托盘

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How does a seesaw work?
::锯锯怎么用?

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It is easier to get a seesaw to move if you push on the board near the end rather than near the middle. This is due to torque - the rotational version of force.
::如果您在板上推近端而不是靠近中间, 获得锯锯更容易移动。 这是因为托盘- 部队的旋转版 。

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The amount of force applied, as well as the location and the direction of the force with respect to the axis of rotation, determines the relative difficulty in causing a rotation.  The longer the length of the seesaw, the easier it is to lift a heavy object. 
::所应用的武力数量,以及部队相对于旋转轴的位置和方向,决定了促成旋转的相对困难,锯越长,举重物体越容易。

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T he  arm or  moment arm  is the between where you apply the force and the axis of rotation. The component of a force applied perpendicular to a lever arm produces a torque . The torque is the product of the length of the lever arm and the force applied perpendicular to the lever arm.  Use the seesaw simulation below to further explore the concept of torque:
::手臂或时钟臂是您应用力力和旋转轴之间的部分。对杠杆臂直接应用的力的组件产生扭力。扭力是杠杆臂长度和杠杆臂垂直应用力的产物。使用下面的锯木模拟进一步探索扭力概念:

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The Direction of Torque
::托克方向

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In  Figure    the force  F is applying a torque.  We define the direction of the torque by noting clockwise (CW) and counterclockwise (CCW) motion of an object as a result of an applied force. Whether the object actually rotates or not is unimportant. We ask how the object would move were it free to do so. For example, in Figure  the force $\begin{align*}F\end{align*}$ would rotate the meter stick in a counterclockwise direction. This is the same direction we turn a jar lid in order to loosen it.    
::在图F 中, F 力是应用一个火把。 我们通过注意一个物体的时钟和反时钟运动来界定它的方向。 对象是否实际旋转并不重要。 我们问一下如果该物体可以自由旋转, 物体会如何移动。 例如, 图F 力F 将计时棍在逆时针方向旋转。 这就是我们用罐头盖来松开它的方向 。

                     

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Example 1
::例1

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If a force $\begin{align*}F^\prime\end{align*}$ was applied parallel to $\begin{align*}F\end{align*}$ , but to the left of $\begin{align*}P\end{align*}$ (see Figure ), in what direction would the meter stick turn?
::如果F`部队与F平行使用,但P左侧(见图),那么测量棒会朝什么方向转?

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Answer: The meter stick would turn clockwise.
::回答:计时棒会顺时针旋转。

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You may be familiar with the expression: “Righty tighty, lefty loosey.”
::你可能熟悉“右紧身,左撇子松懈”的表述。

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We define the counterclockwise direction as positive and the clockwise direction as negative. The sign of the direction is based upon the Right Hand Rule . To understand this rule, hold your right hand with your thumb pointed up and curl your fingers into a fist. Notice that the direction your fingers curl in is counterclockwise (you’re looking down). We define the upward direction in which the thumb points as positive, and the corresponding counterclockwise torque as positive. If you turn your hand such that your thumb now points down and curl your fingers into a fist, you’ll see your fingers turn clockwise. We define the downward direction in which your thumb points as negative, and the corresponding clockwise torque as negative. (For those readers who have grown up using only digital clocks, the term clockwise originated from the rotational direction that the hands of an analog clock move; counterclockwise being the reverse rotational direction.)
::我们把逆时针方向定义为正向,将时钟方向定义为负方向。 方向的标志以右手规则为基础。 要理解这一规则, 请用拇指将右手指向上, 将手指弯成拳头。 请注意, 手指弯曲的方向是逆时针( 你向下看 ) 。 我们定义了拇指指指向正向的上向, 对应的逆时针方向是正向的。 如果您把你的手转向拇指现在指向下, 将手指弯成一拳, 你就会看到手指向右转。 我们定义了拇指指指向下的方向是负的, 相应的时钟指向是负的。 (对于只用数字时钟增长的读者, 术语顺时针源于模拟时钟的旋转方向; 逆时钟是反旋转方向 。 )

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Mathematical Definition of Torque
::托盘数学定义

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We can state the magnitude of the torque in two ways:
::我们可以用两种方式来说明力力的大小:

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(1) The product of the perpendicular distance from the axis of rotation $\begin{align*}r\end{align*}$ (to the applied force) and the perpendicular component of the force $\begin{align*}F \sin \theta\end{align*}$ .
:1) 与旋转轴(至适用力)的垂直距离和Fsin的垂直部分的产物。

