题目
Problem
Three coplanar forces of magnitudes 40 N 40\,\mathrm{N} 40 N , 30 N 30\,\mathrm{N} 30 N and X N X\,\mathrm{N} X N act at a point in the directions shown in the diagram.
Given that the forces are in equilibrium, find the values of θ \theta θ and X X X .
(4)
题目中文翻译
三个大小分别为 40 N 40\,\mathrm{N} 40 N 、30 N 30\,\mathrm{N} 30 N 和 X N X\,\mathrm{N} X N 的共面力,沿图示方向作用于同一点。
已知这些力处于平衡状态,求 θ \theta θ 和 X X X 的值。
(4)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。
Where a numerical value for the acceleration due to gravity (g g g ) is needed, use 10 m s − 2 10\,\mathrm{m\,s^{-2}} 10 m s − 2 .
需要使用重力加速度 g g g 的数值时,取 10 m s − 2 10\,\mathrm{m\,s^{-2}} 10 m s − 2 。
解答
解法一
思路
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把未知力沿水平、竖直方向分解,分别平衡另外两个力。由图可知角为锐角。
答题过程
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Resolving horizontally and vertically gives
X cos θ = 40 , X sin θ = 30. X\cos\theta=40,\qquad X\sin\theta=30. X cos θ = 40 , X sin θ = 30.
Squaring and adding,
X 2 = 40 2 + 30 2 = 2500 , X = 50 N . X^2=40^2+30^2=2500,
\qquad \boxed{X=50\,\mathrm{N}}. X 2 = 4 0 2 + 3 0 2 = 2500 , X = 50 N .
Since 0 < θ < 90 ∘ 0<\theta<90^\circ 0 < θ < 9 0 ∘ ,
tan θ = 30 40 , θ = tan − 1 ( 3 4 ) ≈ 36.9 ∘ . \begin{align*}
\tan\theta=&\,\frac{30}{40},\\[4mm]
\theta=&\,\tan^{-1}\biggl(\frac34\biggr)\approx\boxed{36.9^\circ}.
\end{align*} tan θ = θ = 40 30 , tan − 1 ( 4 3 ) ≈ 36. 9 ∘ .
解法二
思路
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平衡的三个力首尾相接形成闭合直角三角形,直接用勾股定理及三角比。
答题过程
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The three forces form a closed triangle. The 40 N 40\,\mathrm{N} 40 N and 30 N 30\,\mathrm{N} 30 N sides are perpendicular, so
X = 40 2 + 30 2 = 50 N . X=\sqrt{40^2+30^2}
=\boxed{50\,\mathrm{N}}. X = 4 0 2 + 3 0 2 = 50 N .
The acute angle corresponding to the diagram satisfies
tan θ = 30 40 , θ = tan − 1 ( 3 4 ) ≈ 36.9 ∘ . \begin{align*}
\tan\theta=&\,\frac{30}{40},\\[4mm]
\theta=&\,\tan^{-1}\biggl(\frac34\biggr)\approx\boxed{36.9^\circ}.
\end{align*} tan θ = θ = 40 30 , tan − 1 ( 4 3 ) ≈ 36. 9 ∘ .
解法三
思路
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用 Lami 定理,各力除以其余两力夹角的正弦相等。先求角,再求力。
答题过程
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The angles between the other two forces are 90 ∘ 90^\circ 9 0 ∘ , 180 ∘ − θ 180^\circ-\theta 18 0 ∘ − θ and 90 ∘ + θ 90^\circ+\theta 9 0 ∘ + θ , respectively. By Lami’s theorem,
X sin 90 ∘ = 30 sin ( 180 ∘ − θ ) = 40 sin ( 90 ∘ + θ ) . \begin{align*}
\frac{X}{\sin90^\circ}
=&\,\frac{30}{\sin(180^\circ-\theta)}\\[4mm]
=&\,\frac{40}{\sin(90^\circ+\theta)}.
\end{align*} sin 9 0 ∘ X = = sin ( 18 0 ∘ − θ ) 30 sin ( 9 0 ∘ + θ ) 40 .
Thus
30 sin θ = 40 cos θ ⟹ tan θ = 3 4 . \frac{30}{\sin\theta}=\frac{40}{\cos\theta}
\quad\Longrightarrow\quad
\tan\theta=\frac34. sin θ 30 = cos θ 40 ⟹ tan θ = 4 3 .
Therefore
θ = tan − 1 ( 3 4 ) ≈ 36.9 ∘ . \theta=\tan^{-1}\biggl(\frac34\biggr)\approx\boxed{36.9^\circ}. θ = tan − 1 ( 4 3 ) ≈ 36. 9 ∘ .
Since θ \theta θ is acute and tan θ = 3 / 4 \tan\theta=3/4 tan θ = 3/4 , sin θ = 3 / 5 \sin\theta=3/5 sin θ = 3/5 . Thus
X = 30 3 / 5 = 50 N . X=\frac{30}{3/5}=\boxed{50\,\mathrm{N}}. X = 3/5 30 = 50 N .