ × V = 160 kg/m³ × 4.8 × 10⁻⁵ m³ = 0.00768 kg (7.68 g). This light weight reduces centrifugal stress.

叶片体积近似为长度 × 弦长 × 平均厚度。假设平均厚度为 4 mm (0.004 m) 进行计算：V = 0.3 m × 0.04 m × 0.004 m = 4.8 × 10⁻⁵ m³。质量 m = ρ × V = 160 kg/m³ × 4.8 × 10⁻⁵ m³ = 0.00768 kg (7.68 g)。轻量化可降低离心应力。

Kin

Published by TutorHao | Year 9 工程 Revision Series | aleveler.com

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