5.20 产生结果为两部分之和

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  以两个组成部分的总和作为结果

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  You are working in science class on a "weather unit". As part of this class, you are tasked with going out and checking the wind speed each day at a meter behind your school. The wind speed you record for the day is 20 mph at a $\begin{array}{r}E{50}^{\circ }N\end{array}$ trajectory. This means that the wind is blowing at an angle $\begin{array}{r}{50}^{\circ }\end{array}$ taken from the direction that would be due East. At the conclusion of each day, you are supposed to break the wind speed (which is a vector) into two components: the portion that is in a North/South direction and the portion that is in an East/West direction. Can you figure out how to do this?
  ::您正在“ 天气单位 ” 的科学课上工作。 作为这个课的一部分, 您的任务是每天去学校后面的一米处检查风速。 您每天记录的风速是 E50 N 轨迹的20 mph 。 这意味着风从东方方向的50 方向吹起。 在每天结束时, 您应该把风速( 一种矢量) 破碎成两个部分: 北南方向的部分和东西方向的部分。 您知道该怎么做吗 ?

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  Resultant of the Sum of Two Components
  ::两个组成部分之和的计算结果

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  We can look at any vector as the resultant of two perpendicular components. If we generalize some vector $\begin{array}{r}\stackrel{\to }{q}\end{array}$ into perpendicular components, $\begin{array}{r}|\stackrel{\to }{r}|\hat{i}\end{array}$ is the horizontal component of a vector $\begin{array}{r}\stackrel{\to }{q}\end{array}$ and $\begin{array}{r}|\stackrel{\to }{s}|\hat{j}\end{array}$ is the vertical component of $\begin{array}{r}\stackrel{\to }{q}\end{array}$ . Therefore $\begin{array}{r}\stackrel{\to }{r}\end{array}$ is a magnitude , $\begin{array}{r}|\stackrel{\to }{r}|\end{array}$ , times the unit vector in the $\begin{array}{r}x\end{array}$ direction and $\begin{array}{r}\stackrel{\to }{s}\end{array}$ is its magnitude, $\begin{array}{r}|\stackrel{\to }{s}|\end{array}$ , times the unit vector in the $\begin{array}{r}y\end{array}$ direction. The sum of $\begin{array}{r}\stackrel{\to }{r}\end{array}$ plus $\begin{array}{r}\stackrel{\to }{s}\end{array}$ is: $\begin{array}{r}\stackrel{\to }{r}+\stackrel{\to }{s}=\stackrel{\to }{q}\end{array}$ . This addition can also be written as $\begin{array}{r}|\stackrel{\to }{r}|\hat{i}+|\stackrel{\to }{s}|\hat{j}=\stackrel{\to }{q}\end{array}$ .
  ::我们可以查看任何矢量作为两个垂直元件的产物。如果我们将某些矢量 qQQ 推广为垂直元件, ri是矢量 qQQ的横向元件, sj是QQQQ的垂直元件。 因此, r是一个数值, r, 单位矢量在 x 方向上乘倍单位矢量是其大小, s, 单位矢量在 y 方向上乘倍。 r+sqQQQ 的总和是: rsqqqqQ。 添加的值也可以写为 isjqQQ}。

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  If we are given the vector $\begin{array}{r}\stackrel{\to }{q}\end{array}$ , we can find the components of $\begin{array}{r}\stackrel{\to }{q},\stackrel{\to }{r}\end{array}$ , and $\begin{array}{r}\stackrel{\to }{s}\end{array}$ using trigonometric ratios if we know the magnitude and direction of $\begin{array}{r}\stackrel{\to }{q}\end{array}$ .
  ::如果我们得到矢量 q,我们就可以找到 q,r,和s的成分 使用三角比 如果我们知道Q的大小和方向。

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  This is accomplished by taking the magnitude of the vector times the cosine of the vector's angle to find the horizontal component, and the magnitude of the vector times the sine of the vector's angle to find the vertical component.
  ::实现这一目标的方法是,将矢量的大小乘以矢量角度的余弦值以找到水平组件,而矢量的大小乘以矢量角的正弦值以找到垂直组件。

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  Finding the Horizontal and Vertical Components 
  ::查找水平和垂直构件

