题目
Problem
Coplanar forces of magnitudes 30N, 15N, 33N and PN act at a point in the directions shown in the diagram, where tanα=34. The system is in equilibrium.
(a) Show that
(30−P14.4)2+(P+3028.8)2=1.
(4)
(b) Verify that P=6 satisfies this equation and find the value of θ.
(2)
题目中文翻译
大小分别为 30N、15N、33N 和 PN 的共面力作用于同一点,方向如图所示,其中 tanα=34。系统处于平衡状态。
(a) 证明:
(30−P14.4)2+(P+3028.8)2=1.
(4)
(b) 验证 P=6 满足此方程,并求 θ 的值。
(2)
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) is needed, use 10ms−2.
需要使用重力加速度 g 的数值时,取 10ms−2。
解答
(a)
解法一
思路
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先把已知正切转换为精确正弦、余弦,不用舍入角度。分别平衡两方向后,用正弦平方加余弦平方恒等式消去θ。
答题过程
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Since tanα=4/3 and α is acute,
sinα=54,cosα=53.
Horizontal equilibrium gives
(33+15)(53)=(P+30)cosθ,
so (P+30)cosθ=28.8. Vertical equilibrium gives
15(54)+30sinθ=33(54)+Psinθ,
so (30−P)sinθ=14.4. Thus
cosθ=P+3028.8,sinθ=30−P14.4.
Using sin2θ+cos2θ=1 yields
(30−P14.4)2+(P+3028.8)2=1.
(b)
解法一
思路
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明确代入P等于6验证两个平方和为1,再从上一问正弦、余弦分量求锐角。
答题过程
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For P=6, the left-hand side becomes
==(2414.4)2+(3628.8)2(53)2+(54)2259+2516=1.
Thus P=6 satisfies the equation. Also sinθ=3/5 and cosθ=4/5, consistent with an acute angle. Therefore
θ≈36.9∘.