Moments and Equilibrium in GCSE Maths | GCSE 数学：力矩与平衡 考点精讲

📚 Moments and Equilibrium in GCSE Maths | GCSE 数学：力矩与平衡 考点精讲

In GCSE Mathematics (and the mechanics topics that often appear on higher-tier papers), understanding moments and equilibrium is essential for solving problems involving forces, levers, and balance. This article breaks down the key concepts, formulas, and problem-solving strategies you need to master, with plenty of worked examples and exam tips.

在 GCSE 数学（以及更高阶试卷中常出现的力学题目）中，理解力矩与平衡是解决涉及力、杠杆和平衡问题的关键。本文将梳理你必须掌握的核心概念、公式和解题策略，并提供详细的例题与备考技巧。

1. What Is a Moment? | 什么是力矩？

A moment is the turning effect of a force about a point (called the pivot or fulcrum). It depends on two things: the size of the force and the perpendicular distance from the pivot to the line of action of the force. The unit of a moment is the newton-metre (Nm). Moment = Force × Perpendicular distance from pivot.

力矩是力对某一点（称作支点或转轴）产生的转动效应。它取决于两个因素：力的大小以及从支点到力作用线的垂直距离。力矩的单位是牛顿·米（Nm）。力矩 = 力 × 支点到力作用线的垂直距离。

If you push a door near its hinges, it’s hard to open; if you push far from the hinges, it’s easy. A larger force or a longer distance produces a greater turning effect.

如果你在靠近门铰链的位置推门，门很难打开；如果在远离铰链处推门，就很容易。更大的力或更长的距离都会产生更大的转动效应。

2. The Moment Equation and Units | 力矩公式与单位

The formula you must remember is: M = F × d, where M is the moment (Nm), F is the force (N), and d is the perpendicular distance from the pivot to the line of action of the force (m). Always a straight line at 90° to the force direction.

你必须记住的公式是：M = F × d，其中 M 是力矩（Nm），F 是力（N），d 是支点到力作用线的垂直距离（m）。这个距离一定是沿着与力的方向成 90° 的直线。

If the force is not already perpendicular, you must resolve the force or find the perpendicular component. In GCSE questions, the distance given in a diagram is usually the perpendicular distance, but watch out for slanted forces where you might need to use trigonometry to find the perpendicular distance.

如果力并不是垂直的，你需要分解力或找到其垂直分量。在 GCSE 题目中，示意图里给出的距离通常就是垂直距离，但要注意斜向的力，那时可能需要用三角学来求垂直距离。

3. Clockwise and Anticlockwise Moments | 顺时针力矩与逆时针力矩

A moment can cause rotation in two directions: clockwise (CW) and anticlockwise (ACW). In equilibrium problems, we usually assign one direction as positive and the other as negative. It doesn’t matter which you choose, but be consistent.

力矩可以引起两个方向的转动：顺时针（CW）和逆时针（ACW）。在处理平衡问题时，我们通常将一个方向设为正，另一个方向设为负。你选哪个作为正方向并不重要，但要前后一致。

For a see-saw, a weight placed on the left side causes an anticlockwise moment about the centre pivot if it pushes down. A weight on the right side causes a clockwise moment. The principle of moments states that for a system to be in balance, the sum of clockwise moments must equal the sum of anticlockwise moments about any pivot.

以跷跷板为例，放在左侧的重物如果向下压，会对中心支点产生逆时针力矩；右侧的重物会产生顺时针力矩。力矩原理指出，若系统处于平衡，则对于任意支点，顺时针力矩之和等于逆时针力矩之和。

4. The Principle of Moments | 力矩原理

The Principle of Moments: For an object in rotational equilibrium, total clockwise moments = total anticlockwise moments about any given point. This is the key equation you will use to find unknown forces or distances.

力矩原理：对于处于转动平衡的物体，以任意给定点为支点，顺时针力矩的总和等于逆时针力矩的总和。这是你用来求未知力或未知距离的核心方程。

Mathematically: Σ M_cw = Σ M_acw. When setting up the equation, include all forces that produce a moment about your chosen pivot. Forces acting through the pivot produce zero moment, so they can be ignored.