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  1. If $\begin{array}{r}|\stackrel{\to }{q}|=19.6\end{array}$ and its direction is $\begin{array}{r}{73}^{\circ }\end{array}$ , find the horizontal and vertical components.
  ::1. 如果QQ19.6及其方向是73,请找到横向和纵向组成部分。

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  If we know an angle and a side of a right triangle, we can find the other remaining sides using trigonometric ratios. In this case, $\begin{array}{r}\stackrel{\to }{q}\end{array}$ is the hypotenuse of our triangle, $\begin{array}{r}\stackrel{\to }{r}\end{array}$ is the side adjacent to our $\begin{array}{r}{73}^{\circ }\end{array}$ angle, $\begin{array}{r}\stackrel{\to }{s}\end{array}$ is the side opposite our $\begin{array}{r}{73}^{\circ }\end{array}$ angle, and $\begin{array}{r}\stackrel{\to }{r}\end{array}$ is directed along the $\begin{array}{r}x-\end{array}$ axis.
  ::如果我们知道一个角度和一个右三角形的侧面, 我们就可以使用三角比来找到其余的两边。 在这种情况下, q是三角形的下限, r是靠近我们73角的侧面, s是我们73角的对面, r是沿着x- 轴线方向的侧面 。

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  To find $\begin{array}{r}\stackrel{\to }{r}\end{array}$ , we will use cosine and to find $\begin{array}{r}\stackrel{\to }{s}\end{array}$ we will use sine. Notice this is a scalar equation so all quantities are just numbers. It is written as the quotient of the magnitudes, not the vectors.
  ::为了找到 r , 我们将使用 cosine , 并找到 s r正弦 。 请注意这是一个 scalar 方程式, 所以所有数量都是数字 。 它写成星号的商数, 而不是矢量 。

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  $$\begin{array}{rlrl}\cos 73& =\frac{|\stackrel{\to }{r}|}{|\stackrel{\to }{q}|}=\frac{r}{q}& & \sin 73=\frac{|\stackrel{\to }{s}|}{|\stackrel{\to }{q}|}=\frac{s}{q}\\ \cos 73& =\frac{r}{19.6}& & \sin 73=\frac{s}{19.6}\\ r& =19.6\cos 73& & s=19.6\sin 73\\ r& =5.7& & s=18.7\end{array}$$

  
  ::=19.6r=19.6r=19.6cos*73s=19.6sin*73s=5.7s=18.7

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  The horizontal component is 5.7 and the vertical component is 18.7. One can rewrite this in vector notation as $\begin{array}{r}5.7\hat{i}+18.7\hat{j}=\stackrel{\to }{q}\end{array}$ . The components can also be written $\begin{array}{r}\stackrel{\to }{q}=⟨5.7,18.7⟩\end{array}$ , with the horizontal component first, followed by the vertical component. Be careful not to confuse this with the notation for plotted points.
  ::水平部分为5.7, 垂直部分为18.7。 在矢量标记中, 也可以将其重写为5. 7 18.7j q。 也可以写成 q 5. 7, 18.7 , 先写为水平部分, 后写为垂直部分。 注意不要混淆此点与绘图点的标记 。

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  2. If $\begin{array}{r}|\stackrel{\to }{m}|=12.1\end{array}$ and its direction is $\begin{array}{r}{31}^{\circ }\end{array}$ , find the horizontal and vertical components.
  ::2. 如果m12.1及其方向为31,请找到横向和纵向组成部分。

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  To find $\begin{array}{r}\stackrel{\to }{r}\end{array}$ , we will use cosine and to find $\begin{array}{r}\stackrel{\to }{s}\end{array}$ we will use sine. Notice this is a scalar equation so all quantities are just numbers. It is written as the quotient of the magnitudes, not the vectors.
  ::为了找到 r , 我们将使用 cosine , 并找到 s r正弦 。 请注意这是一个 scalar 方程式, 所以所有数量都是数字 。 它写成星号的商数, 而不是矢量 。

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  $$\begin{array}{rlrl}\cos 31& =\frac{|\stackrel{\to }{r}|}{|\stackrel{\to }{m}|}=\frac{r}{m}& & \sin 31=\frac{|\stackrel{\to }{s}|}{|\stackrel{\to }{m}|}=\frac{s}{m}\\ \cos 31& =\frac{r}{12.1}& & \sin 31=\frac{s}{12.1}\\ r& =12.1\cos 31& & s=12.1\sin 31\\ r& =10.37& & s=6.23\end{array}$$

