Graphical Analysis of Forces | 力的图示分析法

📚 Graphical Analysis of Forces | 力的图示分析法

In A-Level mathematics, physics problems involving forces are often solved by drawing careful vector diagrams. This method, called the graphical analysis of forces, provides a direct visual understanding of how forces combine and balance. It is especially useful for checking results obtained by trigonometry and for solving equilibrium problems.

在A-Level数学中,涉及力的物理问题常常通过绘制精确的矢量图来求解。这种被称为力的图示分析法的方法,能够直观地理解力的合成与平衡,尤其适用于检验三角计算的结果以及解决平衡问题。


1. Force as a Vector Quantity | 力作为向量的基本概念

A force is a vector quantity: it has both magnitude and direction. In SI units, force is measured in newtons (N). When we represent a force graphically, we draw an arrow whose length is proportional to the magnitude and whose direction indicates the direction of the force.

力是向量:既有大小又有方向。在国际单位制中,力的单位是牛顿(N)。在图像中表示力时,我们用带箭头的线段表示,线段长度与力的大小成比例,箭头方向表示力的方向。

  • The tail of the arrow is placed at the point where the force acts.

    箭尾位于力的作用点。

  • The arrowhead shows the direction along the line of action.

    箭头表示沿作用线的方向。

  • The point of application is important for rigid bodies, but for a particle all forces can be treated as acting at a single point.

    对于刚体,力的作用点很重要,但对于质点,所有力可视为作用于同一点。

Because a vector has both size and direction, drawing an accurate arrow automatically contains the same information as a numerical expression such as 50 N at 30° to the horizontal.

由于向量同时具有大小和方向,精确绘制的箭头与诸如“大小为50 N、与水平方向成30°”这样的数值表达包含相同的信息。


2. Scaled Force Diagrams | 比例力图示法

To turn a physical force into a drawing, we must choose a sensible scale. For example, 1 cm might represent 10 N, so a 50 N force is drawn as a 5 cm arrow. The scale should be chosen so that the finished diagram is large enough to measure accurately.

为了将力转换为图形,我们必须选择合适的比例尺。例如,1 cm可代表10 N,那么50 N的力就画成5 cm长的箭头。比例尺应使最终图形足够大,以便进行精确测量。

  • Use a sharp pencil, a ruler and a protractor; graph paper helps to maintain straight edges and perpendicular lines.

    使用削尖的铅笔、直尺和量角器;方格纸有助于保持线条平直和垂直。

  • Mark the scale clearly on the diagram, for instance: Scale: 1 cm = 10 N.

    在图上清楚标明比例尺,例如:比例尺:1 cm = 10 N。

  • Measure all lengths from the tail to the tip of the arrow; do not include the arrowhead when measuring.

    测量时应从箭尾量到箭头尖端,不要把箭头本身计入长度。

It is often helpful to draw a small sketch before the final diagram, so that the layout can be planned and unnecessary intersections avoided.

在绘制最终图之前,通常先画一个简图,以便规划布局、避免不必要的交叉。


3. The Parallelogram Law | 平行四边形法则

When two forces P and Q act at the same point with an angle θ between them, their resultant R can be found graphically using the parallelogram law. Draw both forces from the same point as adjacent sides of a parallelogram; the diagonal starting from that point represents the resultant.

当两个力P和Q作用于同一点且夹角为θ时,其合力R可用平行四边形法则通过作图求得。从同一点画出两个力作为平行四边形的邻边,从该点出发的对角线即表示合力。

From the cosine rule, the magnitude of the resultant is given by:

由余弦定理,合力的大小为:

R² = P² + Q² + 2PQ cos θ

The direction of R can be described by the angle α it makes with P, where:

合力R与P之间的夹角α满足:

tan α = Q sin θ / (P + Q cos θ)

  • If θ = 0°, the forces act in the same direction and R = P + Q.

    当θ = 0°时,两力同向,R = P + Q。

  • If θ = 180°, the forces act in opposite directions and R = |P − Q|.