数学表达为：Σ M_cw = Σ M_acw。列方程时，将你所选支点周围所有产生力矩的力都考虑进去。作用线通过支点的力产生的力矩为零，可以忽略。

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5. Conditions for Complete Equilibrium | 完全平衡的条件

For a rigid body to be in complete equilibrium, two conditions must be met: (1) The resultant force in any direction must be zero (translational equilibrium). (2) The resultant moment about any point must be zero (rotational equilibrium). In many GCSE moments problems, you only need the moment condition, but sometimes vertical force balance is also required to find a reaction force at the pivot.

要使刚体处于完全平衡，必须满足两个条件：（1）任何方向上的合力都必须为零（平动平衡）；（2）对任意点的合力矩都必须为零（转动平衡）。在许多 GCSE 力矩问题中，你只需要使用力矩条件，但有时也需要通过竖直方向的力平衡来求支点处的反作用力。

For example, a uniform beam supported at two ends: the upward reaction forces must sum to equal the total downward weight. Then you can take moments about one support to find the other reaction.

例如，一根在两端都有支撑的均匀横梁：向上的支反力之和必须等于向下的总重量。然后你可以对其中一个支点取矩，求出另一个支反力。

6. Choosing a Pivot | 如何选择支点

You can take moments about any point you like. A clever choice simplifies the problem. Pick a point where an unknown force acts: that force will have zero distance from the pivot, so its moment is zero, and it disappears from the equation.

你可以对任意一点取矩。巧妙地选择支点能简化问题。选择有未知力作用的点：该力与支点的距离为零，因此其力矩为零，在方程中就会消失。

If a beam has two unknown reaction forces, take moments about the point where one reaction acts. This eliminates that unknown, allowing you to solve for the other reaction directly.

如果横梁有两个未知的支反力，对其中一个支反力作用点取矩，就能消去这个未知量，从而直接求出另一个支反力。

7. Uniform Rods, Beams, and Centre of Mass | 均匀杆、横梁与质心

For a uniform rod or beam, the weight acts at its centre. This is its centre of mass. In diagrams, the weight is drawn as a downward arrow from the midpoint. If the beam is non-uniform, the question might give you the position of the centre of mass or ask you to find it.

对于均匀杆或横梁，重力作用在其中心点，这就是质心。在示意图中，重力被画为一个从中点向下的箭头。如果横梁不是均匀的，题目可能会给出质心的位置，或者要求你求出它。

A common problem: a uniform beam of length L and weight W rests on two supports. The weight acts at L/2 from one end. Use this to take moments.

常见问题：一根长度为 L、重量为 W 的均匀横梁架在两个支座上。重力作用在距一端 L/2 处。利用这一点来列力矩方程。

8. Worked Example 1 – See-Saw | 例题 1——跷跷板

A uniform see-saw is 4 m long and pivoted at its centre. A child of weight 300 N sits 1.5 m to the left of the pivot. Where must a second child of weight 400 N sit on the right side to balance the see-saw?

一个均匀的跷跷板长 4 米，支点在中心。一个体重 300 N 的儿童坐在支点左侧 1.5 米处。另一个体重 400 N 的儿童必须坐在右侧什么位置才能使跷跷板平衡？

Solution: Let the distance from pivot to second child be x m. Anticlockwise moment = 300 × 1.5 = 450 Nm. Clockwise moment = 400 × x. For equilibrium: 400x = 450 ⇒ x = 1.125 m. The second child sits 1.125 m to the right of the pivot.

解答：设第二个儿童到支点的距离为 x 米。逆时针力矩 = 300 × 1.5 = 450 Nm。顺时针力矩 = 400 × x。根据平衡条件：400x = 450 ⇒ x = 1.125 m。第二个儿童应坐在支点右侧 1.125 米处。

9. Worked Example 2 – Beam with Two Supports | 例题 2——有两个支座的横梁

A uniform beam AB of length 6 m and weight 200 N rests on two supports at A and C, where C is 1 m from B. Find the reaction forces at A and C.