  
  ::=12.1r=12.1r=12.1r=12.1cos*31s=12.1sin31r=10.37s=6.23

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  Finding the Resultants 
  ::寻找结果

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  If $\begin{array}{r}|\stackrel{\to }{r}|=15\end{array}$ and $\begin{array}{r}|\stackrel{\to }{s}|=11\end{array}$ , find the resultant vector length and angle.
  ::如果 \ \ 15 和 \ \ \ \ \ \ \ \ 11, 找到相应的矢量长度和角度 。

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  We can view each of these vectors on the coordinate system here:
  ::我们可以在协调系统上看到其中的每一个矢量:

  

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  Each of these vectors then serves as sides in a right triangle. So we can use the Pythagorean Theorem to find the length of the resultant:
  ::这些矢量中的每一矢量 然后作为右三角形的侧边。 所以我们可以使用 Pythagorean 定理来找到结果的长度 :

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  $$\begin{array}{r}{c}^2=a^2+b^2\\ c^2={15}^2+{11}^2\\ c^2=225+121\\ c^2=346\\ c=\sqrt{346}\approx 18.60\end{array}$$

  
  ::c2=a2+b2c2=152+112c2=225+121c2=346c=346c=34618.60

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  The angle of rotation that the vector makes with the "x" axis can be found using the tangent function:
  ::矢量与“x”轴的旋转角度,可使用正切函数找到:

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  $$\begin{array}{r}\tan \theta =\frac{opposite}{adjacent}\\ \tan \theta =\frac{11}{15}\\ \tan \theta =.73\\ \theta ={\tan }^{-1}.73\\ \theta \approx 36.13\end{array}$$

  
  ::1115tan 73 tan -1 73 36.13

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to break down vectors into their individual components. 
  ::早些时候,有人要求你将矢量分解成它们的各个组成部分。

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  From this section you've learned how to take a vector and break it into components using trig functions. If you draw the wind speed you recorded as a vector:
  ::您从此节中学习了如何使用 trig 函数将矢量带入并折成组件。 如果您绘制了风速, 您将记录为矢量 :

  

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  You can find the "x" and "y" components. These are the same as the part of the wind that is blowing to the East and the part of the wind that is blowing to the North.
  ::你可以找到“x”和“y”组件。它们与吹向东方的风部分和吹向北方的风部分相同。

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  East component: $\begin{array}{r}\cos {50}^{\circ }=\frac{x}{20}\end{array}$
  ::东部分: cos50x20

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  $\begin{array}{r}20\cos {50}^{\circ }=x\end{array}$
  ::20cos50x

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  x = 12.86 mph
  ::x=12.86米时

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  North component: $\begin{array}{r}\sin {50}^{\circ }=\frac{y}{20}\end{array}$
  ::北块: sin50y20

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  $\begin{array}{r}20\sin {50}^{\circ }=y\end{array}$
  ::20岁50岁

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  y = 15.32 mph
  ::y = 15.32毫秒

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  Example 2
  ::例2

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  Find the magnitude of the horizontal and vertical components of the following vector if the resultant vector’s magnitude and direction are given as $\begin{array}{r}\text{magnitude}=75\text{direction}={35}^{\circ }\end{array}$ .
  ::查找以下矢量的水平和垂直组成部分的大小,如果由此得出的矢量的大小和方向被定为 = 75 direction= 35 。

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    $\begin{array}{r}\cos {35}^{\circ }=\frac{x}{75},\sin {35}^{\circ }=\frac{y}{75},x=61.4,y=43\end{array}$
  ::=61.4,y=43 =43 =61.4,y=43

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  Example 3
  ::例3

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  Find the magnitude of the horizontal and vertical components of the following vector if the resultant vector’s magnitude and direction are given as $\begin{array}{r}\text{magnitude}=3.4\text{direction}={162}^{\circ }\end{array}$ .
  ::查找下列矢量的水平和垂直组成部分的大小,如果由此得出的矢量的大小和方向被划为数量=3.4 方向=162。