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(2) The product of the perpendicular distance $\begin{align*}r \sin \theta\end{align*}$ to the direction in which the force acts, and $\begin{align*}F\end{align*}$ .
:2) 与部队行动方向的垂直距离(rsin)和F的产物。

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Both (1) and (2) are equivalent to the product of $\begin{align*}r, F\end{align*}$ , and the sine of the angle between them. The symbol for torque is the Greek letter tau $\begin{align*}(\tau)\end{align*}$ . Thus, we can write $\begin{align*}\tau=rF \sin \theta\end{align*}$ , where the angle $\begin{align*}\theta\end{align*}$ is the angle between $\begin{align*}r\end{align*}$ and $\begin{align*}F\end{align*}$ . If the angle between $\begin{align*}r\end{align*}$ and $\begin{align*}F\end{align*}$ is $\begin{align*}90^\circ\end{align*}$ then the torque is simply $\begin{align*}\tau=rF\end{align*}$ . We can see by the definition of the torque that the units of torque are m*N though they are usually expressed as N*m but never as joules even though they are dimensionally equivalent.
:1)和(2)都相当于r、F和它们之间角的正弦。 托克的符号是希腊字母 Tau (- ) 。 因此, 我们可以写 rFsin , 角是r和F之间的角。 如果 r和F 之间的角是 90 , 托克就是 rF 。 我们从托克的定义可以看出, 托克的单位是 m * N, 尽管通常以N*m 表示, 但从不以 jouls 表示, 即使它们与尺寸相当 。

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Example 2
::例2

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An engineer is trying to turn a difficult nut.  Using a long wrench, he applies a 105 N force at a distance 30.0-cm from the nut shown in Figure . What is the magnitude of the torque applied by the plumber?
::工程师试图将一个困难的坚果变成一个坚果。用长扳手,他用105N力,距离图中显示的坚果30.0厘米。水管工使用的扭力有多大?

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Answer: The lever arm is 30 cm and the angle between the force and the lever arm is 90-degrees. $\begin{align*}\tau = Fr \sin \theta=(105 \text{N})(0.30\text{m})\sin 90^\circ \rightarrow 31.5 \ \text{N} \cdot \text{m}\end{align*}$
::回答:杠杆臂为30厘米,杠杆臂与杠杆臂之间的角为90度。

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Example 3
::例3

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If the minimum torque required turning the pipe in Figure is 31.5 N·m,, could a force smaller than 105 N be used?
::如果在图31.5N-m中,需要最小的硬力把管子转动,是否可使用小于105N的力?

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Answer: Yes. The workman could extend the length of the handle of the tool he’s using to turn the pipe. The longer the arm of the tool, the smaller the force required. You have probably experienced this phenomenon yourself. If you have ever had a problem turning a , increasing the thickness of the screw driver handle enables you to provide the same torque with a smaller force. The fatter the handle, the farther your hand is from the axis of rotation, and so the smaller the force needed to turn the screw. This is how a lever works. For example, the longer the lever arm on a bottle opener, the smaller the force needed to pry the bottle cap open. You may have heard the famous dictum of Archimedes: "Give me a place to stand on and with a lever I will move the world." In other words, with a long enough lever arm, even the weakest person can move a tremendous .
::答案是:是的。 工人可以延长他用来打开管子的工具柄的长度。 工具臂的长度越长, 需要的力量就越小。 您可能亲身体验过这种现象。 如果您曾经遇到过问题, 增加螺旋驱动器柄的厚度, 使您能够用更小的力量提供同样的手柄。 把手越肥, 你的手就越离旋转轴越远, 使螺旋转动所需的力量越小。 这就是杠杆的作用方式。 比如, 瓶开器的杠杆臂越长, 开瓶盖所需的力量就越小。 您可能听过Archimedes 的著名口语 : “ 给我一个位置, 用一个杆子我就能移动世界。 ”换句话说, 用一个足够长的杠杆臂, 即使最弱的人也能移动一个巨大的动作。 ”

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Example 4
::例4

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A door of width 0.810 m requires a minimum torque of 14.47 m*N in order to open. The door knob is positioned 5.70 cm from the left edge of the door.
::宽度为0.810米的门要打开,至少需要14.47毫 *N的托盘。门把手在门左边缘位置为5.70厘米。

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(a) What minimum force is required to open the door?
:a) 开门需要何种最低限度的力量?