一根长 6 米、重 200 N 的均匀横梁 AB 架在 A 和 C 两个支座上，C 距 B 为 1 米。求 A 和 C 处的支反力。

Draw the beam: A at left end, B at right end. Support C is at 5 m from A (since 6 – 1 = 5). Weight acts at centre, 3 m from A. Let reactions be R_A (up) and R_C (up). Take moments about A: clockwise moments = (weight × 3) + (R_C × 5)? No, careful: R_C at C is upward, it creates an anticlockwise moment about A if we consider CW=positive? Better to state: Sum of ACW moments = Sum of CW moments about A. Weight (200 N) tries to rotate clockwise about A, R_C tries to rotate anticlockwise. So 200 × 3 = R_C × 5. Solve: R_C = (200×3)/5 = 120 N. Then vertical equilibrium: R_A + R_C = 200 ⇒ R_A = 80 N.

画出横梁：A 在左端，B 在右端。支座 C 位于距 A 5 米处（因为 6 – 1 = 5）。重力作用在中心，距 A 3 米。设支反力为 R_A（向上）和 R_C（向上）。对 A 点取矩：重力 200 N 关于 A 产生顺时针力矩，R_C 产生逆时针力矩。根据力矩平衡：200 × 3 = R_C × 5。解得：R_C = (200×3)/5 = 120 N。再根据竖直方向力平衡：R_A + R_C = 200 ⇒ R_A = 80 N。

Always check: take moments about C to verify: R_A × 5 (anticlockwise) vs weight × (distance from C to weight = 2 m) clockwise? Actually weight is 200 N acting 2 m to left of C, so its moment about C = 200 × 2 = 400 Nm ACW? Wait: if we take C, R_A is upward at A, distance 5 m left, so R_A creates clockwise or anticlockwise? Imagine pivot at C, force R_A upward at left end → it would rotate the beam anticlockwise. Weight acts downward 2 m to left of C → it would rotate the beam clockwise. So check: R_A × 5 = 80 × 5 = 400 Nm ACW, weight moment = 200 × 2 = 400 Nm CW. They balance, correct.

验证时对 C 取矩：R_A × 5（逆时针）= 80 × 5 = 400 Nm；重力 × 2（顺时针）= 200 × 2 = 400 Nm。平衡正确。

10. Dealing with Slanted Forces | 处理倾斜的力

When a force is applied at an angle, you must use the perpendicular distance from the pivot to the line of action. Either find that perpendicular distance using trigonometry (e.g., d = lever length × sin θ), or resolve the force into components and then multiply the perpendicular component by the lever distance.

当施加的力有角度时，你必须使用从支点到力作用线的垂直距离。要么用三角学找到这个垂直距离（如 d = 杆长 × sin θ），要么将力分解为分量，然后用垂直分量乘以杠杆距离。

For example, a door handle is pulled with a force F at an angle to the door. Only the component perpendicular to the door creates a moment about the hinges. That component = F × sin (angle between force and door).

例如，用一个与门板成一定角度的力 F 拉门把手。只有垂直于门板的分量才会对铰链产生力矩。该分量 = F × sin（力与门板之间的夹角）。

11. Common Exam Pitfalls | 常见考试陷阱

• Forgetting to use perpendicular distance. When the force is not perpendicular, you must calculate the perpendicular distance.
• Confusing units: Moments are in Nm, not Joules.
• Not including the weight of the beam itself when it is uniform (acts at centre).
• Sign errors: Be consistent with clockwise as positive and anticlockwise as negative, or equate sums directly.
• Ignoring reaction forces when taking moments about a point that eliminates them – that’s fine, but then use vertical force balance to find the eliminated reaction later.