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  $\begin{array}{r}\cos {162}^{\circ }=\frac{x}{3.4},\sin {162}^{\circ }=\frac{y}{3.4},x=3.2,y=1.1\end{array}$
  ::COS162x3.4,sin162_Y3.4,x=3.2,y=1.1

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  Example 4
  ::例4

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  Find the magnitude of the horizontal and vertical components of the following vector if the resultant vector’s magnitude and direction are given as $\begin{array}{r}\text{magnitude}=15.9\text{direction}={12}^{\circ }\end{array}$ .
  ::查找下列矢量的水平和垂直组成部分的大小,如果由此得出的矢量的大小和方向被划为数量=15.9 方向=12。

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  $\begin{array}{r}\cos {12}^{\circ }=\frac{x}{15.9},\sin {12}^{\circ }=\frac{y}{15.9},x=15.6,y=3.3\end{array}$
  ::12x15.9,sin12\y15.9,x=15.6,y=3.3

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  Review
  ::回顾

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  Find the horizontal and vertical components of the following vectors given the resultant vector’s magnitude and direction.
  ::查找下列矢量的横向和纵向组成部分,并参照由此得出的矢量的大小和方向。

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  1. $\begin{array}{r}\text{magnitude}=65\text{direction}={22}^{\circ }\end{array}$ .
     ::等值=65 方向=22\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\。\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
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  2. $\begin{array}{r}\text{magnitude}=34\text{direction}={15}^{\circ }\end{array}$ .
     ::范围=34 方向=15 。
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  3. $\begin{array}{r}\text{magnitude}=29\text{direction}={160}^{\circ }\end{array}$ .
     ::范围=29 方向=160\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\。
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  4. $\begin{array}{r}\text{magnitude}=100\text{direction}={320}^{\circ }\end{array}$ .
     ::范围=100 方向= 320 。
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  5. $\begin{array}{r}\text{magnitude}=320\text{direction}={200}^{\circ }\end{array}$ .
     ::等距= 320 方向= 200\\\\ 。
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  6. $\begin{array}{r}\text{magnitude}=15\text{direction}={110}^{\circ }\end{array}$ .
     ::范围=15 方向=110 。
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  7. $\begin{array}{r}\text{magnitude}=10\text{direction}={80}^{\circ }\end{array}$ .
     ::等距= 10 方向= 80\\\ 。
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  8. $\begin{array}{r}\text{magnitude}=90\text{direction}={290}^{\circ }\end{array}$ .
     ::范围=90 方向=290\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
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  9. $\begin{array}{r}\text{magnitude}=87\text{direction}={10}^{\circ }\end{array}$ .
     ::范围=87 方向=10 。
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  10. $\begin{array}{r}\text{magnitude}=42\text{direction}={150}^{\circ }\end{array}$ .
      ::范围=42 方向=150\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  11. If $\begin{array}{r}|\stackrel{\to }{r}|=12\end{array}$ and $\begin{array}{r}|\stackrel{\to }{s}|=8\end{array}$ , find the resultant vector magnitude and angle.
      ::如果\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\去\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\去去去去去去\\\\\\\\\\\\\\\\\\去去去去去去去去去去去矢向的矢向向向向向向向向向向
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  12. If $\begin{array}{r}|\stackrel{\to }{r}|=14\end{array}$ and $\begin{array}{r}|\stackrel{\to }{s}|=6\end{array}$ , find the resultant vector magnitude and angle.
      ::如果\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\去的矢的矢的矢的矢的矢的矢的矢的矢的矢的矢的矢的矢的矢的矢的矢的矢的矢的矢体的矢的矢的矢的矢体的矢体的矢体大小和角度的 和角度的矢体大小和角度和角度和角度和角度和角度和角度和角度。去
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  13. If $\begin{array}{r}|\stackrel{\to }{r}|=9\end{array}$ and $\begin{array}{r}|\stackrel{\to }{s}|=24\end{array}$ , find the resultant vector magnitude and angle.
      ::如果 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去
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  14. Will cosine always be used to find the horizontal component of a vector?
      ::是否总是使用余弦来找到矢量的横向组成部分 ?
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  15. If you know the component form of a vector, how can you find its magnitude and direction?
      ::如果您知道矢量的构成形式, 您如何找到它的大小和方向 ?

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  Review (Answers)
  ::回顾(答复)

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  Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
  ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。