    当θ = 180°时,两力反向,R = |P − Q|。

  • If θ = 90°, the parallelogram becomes a rectangle and R = √(P² + Q²).

    当θ = 90°时,平行四边形变为矩形,R = √(P² + Q²)。

Graphically, after constructing the parallelogram, measure the diagonal with a ruler and the angle with a protractor; then convert back using the scale.

作图时,在构造平行四边形后,用直尺量取对角线长度、用量角器测量角度,然后按比例尺换算回数值。


4. The Triangle of Forces | 力的三角形法则

The triangle of forces is another graphical construction for two forces. Instead of drawing both from the same point, we draw the second force starting from the tip of the first. The resultant is then represented by the closing side from the tail of the first force to the tip of the second force.

力的三角形法则是另一种用于两个力的作图方法。我们不从同一点画两个力,而是将第二个力的起点放在第一个力的箭头末端。于是,从第一个力的箭尾到第二个力的箭头的闭合边就表示合力。

This method is equivalent to the parallelogram law but often simpler when more than two forces are involved.

这种方法与平行四边形法则等价,但在涉及多个力时往往更简单。

For three forces acting on a particle to be in equilibrium, the three vector arrows must form a closed triangle when drawn head-to-tail. That is, the first force starts at a point, the second starts where the first ends, the third starts where the second ends, and the tip of the third returns to the starting point.

若三个力作用在质点上并保持平衡,则这三个矢量首尾相接时必须构成一个闭合三角形。也就是说,第一个力从某点出发,第二个力从第一个力的末端开始,第三个力从第二个力的末端开始,且第三个力的箭头恰好回到起点。

A + B + C = 0 ⟺ closed vector triangle

A + B + C = 0 ⟺ 闭合矢量三角形

In practice, if you know two forces and the fact that the system is in equilibrium, the third force can be found by completing the triangle and measuring the missing side.

实际应用中,若已知两个力并且系统处于平衡,则第三个力可通过补全三角形并测量缺失的边来求得。


5. Resolution of a Force | 力的分解

Just as two forces can be combined into one resultant, a single force can be split into two components. The most common split is into perpendicular components along chosen axes, usually horizontal and vertical.

正如两个力可以合成为一个合力,一个力也可以分解为两个分力。最常见的分解是沿选定坐标轴(通常是水平轴和竖直轴)分解为互相垂直的分量。

For a force F making an angle θ with the positive x-axis, the components are:

对于与x轴正方向成θ角的力F,其分量为:

Fₓ = F cos θ

Fᵧ = F sin θ

  • The component Fₓ is the projection of F onto the x-axis; it is drawn as the adjacent side of a right-angled triangle.

    分力Fₓ是F在x轴上的投影,画成直角三角形中与θ相邻的边。

  • The component Fᵧ is the projection onto the y-axis, drawn as the opposite side.

    分力Fᵧ是F在y轴上的投影,画成直角三角形中与θ相对的边。

  • The original force F is the diagonal of the rectangle formed by Fₓ and Fᵧ.

    原力F是Fₓ与Fᵧ构成的矩形的对角线。

Graphically, to resolve a force, draw the vector to scale, then construct a rectangle with the vector as diagonal; the sides of the rectangle are the components. This is extremely useful when setting up equations for equilibrium or acceleration.

作图时,要分解一个力,先按比例画出矢量,再以该矢量为对角线构造矩形,矩形的两边就是分力。这在建立平衡方程或加速度方程时非常有用。


6. Polygon Law for Concurrent Forces | 共点力系的多边形法则

When three or more forces act at a point, the resultant can be found by extending the triangle method into a polygon. Draw each force in turn, head-to-tail; the resultant is the vector from the tail of the first force to the tip of the last force.

当三个或更多个力作用于一点时,可将三角形法则推广为多边形法则。依次将各力首尾相接,合力就是从第一个力的箭尾指向最后一个力的箭头的矢量。

If the polygon closes exactly, the vector sum is zero and the forces are in equilibrium.