• 忘记使用垂直距离。当力不垂直时，必须计算垂直距离。
• 混淆单位：力矩的单位是 Nm，而不是焦耳。
• 当横梁均匀时忘记考虑横梁自身的重量（重力作用在中心）。
• 符号错误：要一致地设定顺时针为正、逆时针为负，或直接列出等式。
• 在对某点取矩时忽略了那些被消去的支反力——这没问题，但要记得之后用竖直力平衡求出被消去的支反力。

12. Exam-Style Practice Question | 考试式练习题

A non-uniform plank AB of length 5 m is balanced on a pivot placed at its centre. When a 10 kg mass is placed at A, the plank balances. When the mass is moved to B, an additional 6 kg must be added at A to maintain balance. Find the distance from A to the centre of mass of the plank. (Take g = 10 N/kg to convert masses to weights.)

一块长 5 米的非均匀木板 AB，支点置于其中心位置，当将一个 10 kg 的重物放在 A 端时，木板平衡。当将该重物移到 B 端时，需在 A 端再加 6 kg 才能保持平衡。求木板质心到 A 端的距离。（取 g = 10 N/kg 将质量转换为重量。）

Let the distance from A to centre of mass be x m. Since pivot is at centre (2.5 m from A), the distance from pivot to centre of mass is |2.5 – x|. Convert mass to weight: 10 kg → 100 N, 6 kg → 60 N. Let weight of plank be W (unknown). First case: 100 N at A (2.5 m from pivot) produces anticlockwise moment? If A is left end, pivot in middle, weight at A is left of pivot, so it produces an anticlockwise moment. The plank’s weight W acts at centre of mass (x from A, so distance from pivot = 2.5 – x if x < 2.5). Moment balance: 100 × 2.5 = W × (2.5 - x). Second case: mass 100 N at B (2.5 m right of pivot) – now it produces clockwise moment. Extra 60 N at A (left, anticlockwise). So: (60 × 2.5) + anticlockwise from W? Wait, be systematic. We take moments about pivot. In first scenario: 100 N at A (anticlockwise), W (if x < 2.5) produces clockwise moment (since centre of mass is to left of pivot? Actually if x < 2.5, then centre of mass is left of pivot, weight W acts downward, so it creates an anticlockwise moment? Imagine pivot at centre; a weight left of pivot, pulling down, would tip the beam down on left → that's anticlockwise rotation. So both 100 N and W act anticlockwise? That would not balance. Thus x must be > 2.5, so centre of mass is right of pivot, producing clockwise moment. Let’s assume x > 2.5, then distance from pivot to c.o.m. = x – 2.5. First case: 100 N left (ACW), W right (CW). Balance: 100 × 2.5 = W × (x – 2.5). Second case: 100 N right (CW), 60 N left (ACW), W right (CW). Then: 60 × 2.5 + ??? Actually we need both sides: ACW total = 60 × 2.5, CW total = (100 + W) × 2.5? No, distances: 100 N is at B, distance 2.5 m from pivot; W acts at (x – 2.5) from pivot. So equation: 60 × 2.5 = 100 × 2.5 + W × (x – 2.5). Simplify: 150 = 250 + W(x – 2.5) → W(x – 2.5) = -100, impossible. So we must interpret differently. Rethink: When mass moved to B, additional 6 kg at A to maintain balance. That means the 10 kg at B (right) is too heavy, so need extra left weight. So W must be left of centre? Let’s set x < 2.5. First case: 100 N at A (left), W (left) – both ACW, can't balance unless pivot is not centre? The problem says "balanced on a pivot placed at its centre", so pivot is fixed at centre, plank balanced when 10 kg at A, meaning the weight of plank must be on right side. So x > 2.5, but then first case: 100 left, W right: 100×2.5 = W×(x-2.5). Second case: 100 right, W right, 60 left: 60×2.5 = 100×2.5 + W(x-2.5). Plug in W(x-2.5) = 250 from first? Actually from first: W(x-2.5) = 250. Then second: 150 = 250 + 250? No, that’s wrong, 60×2.5=150, RHS: 100×2.5=250 + (W(x-2.5)=250) = 500, not equal. This contradiction means my sign convention is flawed. Let’s assign ACW positive, CW negative. Take moments about pivot. Let anticlockwise positive. First scenario: 100 N at A (distance 2.5 m left), moment = +100×2.5. Plank weight W acts at distance (x-2.5) from pivot, if x>2.5, it’s to the right, force down → that produces clockwise (negative) moment: -W(x-2.5). Sum = 0 → 250 – W(x-2.5) = 0 → W(x-2.5) = 250. Second scenario: 10 kg at B (right, 2.5 m) → moment = -100×2.5 = -250. Additional 6 kg at A → +60×2.5 = +150. Plank weight still -W(x-2.5). Sum = 150 – 250 – W(x-2.5) = 0 → -100 – W(x-2.5)=0 → W(x-2.5) = -100, impossible. So x must be < 2.5. Then in first scenario: 100 N at A (left) + ACW. Plank weight W at distance (2.5 - x) from pivot, left side → both ACW, cannot sum to zero unless W negative. So there must be a different pivot condition? Wait, "balanced on a pivot placed at its centre" might mean the pivot is at the centre of the plank (2.5 m from ends), but it's not the balance point originally; the plank alone might not balance at centre because it's non-uniform. By adding 10 kg at A, it balances. So initially, without added mass, the plank would tip. So the centre of mass is not at centre. Let x be distance from A to c.o.m. The pivot is fixed at 2.5 m from A. When 10 kg at A, it balances, so the moment of plank weight about pivot must be opposite to the moment of added weight. Let's take pivot as reference. Suppose c.o.m. is to the right of pivot (x > 2.5). Then plank weight W creates clockwise moment. 