题目
Problem
Four coplanar forces of magnitudes 20 N 20\,\mathrm{N} 20 N 30 N 30\,\mathrm{N} 30 N 15 N 15\,\mathrm{N} 15 N 25 N 25\,\mathrm{N} 25 N P P P S N S\,\mathrm{N} S N α ∘ \alpha^\circ α ∘
(a) Find the value of S S S α \alpha α
(6)
A small ring of mass 0.6 k g 0.6\,\mathrm{kg} 0.6 kg 3 s 3\,\mathrm{s} 3 s 2 m 2\,\mathrm{m} 2 m
(b) Find the coefficient of friction between the ring and the wire.
(5)
题目中文翻译
四个大小分别为 20 N 20\,\mathrm{N} 20 N 30 N 30\,\mathrm{N} 30 N 15 N 15\,\mathrm{N} 15 N 25 N 25\,\mathrm{N} 25 N P P P S N S\,\mathrm{N} S N α ∘ \alpha^\circ α ∘
(a) 求 S S S α \alpha α
(6)
一个质量为 0.6 k g 0.6\,\mathrm{kg} 0.6 kg 2 m 2\,\mathrm{m} 2 m 3 s 3\,\mathrm{s} 3 s
(b) 求小环与金属丝之间的摩擦系数。
(5)
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 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 10 m s − 2 10\,\mathrm{m\,s^{-2}} 10 m s − 2
解答
(a)
解法一
思路
展开
先取右为正、下为正求水平与向下分量，再用平方和与正切求合力。
答题过程
展开
Let X X X Y Y Y
X = 30 cos 30 ∘ + 15 cos 45 ∘ − 20 cos 20 ∘ = 17.793511416 … N , \begin{align*}
X=&\,30\cos30^\circ+15\cos45^\circ-20\cos20^\circ\\[4mm]
=&\,17.793511416\ldots\text{ N},
\end{align*} X = = 30 cos 3 0 ∘ + 15 cos 4 5 ∘ − 20 cos 2 0 ∘ 17.793511416 … N , Y = 25 + 15 sin 45 ∘ − 20 sin 20 ∘ − 30 sin 30 ∘ = 13.766198851 … N . \begin{align*}
Y=&\,25+15\sin45^\circ-20\sin20^\circ-30\sin30^\circ\\[4mm]
=&\,13.766198851\ldots\text{ N}.
\end{align*} Y = = 25 + 15 sin 4 5 ∘ − 20 sin 2 0 ∘ − 30 sin 3 0 ∘ 13.766198851 … N . Therefore
S = X 2 + Y 2 = 22.497050458 … ≈ 22.5 . \begin{align*}
S=&\,\sqrt{X^2+Y^2}=22.497050458\ldots\\[4mm]
\approx&\,\boxed{22.5}.
\end{align*} S = ≈ X 2 + Y 2 = 22.497050458 … 22.5 . Both X X X Y Y Y
α = tan − 1 ( Y X ) = 37.727829 … ∘ ≈ 37.7 ∘ . \alpha=\tan^{-1}\biggl(\frac YX\biggr)
=37.727829\ldots^\circ\approx\boxed{37.7^\circ}. α = tan − 1 ( X Y ) = 37.727829 … ∘ ≈ 37. 7 ∘ .
(b)
解法一
思路
展开
四个外力的合竖直分量还要加小环重量才是支持力；水平方向用运动学求加速度，再算摩擦。
答题过程
展开
From rest, the constant acceleration satisfies
2 = 1 2 a ( 3 ) 2 , a = 4 9 m s − 2 . 2=\frac12a(3)^2,
\qquad a=\frac49\text{ m s}^{-2}. 2 = 2 1 a ( 3 ) 2 , a = 9 4 m s − 2 . Using the unrounded components from (a), vertical equilibrium gives
R = Y + 0.6 ( 10 ) = 19.766198851 … N . R=Y+0.6(10)=19.766198851\ldots\text{ N}. R = Y + 0.6 ( 10 ) = 19.766198851 … N . Horizontally,
X − F = 0.6 a , F = X − 4 15 . X-F=0.6a,
\qquad F=X-\frac4{15}. X − F = 0.6 a , F = X − 15 4 . Hence
μ = F R = 17.793511416 … − 4 / 15 19.766198851 … = 0.886707904 … ≈ 0.887 . \begin{align*}
\mu=&\,\frac{F}{R}=\frac{17.793511416\ldots-4/15}{19.766198851\ldots}\\[4mm]
=&\,0.886707904\ldots\approx\boxed{0.887}.
\end{align*} μ = = R F = 19.766198851 … 17.793511416 … − 4/15 0.886707904 … ≈ 0.887 .
解法二
思路
展开
也可由位移与时间求末速，再用水平合力做功求摩擦，支持力仍由竖直平衡决定。
答题过程
展开
Since the motion has constant acceleration from rest,
2 = 0 + v 2 ( 3 ) , v = 4 3 m s − 1 . 2=\frac{0+v}{2}(3),
\qquad v=\frac43\text{ m s}^{-1}. 2 = 2 0 + v ( 3 ) , v = 3 4 m s − 1 . The vertical contact force is R = Y + 6 R=Y+6 R = Y + 6 Y Y Y
2 X − 2 F = 1 2 ( 0.6 ) ( 4 3 ) 2 = 8 15 . 2X-2F=\frac12(0.6)\biggl(\frac43\biggr)^2=\frac8{15}. 2 X − 2 F = 2 1 ( 0.6 ) ( 3 4 ) 2 = 15 8 . Thus F = X − 4 / 15 F=X-4/15 F = X − 4/15
μ = X − 4 / 15 Y + 6 ≈ 0.887 (3 s.f.) . \mu=\frac{X-4/15}{Y+6}\approx\boxed{0.887}\text{ (3 s.f.)}. μ = Y + 6 X − 4/15 ≈ 0.887 (3 s.f.) . 