若多边形恰好闭合,则矢量和为零,各力处于平衡状态。

R = A + B + C + …

R = A + B + C + …

  • The order in which the forces are drawn does not affect the resultant.

    绘制各力的顺序并不影响合力。

  • If the polygon is closed, the resultant is zero and the particle is in equilibrium.

    若多边形闭合,则合力为零,质点处于平衡。

  • If the polygon is not closed, the missing side represents the resultant (or the additional force needed for equilibrium).

    若多边形不闭合,缺失的边就表示合力(或所需添加的平衡力)。

The polygon method is particularly helpful when dealing with more than three forces, since it avoids repeated pairwise combinations.

多边形法在处理三个以上力时尤其方便,因为它避免了反复的两两合成。


7. Equilibrium and Closed Vector Diagrams | 平衡与闭合矢量图

A particle is in equilibrium when the vector sum of all forces acting on it is zero. Graphically, this means that if all the force vectors are drawn head-to-tail, the tip of the last vector returns to the tail of the first, forming a closed polygon.

当作用在质点上的所有力的矢量和为零时,质点处于平衡状态。从图形上看,这意味着如果将所有的力矢量首尾相接,最后一个矢量的箭头恰好回到第一个矢量的箭尾,形成一个闭合多边形。

For three forces in equilibrium, the closed polygon is a triangle. This leads to a convenient visual test: three known forces are in equilibrium if their vector diagram closes.

对于三个力平衡的情形,闭合多边形就是三角形。这给出了一个方便的直观判断:若三个力的矢量图闭合,则它们平衡。

The sine rule can be applied to a triangle of forces. For forces A, B and C in equilibrium, with angles α, β and γ opposite the corresponding forces:

在力的三角形中可以应用正弦定理。对于平衡的力A、B、C,设α、β、γ分别是与各力相对的角:

A / sin α = B / sin β = C / sin γ

This is essentially Lami’s theorem, which can be read directly from a carefully drawn vector triangle.

这本质上就是拉密定理,可以直接从精确绘制的矢量三角形中读出。

When drawing an equilibrium diagram, always check that angles are measured from the correct directions. In a closed triangle, the direction arrows follow one another around the triangle, either all clockwise or all anticlockwise.

绘制平衡图时,务必检查角度是否从正确的方向测量。在闭合三角形中,各个方向箭头沿三角形依次排列,要么全部顺时针,要么全部逆时针。


8. Drawing Technique and Accuracy | 绘图技巧与精度

The accuracy of a graphical solution depends entirely on the accuracy of the drawing and the measuring instruments. A small mistake in angle or length can lead to a large error in the final answer.

图解法的精度完全依赖于作图和测量仪器的精度。角度或长度上的一个小错误会导致最终答案产生较大误差。

  • Select a scale that makes the diagram as large as possible within the space available, while keeping it convenient to work with.

    在可用空间内选择尽可能大的比例尺,并保持方便使用。

  • Use a solid, sharp pencil and draw lines carefully; a 2H pencil often gives thinner, more accurate lines.

    使用实心且削尖的铅笔,仔细画线;2H铅笔通常线条更细、更精确。

  • When measuring angles with a protractor, place its centre exactly at the vertex and read the scale without parallax.

    用量角器测量角度时,将中心准确放在顶点,读取刻度时要避免视差。

  • Always state the scale and provide the measured lengths; this helps in checking work and earning method marks.

    始终写明比例尺和量得的长度,这有助于检查过程并获得方法分。

For examination answers, a graphical solution is usually accepted with a small tolerance, for instance ±2 N in magnitude or ±2° in direction. Always draw the final answer to two significant figures.

在考试中,图解答案通常允许有较小的误差范围,例如大小在±2 N、角度在±2°以内。最终答案尽量保留两位有效数字。


9. Worked Example 1 – Resultant of Two Forces | 例题1:求两个力的合力

Two forces of 50 N and 30 N act at a point, with an angle of 60° between them. Use a graphical method to find the resultant force.