10 kg at A (left) creates anticlockwise moment. For balance: 100×2.5 = W×(x-2.5). That’s equation (1). Now when 10 kg moved to B (right), it adds to clockwise moment, so we need extra anticlockwise moment, which is provided by additional 6 kg at A. So total anticlockwise = 60×2.5 = 150, total clockwise = 100×2.5 + W(x-2.5). Equation (2): 150 = 250 + W(x-2.5). But from (1), W(x-2.5)=250, so 150 = 250+250=500, impossible. So my interpretation that c.o.m. is right leads to inconsistency. Try c.o.m. to the left of pivot (x < 2.5). Then first scenario: 10 kg at A (left), both W and 100 N are left of pivot → both produce anticlockwise moments (if we consider pivot), cannot sum to zero unless there is a clockwise moment from something else. But there isn't. So x cannot be <2.5. Wait, what if we consider the pivot at centre, and the plank's weight acts at its centre of mass, which might be to the left, but then the plank alone would tip left, so adding 10 kg at A would increase left tipping, not balance. So the 10 kg must counteract an existing imbalance. If the plank's c.o.m. is to the right, it tips right. Adding 10 kg left balances it. That works: W (right) vs 100 (left). So x>2.5. Then second scenario: move 10 kg to right → now both W and 10 kg are right, total clockwise increases, so we need extra left weight. We add 6 kg left. That gives 60 left vs. (100 + W) right. Equation: 60×2.5 = (100+W)×2.5? But W acts at (x-2.5), not 2.5. Correct: total clockwise moment = 100×2.5 + W(x-2.5), total ACW = 60×2.5. So 150 = 250 + W(x-2.5) => W(x-2.5) = -100. Impossible. So maybe the additional 6 kg is added at A together with the 10 kg still at B? The phrasing: “When the mass is moved to B, an additional 6 kg must be added at A to maintain balance.” So now we have 10 kg at B and an extra 6 kg at A. So left side: 6 kg = 60 N at A, right side: 10 kg = 100 N at B, plus plank weight W. ACW from 60 N: 60×2.5 = 150. CW from 100 N: 100×2.5 = 250, CW from W: W(x-2.5). Total CW = 250 + W(x-2.5). Balance: 150 = 250 + W(x-2.5) -> W(x-2.5) = -100. Still negative. That would imply W is negative or x-2.5 negative, i.e., c.o.m. left of pivot. So let’s try x<2.5. Then in first scenario, 10 kg at A (left) produces ACW 250. W at (2.5-x) left of pivot also ACW. Sum ACW = 250 + W(2.5-x). No CW, so can't balance. Unless pivot is not centre? The question says "balanced on a pivot placed at its centre", so pivot is fixed at centre. A non-uniform plank on a central pivot will tip unless extra weight added. So to balance, the extra weight must provide a moment opposite to that of the plank's weight. So if plank's c.o.m. is to the right, extra weight left balances. That we tried. But second condition contradictions. Let's numerically solve without assuming sign. Let x be distance from A to c.o.m. Pivot at 2.5 m from A. Let anticlockwise positive. Moment of a force = F × (distance from pivot), sign determined by direction. I'll write equation: Sum of moments = 0. Force downward, so if force is to left of pivot, it tends to rotate anticlockwise (positive); if right of pivot, clockwise (negative). For plank weight W (unknown), its moment = W × (2.5 - x) if x<2.5? Actually distance from pivot to c.o.m. = |2.5 - x|. Sign: if x<2.5 (c.o.m. left), moment positive; if x>2.5, moment negative. So moment from W = W × (2.5 – x). This expression will be positive if x<2.5, negative if x>2.5. That’s neat. Similarly, 10 kg at A (distance from pivot = 2.5 m, left) => moment = +100×2.5 = +250. 10 kg at B (right, 2.5 m) => -100×2.5 = -250. Additional 6 kg at A => +60×2.5 = +150.
First condition: only 10 kg at A. Sum: +250 + W(2.5-x) = 0 => W(2.5-x) = -250. (A)
Second condition: 10 kg at B, 6 kg at A. Sum: +150 -250 + W(2.5-x) = 0 => -100 + W(2.5-x) = 0 => W(2.5-x) = +100. (B)
From (A) and (B): -250 = 100, impossible. So no solution? Maybe the pivot position changes? The phrase “non-uniform plank AB of length 5 m is balanced on a pivot placed at its centre.” That’s clear. Maybe I misinterpreted the second condition: “When the mass is moved to B, an additional 6 kg must be added at A to maintain balance.” Perhaps the additional 6 kg is added at A instead of the 10 kg? No, “moved to B”, so 10 kg goes to B, and then we add 6 kg at A. So both present. Then we get contradictory equations, meaning no such plank? But typical exam problems have solutions. Let’s re-read: “When a 10 kg mass is placed at A, the plank balances. When the mass is moved to B, an additional 6 kg must be added at A to maintain balance.” Could mean: first, plank alone is unbalanced. Place 10 kg at A -> balances. Then, start over, move the 10 kg to B (so now 10 kg at B), and to balance you now need an additional 6 kg at A (so total at A is 6 kg). That’s what I did. If that leads to no solution, maybe the “additional 6 kg” is added together with the 10 kg still at A? No, “moved to B” implies it’s no longer at A. Hmm.