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(b) If the doorknob is moved to the center of the door, what is the minimum force required in opening the door? Assume that the force acts perpendicular to the plane of the door. See Figure .
:b) 如果门把手被移到门中央,打开门所需的最小力是多少?

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Answers:
::答复:

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(a.) The force is perpendicular to the lever arm so $\begin{align*}\tau=rF=14.47 \ m*N\end{align*}$ . Since the doorknob is 5.7 cm from the left edge of the door, the distance from the axis of rotation is $\begin{align*}r=0.81m-0.057m=0.753m\end{align*}$ . The force required is then:
::由于门把手离门左边缘5.7厘米,与旋转轴的距离为r=0.81m-0.057m=0.753米。

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$\begin{align*}F=\frac{14.47 \ m*N}{0.753 \ m}=19.216 \rightarrow 19.2 \ N\end{align*}$
::F=14.47 m*N0.753 m=19.21619.2 N

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(b.) Since the doorknob is now in the center of the door, it is $\begin{align*}\frac{0.81 \ m}{2}=0.405 \ m\end{align*}$ from the axis of rotation. Therefore, the force required is  $\begin{align*}\frac{14.47 \ m*N}{0.405 \ m}=35.73 \rightarrow 35.7 \ N\end{align*}$
:b) 由于门把手现在在门中央,离旋转轴0.81平方米=0.405米,因此,需要的警力是14.47毫 *N0.405毫=35.73-35.7毫

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How much force would be necessary to open the door if the doorknob were placed along the axis of rotation?
::如果把门把手放在旋转轴心上,打开大门需要多少武力?

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Example 5
::例5

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In Figure , a force $\begin{align*}F\end{align*}$ of 95.0 N is applied to a hinged rod of length $\begin{align*}r = 2.2 \ m\end{align*}$ . The angle between $\begin{align*}F\end{align*}$ and $\begin{align*}r\end{align*}$ is 130-degrees. Find the magnitude of the torque that the force $\begin{align*}F\end{align*}$ exerts upon the rod.
::在图中,F型威力为95.0牛顿,用于长度为r=2.2米的断链杆。F型和r型之间的角为130度。找出F型威力对棒施加的力力强度。

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Answer:
::答复:

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Once again, $\begin{align*}\tau=r(F \sin \theta)\end{align*}$ , but this time since $\begin{align*}\theta \neq 90^\circ\end{align*}$ , the equation does not reduce to $\begin{align*}\tau=rF\end{align*}$ .
::再说一遍,"(Fsin)", 但这次从"90"开始, 方程式并没有降低到"rF"。

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Therefore, $\begin{align*}\tau=(2.20m)(95.0N)\sin 130^\circ=160.1 \ m*N\end{align*}$
::因此,(2.20m)(95.0N)sin 130160.1 m*N

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The drawbridge in the simulation below has a cable that pulls on the bridge in order to rotate it counter-clockwise. Try to adjust the sliders so that all the forces in the bridge system are in equilibrium. Then, play around and see what conditions cause the bridge’s cable to snap:
::下面模拟中的吊桥有一条电缆拉动桥,以便逆时针旋转。 试着调整滑轮,以使桥系统的所有力量都处于平衡状态。 然后,玩一玩,看看是什么条件导致桥电缆断裂:

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Further Reading
::继续阅读

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Summary
::摘要

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• Torque is the component of a force applied perpendicular to a lever arm. It can have a clockwise or counter-clockwise direction.
  ::托克是伸展到杠杆臂的垂直力的组成部分。 它可以有一个顺时针或逆时针方向 。
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• Torque can be calculated by taking the product of the length of a lever arm and the force applied perpendicular to a lever arm. 
  ::可将杠杆臂长度的产物和对杠杆臂垂直施加的威力乘以杠杆臂,即可计算毒性。

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Review
::回顾

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1. Why do we not calculate the torque using the cosine of the angle between $\begin{align*}r\end{align*}$ and $\begin{align*}F\end{align*}$ ?
::1. 为什么我们不使用r和F之间的角的余弦计算火力?

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2. Mathematically, the definition of torque can also be expressed as $\begin{align*}\tau=(r \sin \theta)F\end{align*}$ . In other words, we can always assume that $\begin{align*}F\end{align*}$ , rather than a component of the force, is responsible for rotation. How can this be?
::2. 从数学角度讲, " 硬力 " 的定义也可以表述为 " (rsin)F " 。 换句话说,我们总是可以假定,由F而不是部队的一部分负责轮换,这怎么可能?