Year 7 CIE Engineering: Interdisciplinary Integrated Question Practice | Year 7 CIE 工程：跨学科综合题型训练

📚 Year 7 CIE Engineering: Interdisciplinary Integrated Question Practice | Year 7 CIE 工程：跨学科综合题型训练

In Year 7 CIE Engineering, you are often asked to solve problems that combine knowledge from different subjects like mathematics, physics, and design. These interdisciplinary questions test your ability to apply concepts in real-world contexts, rather than just recalling isolated facts. This article provides a series of integrated practice questions with step-by-step guidance, helping you build the skills needed to think like an engineer.

在 Year 7 CIE 工程课程中，你经常需要解决那些融合了数学、物理和设计等多学科知识的题目。这类跨学科综合题考查的是你在现实情境中应用概念的能力，而不仅仅是记忆孤立的事实。本文提供一系列综合训练题型并给出分步指导，帮助你培养工程师的思维方式。

---

1. Calculating Moments with Real-Life Context | 结合生活情境的力矩计算

A seesaw has a child weighing 300 N sitting 2.0 m to the left of the pivot. You need to apply a force on the right side at a distance of 1.5 m from the pivot to balance the seesaw horizontally. What force must you apply? (Ignore the weight of the seesaw.)

一个跷跷板左侧有一个重 300 N 的小孩，距离支点 2.0 m。你需要在右侧距支点 1.5 m 处施加一个力使跷跷板水平平衡。你需要施加多大的力？（忽略跷跷板自重。）

For equilibrium, the clockwise moment must equal the anticlockwise moment. The moment of a force is calculated as Moment = Force x perpendicular distance from the pivot.

为达到平衡，顺时针力矩必须等于逆时针力矩。力矩的计算公式为 力矩 = 力 × 到支点的垂直距离。

The child’s weight is 300 N and distance is 2 m, so anticlockwise moment = 300 N × 2 m = 600 N m.

小孩体重 300 N，距离 2 m，因此逆时针力矩 = 300 N × 2 m = 600 N m。

Let the required force on the right be F. Its clockwise moment = F × 1.5 m.

设右侧所需力为 F。其顺时针力矩 = F × 1.5 m。

Equating the moments: 600 = 1.5 F, therefore F = 600 ÷ 1.5 = 400 N.

令力矩相等：600 = 1.5 F，因此 F = 600 ÷ 1.5 = 400 N。

Notice that a smaller distance (1.5 m < 2.0 m) requires a larger force (400 N > 300 N) to balance, which makes sense mechanically.

注意，较小的距离（1.5 m < 2.0 m）需要较大的力（400 N > 300 N）来平衡，这在力学上是合理的。

---

2. Material Selection Based on Properties | 基于性能的材料选择

Your engineering team must choose a material for a bicycle frame that needs to be lightweight, strong, and stiff enough to withstand riding loads. The table below gives data for three candidate materials.

你的工程团队需要为自行车车架选择一种材料，要求轻便、坚固且足够的刚性以承受骑行载荷。下表给出了三种备选材料的数据。

Material	Density (kg/m³)	Tensile Strength (MPa)	Relative Cost
Mild Steel	7850	400	Low
Aluminium Alloy	2700	300	Medium
Carbon Fibre Composite	1600	600	High

Using the data, recommend a material. Justify your choice based on the engineering requirements.

使用这些数据，推荐一种材料。根据工程要求说明你的选择理由。

First, consider ‘lightweight’: a lower density gives a lighter frame for the same volume. Carbon fibre has the lowest density (1600 kg/m³), followed by aluminium (2700 kg/m³), then steel (7850 kg/m³).

首先考虑”轻便”：相同体积下，密度越低车架越轻。碳纤维密度最低（1600 kg/m³），其次是铝（2700 kg/m³），钢最重（7850 kg/m³）。

Next, ‘strength’: tensile strength indicates how much pulling force the material can withstand before breaking. Carbon fibre is strongest (600 MPa), steel is 400 MPa, aluminium is 300 MPa.

其次”强度”：抗拉强度表示材料断裂前能承受的拉力。碳纤维最强（600 MPa），钢为 400 MPa，铝为 300 MPa。

Aluminium offers a good balance: it is much lighter than steel and has adequate strength for a bicycle frame. Carbon fibre is the best performer but its high cost often limits its use for mass production.

铝提供了良好的平衡：它比钢轻得多，且具有足够的强度用于自行车车架。碳纤维性能最佳，但其高成本通常限制了在大规模生产中的使用。

Therefore, aluminium alloy is a sensible choice, combining acceptable strength, low weight, and reasonable cost.