两个大小为50 N和30 N的力作用于同一点,它们之间的夹角为60°。用图解法求合力。

Step 1 – Choose a scale. Let 1 cm = 10 N. Then 50 N is drawn as 5 cm and 30 N as 3 cm.

第一步——选择比例尺。设1 cm = 10 N,那么50 N画成5 cm,30 N画成3 cm。

Step 2 – Draw the two vectors from a common point O with 60° between them.

第二步——从公共点O画出两个矢量,它们之间的夹角为60°。

Step 3 – Complete the parallelogram by drawing parallel lines through the tips of the vectors.

第三步——通过两个矢量的末端作平行线,补全平行四边形。

Step 4 – Draw the diagonal from O to the opposite corner. Measure its length.

第四步——从点O到对角画对角线。量取其长度。

Step 5 – Convert to actual force. The measured diagonal is about 7.0 cm, so R ≈ 7.0 × 10 = 70 N. The angle with the 50 N force is measured as approximately 21.8°.

第五步——换算为实际力。量得对角线约为7.0 cm,因此R ≈ 7.0 × 10 = 70 N。与50 N力之间的夹角约为21.8°。

Check analytically using the cosine law:

用余弦定理进行解析检验:

R = √(50² + 30² + 2 × 50 × 30 × cos 60°) = √(2500 + 900 + 1500) = √4900 = 70 N

The graphical result agrees perfectly with the analytic value in this case.

本例中图解结果与解析计算完全一致。


10. Worked Example 2 – Three Forces in Equilibrium | 例题2:三力平衡问题

A particle is in equilibrium under three forces. Two of them are 40 N acting horizontally and 30 N acting vertically upward. Find the third force graphically.

一个质点在三个力作用下处于平衡。其中两个力分别是水平向右的40 N和竖直向上的30 N。用图解法求第三个力。

Step 1 – Choose a scale, e.g., 1 cm = 10 N. Draw the 40 N force as 4 cm to the right, then from its tip draw the 30 N force as 3 cm upward.

第一步——选择比例尺,例如1 cm = 10 N。先向右画4 cm代表40 N的力,然后从它的箭头末端向上画3 cm代表30 N的力。

Step 2 – Complete the triangle. The third side, from the tail of the 40 N force to the tip of the 30 N force, represents the third force needed for equilibrium.

第二步——补全三角形。从40 N力的箭尾到30 N力箭头的第三边,就表示平衡所需的第三个力。

Step 3 – Measure the side. Its length is 5.0 cm, so the third force has magnitude 50 N. The angle below the horizontal is measured as approximately 36.9°.

第三步——量取该边。长度为5.0 cm,因此第三个力的大小为50 N。与水平方向下方的夹角约为36.9°。

Analytic check: since the 40 N and 30 N forces are perpendicular,

解析检验:由于40 N与30 N两个力互相垂直,

|F₃| = √(40² + 30²) = √(1600 + 900) = √2500 = 50 N

If the third force is found to be opposite in direction to the resultant of the first two, the particle is in equilibrium. The arrow of the third force must point back toward the starting point in the closed triangle.

如果第三个力的方向与前两个力的合力相反,则质点处于平衡。在闭合三角形中,第三个力的箭头必须指向起点。


11. Comparing Graphical and Analytical Methods | 图解与解析法的比较

Graphical methods offer an immediate visual picture of the forces and are often quicker for simple systems. They also help to prevent sign errors, because the direction of the resultant is visible from the diagram.

图解法提供了对力的直观视觉图像,对于简单系统通常更快,还能帮助避免符号错误,因为合力的方向在图中一目了然。

Analytical methods, using trigonometry or vector components, are more precise and are essential when forces are not perpendicular or when many forces are involved. They are also easier to automate in calculations.

解析法使用三角或矢量分量,更加精确,在处理非垂直力或多个力时至关重要,也更容易进行数值计算。

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