Let’s try another interpretation: The plank is balanced on a pivot at its centre. Initially, maybe it is balanced by itself? “non-uniform plank … is balanced on a pivot placed at its centre.” That could mean when placed on a pivot at its centre, it is balanced (i.e., its centre of mass is exactly at the centre). But then it says “When a 10 kg mass is placed at A, the plank balances.” But if centre of mass is at centre, placing 10 kg at A would tip it, not balance. So “balanced on a pivot placed at its centre” might just mean the pivot is placed at the geometric centre, and the plank is non-uniform so it does NOT balance on its own. Then they add mass to achieve balance. So my initial interpretation stands. Let’s solve with x as distance from A to c.o.m., but maybe I need to consider that the pivot is not fixed at a point; we can move the pivot? No, “pivot placed at its centre” indicates fixed. So maybe the error is in assuming the direction of moments for the plank weight in the first case. If x<2.5, then W is left, 10 kg left, both ACW -> sum ACW = 100×2.5 + W(2.5-x). That must be zero, impossible unless W negative. So x>2.5 is necessary for first case to balance. Then first: 250 – W(x-2.5) = 0 => W(x-2.5) = 250. Second: with 10 kg right, 6 kg left: ACW = 60×2.5=150, CW = 100×2.5 + W(x-2.5)=250+250=500, net CW=350, doesn’t balance. So maybe the additional 6 kg is added at A when the mass is moved to B, but the 10 kg is still there at B, and the plank also has some extra weight? Let’s assume the “additional 6 kg” means total mass at A becomes 6 kg (i.e., we replace the 10 kg with 6 kg? No, “additional” means extra. So perhaps the second condition is: Starting from the balanced state with 10 kg at A, we move that 10 kg to B, then the plank tips. To bring it back to balance, we add an extra 6 kg at A. So now we have 10 kg at B and 6 kg at A. That gives equations (1) 250 = W(x-2.5) and (2) 150 = 250 + W(x-2.5). Substitute (1) into (2): 150 = 250 + 250 = 500, no.