因此，铝合金是一个明智的选择，兼具可接受的强度、低重量和合理的成本。

---

3. Simple Electrical Circuit Calculations | 简单电路计算

A 3.0 V battery is connected to a single 10 Ω resistor. Calculate the current flowing in the circuit. Then, a second 5.0 Ω resistor is connected in series with the first. What is the new current?

一个 3.0 V 的电池连接到一个 10 Ω 的电阻上。计算电路中的电流。然后，在第一个电阻上串联一个 5.0 Ω 的电阻。新的电流是多少？

Ohm’s law relates voltage (V), current (I) and resistance (R): I = V ÷ R.

欧姆定律将电压 (V)、电流 (I) 和电阻 (R) 联系起来：I = V ÷ R。

For the initial circuit, I = 3.0 V ÷ 10 Ω = 0.3 A.

对于初始电路，I = 3.0 V ÷ 10 Ω = 0.3 A。

When resistors are in series, total resistance R_total = R₁ + R₂ = 10 Ω + 5.0 Ω = 15 Ω.

当电阻串联时，总电阻 R_total = R₁ + R₂ = 10 Ω + 5.0 Ω = 15 Ω。

The new current I = 3.0 V ÷ 15 Ω = 0.2 A. Adding more resistance reduces the current.

新的电流 I = 3.0 V ÷ 15 Ω = 0.2 A。增加电阻会减小电流。

---

4. Energy Efficiency of a Ramp System | 斜面系统的能量效率

A ramp is used to lift a 200 N crate into a truck bed 1.0 m high. The ramp is 4.0 m long, and the worker must push with a constant force of 75 N. Determine the useful work output, the work input, and the efficiency of the ramp.

使用一个斜面将一个 200 N 的板条箱提升到 1.0 m 高的卡车车厢。斜面长 4.0 m，工人必须用 75 N 的恒力推动。求有用输出功、输入功及斜面的效率。

Useful work output (against gravity) = weight × vertical height = 200 N × 1.0 m = 200 J.

有用输出功（克服重力）= 重量 × 垂直高度 = 200 N × 1.0 m = 200 J。

Work input (work done by the worker) = applied force × ramp length = 75 N × 4.0 m = 300 J.

输入功（工人所做的功）= 施加的力 × 斜面长度 = 75 N × 4.0 m = 300 J。

Efficiency = (useful output ÷ input) × 100% = (200 J ÷ 300 J) × 100% ≈ 66.7%.

效率 = （有用输出 ÷ 输入）× 100% = （200 J ÷ 300 J）× 100% ≈ 66.7%。

Some energy is wasted, probably due to friction between the crate and the ramp surface.

一些能量被浪费了，可能是由于箱子与斜面之间的摩擦所致。

---

5. Designing a Triangular Truss Bridge | 设计三角形桁架桥

You need to design a simple bridge to span a 2 m gap using wooden rods and string. The bridge must support a central load of 500 N without collapsing. Sketch (in your mind) a truss structure and explain why triangular shapes are used.

你需要用木棒和绳子设计一座简易桥，跨越 2 m 的缺口。该桥必须能承受中央 500 N 的载荷而不垮塌。在脑海中草绘一个桁架结构，并解释为什么要使用三角形形状。

In a truss bridge, triangular units are assembled because a triangle is the only polygon that cannot be deformed without changing the length of its sides. This makes the structure rigid.

在桁架桥中，组装三角形单元是因为三角形是唯一一种不改变边长就无法变形的多边形。这使得结构具有刚性。

When a load is applied at the top joint of a triangle, the two sloping members experience compression, pushing outward against the supports while the bottom horizontal member is in tension, preventing the legs from spreading.

当载荷施加在三角形顶部节点时，两个斜杆承受压力，向外推挤支座，而底部水平杆承受拉力，防止两边张开。

Using multiple connected triangles, the forces are distributed evenly throughout the structure, making it much stronger than a simple beam of the same material.

通过使用多个相连的三角形，力被均匀分布到整个结构中，使其比相同材料的简支梁坚固得多。

For a 2 m span, a Warren truss (a row of equilateral triangles) with a roadway on top or bottom would be an efficient design.

对于 2 m 的跨度，一个顶部或底部带桥面的华伦式桁架（一排等边三角形）将是一个高效的设计。

---

6. Heat Transfer and Insulation in a Flask | 保温瓶中的热传递与隔热

A thermos flask is designed to keep a drink hot. Explain how its features reduce heat loss by conduction, convection, and radiation. Use engineering vocabulary.

保温瓶的设计是为了保持饮品的热量。请解释它的特征如何减少传导、对流和辐射造成的热损失。使用工程词汇。

The flask has a double-walled container with a vacuum between the walls. Since conduction and convection require particles, the vacuum largely stops heat transfer by these two mechanisms.

保温瓶有一个双层壁容器，两壁之间为真空。由于传导和对流都需要粒子，真空在很大程度上阻止了这两种机制的热传递。

The inner and outer surfaces are often silvered (shiny). Shiny surfaces are poor radiators and poor absorbers of thermal radiation, so the inner wall reflects heat back into the liquid, and the outer wall reflects external heat away.

内壁和外壁通常镀银（光亮）。光亮的表面是热辐射的不良发射体和吸收体，因此内壁将热量反射回液体，外壁将外部热量反射走。

The stopper (usually plastic or cork) is an insulator that reduces heat loss by conduction and convection at the opening. Plastic and cork have low thermal conductivity.

瓶塞（通常是塑料或软木）是绝缘体，可减少开口处的传导和对流热损失。塑料和软木的热导率很低。

All these features work together to maintain the liquid’s temperature for a long time.

所有这些特征共同作用，使液体长时间保持温度。

---

7. Mechanical Advantage of a Pulley System | 滑轮系统的机械效益

A simple block-and-tackle pulley system has 4 rope sections supporting the lower moving block. You want to lift a load of 200 N. Ignoring friction and rope weight, what is the minimum effort force you need to pull?

一个简单的差动滑轮系统有 4 段绳子承载下部的动滑轮。你想提起一个 200 N 的负载。忽略摩擦和绳子重量，你至少需要施加多大的拉力？

The mechanical advantage (MA) of an ideal pulley system equals the number of rope sections supporting the load. Here, MA = 4.

理想滑轮系统的机械效益 (MA) 等于承载负载的绳子段数。此处 MA = 4。

Mechanical advantage is defined as MA = Load / Effort. Rearranging: Effort = Load / MA.

机械效益定义为 MA = 负载 / 拉力。重排后得：拉力 = 负载 / MA。

Therefore, Effort = 200 N ÷ 4 = 50 N. You only need to pull with 50 N of force to lift the 200 N load.

因此，拉力 = 200 N ÷ 4 = 50 N。你只需用 50 N 的力就可以提起 200 N 的负载。

In reality, friction reduces the actual MA, so the effort will be slightly larger, but the design greatly reduces the force required.

实际上，摩擦会降低实际的机械效益，因此拉力会稍大一些，但该设计大大减少了所需的力量。

---

8. Integrated Mini-Project: Egg Drop Lander | 综合小项目：落蛋着陆器

Design an ‘egg drop lander’ that will protect a raw egg from breaking when dropped from a height of 3 m. You are allowed to use only 10 straws, 1 m of adhesive tape, and one sheet of newspaper. Explain your design using concepts from science and engineering.

设计一个”落蛋着陆器”，要求从 3 m 高处落下时保护生鸡蛋不破裂。你只能使用 10 根吸管、1 m 胶带和一张报纸。运用科学与工程概念解释你的设计。

A successful design must reduce the impact force on the egg. From physics, we know that impact force is related to the change in momentum and the time over which the egg stops: force = change in momentum ÷ time.

成功的设计必须减小鸡蛋所受的冲击力。从物理学中我们知道，冲击力与动量的变化以及鸡蛋停止所需的时间有关：力 = 动量变化 ÷ 时间。

By increasing the stopping time, the impact force decreases. You can achieve this by building a crumple zone – a structure that deforms on impact. Paper crumpled into a cone or ball and held by straws can absorb energy by collapsing slowly.

通过增加停止时间，冲击力会减小。你可以通过构建一个”压缩区”来实现这一点——一种在撞击时变形的结构。报纸揉成锥形或球形并由吸管支撑，可以通过缓慢坍塌来吸收能量。

Straws can be used to build a strong, lightweight protective cage that holds the egg in the centre. A triangular pyramid (tetrahedron) frame distributes impact forces along multiple members. The egg should be placed in the centre, surrounded by crumpled paper to cushion it in all directions.

吸管可用来建造一个坚固、轻便的保护笼，将鸡蛋固定在中央。三角形金字塔（四面体）框架将冲击力沿多个杆件分散。鸡蛋应放置在中央，周围用皱纸包围，提供全方位的缓冲。

This design integrates material properties (paper for energy absorption, straws for rigidity), structural geometry (triangles for stability), and physics (impulse and force reduction), demonstrating true interdisciplinary thinking.

这个设计整合了材料特性（纸张吸能、吸管刚性）、结构几何（三角形稳定性）和物理学（冲量与减力），展示了真正的跨学科思维。

Published by TutorHao | Engineering Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）