Wait, maybe the pivot is not at the centre in the second scenario? But it says “a pivot placed at its centre”. That’s fixed. Could the pivot be moved in the second? No.

Let’s check a typical problem: a non-uniform plank balanced at centre by adding weights at ends. The method often involves finding the distance of centre of mass. Let me set up: Let centre of mass be at distance d from A. Pivot at L/2 = 2.5 m. First condition: 10g at A (left), plank weight Mg at centre of mass, which is x m from A. Taking moments about pivot: 10g × 2.5 = Mg × (x – 2.5) if x > 2.5. (1). Second condition: 10g moved to B, so clockwise: Mg(x-2.5) + 10g×2.5 = additional mass at A, let’s say m at A, so m g × 2.5 (ACW). m = 6? Then: m g × 2.5 = Mg(x-2.5) + 10g×2.5. Using (1) Mg(x-2.5) = 250. Then m×2.5 = 250 + 250 = 500 => m = 200 N / 2.5 = 80 N = 8 kg. So m would be 8 kg, not 6 kg. So if the problem stated 6 kg, might be a typo, or the answer is 8 kg? But our job is to write article, perhaps avoid this flawed question. I’ll modify to a simpler one or correct numbers. In my article, I can use a similar but well-posed problem. For the practice question, I’ll adjust to make it work: maybe the pivot is at a point not at centre? But I want it balanced at centre. Let’s make the plank length 6 m, pivot at centre 3 m from A. Mass 10 kg at A balances; when mass moved to B, need 5 kg at A? Then solve. Let’s do that: Uniform? Non-uniform. Let x be distance from A to c.o.m. Pivot at 3 m. First: 10 kg left: 10g × 3 = Mg(x-3) if x>3. Second: 10 kg right, m kg left: m g × 3 = 10g×3 + Mg(x-3) => m×3 = 10×3 + 10×3 = 60 => m = 20 kg. That’s too big. If I want m=6, then 6g×3 = 10g×3 + Mg(x-3) => 18 = 30 + Mg(x-3) => Mg(x-3) = -12, impossible. So not that either.

Better: I’ll use a classic beam problem. I’ll replace with: A non-uniform bar of length 4 m is balanced at a point 1.5 m from A when a 5 kg mass hangs at A. When the mass is moved to B, a 3 kg mass must hang at A to keep balance. Find the distance from A to the centre of mass. This works? Let’s test. Let pivot at P, distance from A = 1.5 m? Maybe. I’ll design a simple one that yields integer answer. I’ll edit the article to have a clean worked example. Since I’m writing the article, I can just provide a well-posed problem. I’ll replace the whole practice question with a straightforward one: a uniform beam with a child and a parent on a see-saw, or a beam supported by two strings. That avoids confusion. I’ll change the practice question to: “A uniform plank of length 4 m and weight 150 N rests on two trestles placed 0.5 m from each end. A workman of weight 600 N stands at one end. Find the reaction forces from the trestles

Published by TutorHao | GCSE Mathematics Revision Series | aleveler.com

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