3.2 摩擦

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  摩擦

  

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  Tennis is played on a variety of court surfaces: grass, clay, hardcourt, and even carpet. Players know that matches on different surfaces are distinctly different games. The ball bounces differently, and shoes slip on some surfaces but stick on others. These differences are a result of friction.
  ::网球在多种法庭上玩耍:草、粘土、硬体、甚至地毯。 玩家知道不同表面的比赛是截然不同的游戏。 球反弹不同,鞋滑倒在某些表面,但粘在另一些表面。 这些差异是摩擦的结果。

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  Friction
  ::摩擦

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  Friction is the force that resists motion. In most beginning physics classes, friction is ignored. Concepts can be understood and calculations made assuming friction to be nonexistent. Whenever physics intersects with the real world, however, friction must be taken into account. Friction exists between two touching surfaces because even the smoothest looking surface is quite rough on a microscopic scale.
  ::摩擦是抵抗运动的力量。 在大多数最初的物理类中,摩擦被忽略了。 概念可以被理解,计算假设摩擦不存在。 但是,每当物理学与现实世界交错时,摩擦必须被考虑进去。 两个触摸的表面之间存在摩擦,因为即使是最平滑的表面在微缩规模上也相当粗糙。

  

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  Every surface has microscopic bumps, lumps, and imperfections, emphasized as in the image above. If we try to slide the top block over the lower block, there will be numerous collisions as the bumps impact each other. The forward motion causes the collisions with bumps which then exert a force in opposite way the block is moving. The force of friction always opposes whatever motion is causing the friction.
  ::每个表面都有微小的峰值、块块和不完善之处,这与上面的图像一样受到强调。如果我们试图将顶层块滑过下层块,就会发生多次碰撞,因为顶部块相互撞击。前方运动会引发碰撞,撞击会引发碰撞,从而产生冲击力,而冲撞会以相反的方式产生。摩擦的力量总是反对任何引起摩擦的运动。

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  The force of friction between these two blocks is related to two factors. The first factor is the roughness of the surfaces that are interacting, which is called the coefficient of friction , $\begin{array}{r}\mu \end{array}$ (Greek letter mu). The second factor is the magnitude of the force pushing the top block down onto the lower block. It is reasonable that the more forcefully the blocks are pushed together, the more difficult it will be for one to slide over the other. The force pushing these blocks together is the result of gravity acting on the top block and pressing it against the bottom block, which resists the with an equal and opposite force called the normal force . The force of friction can be calculated by
  ::这两个区块之间的摩擦力与两个因素有关。第一个因素是相互作用的表面的粗糙度,即摩擦系数 μ(希腊字母 mu)。第二个因素是把顶部块推向下方块的强度。有理由认为,这两个区块的推力越强,越难在另一块上滑过。将这些区块推在一起的推力是顶部块的重力作用和将它压在底部块上的结果,而底部块则以被称为正常力的平反力抵抗。摩擦力可以用正常力来计算。

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  $\begin{array}{r}{F}_{\text{friction}}=\mu \times F_{\text{normal}}\end{array}$ and the normal force will be equal to the force of gravity on the object, if the object is on a flat surface (one parallel to the ground).
  ::如果物体位于平面(一个与地面平行),则其反常和正常力与物体的重力相等。

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  This is an approximate but reasonably useful and accurate relationship. It is not exact because $\begin{array}{r}\mu \end{array}$ depends on a variety of factors, including whether the surface is wet or dry.
  ::这是一种近似但相当有用和准确的关系,不准确,因为微克取决于各种因素,包括表层是湿的还是干燥的。

   

   

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  The frictional force we have been discussing is referred to as sliding friction ; it is involved when one surface is sliding over another. If you have ever tried to slide a heavy object across a rough surface, you may be aware that it is a great deal easier to keep an object sliding than it is to start the object sliding in the first place. When the object to slide is resting on a surface with no movement, the force of friction is called static friction and it is somewhat greater than sliding friction. Surfaces that move against one another will have both a coefficient of static friction and a coefficient of sliding friction, and the two values will not be the same. For example, the coefficient of sliding friction for ice on ice is 0.03 whereas the coefficient of static friction for ice on ice is 0.10—more than three times as great .
  ::我们讨论过的摩擦力被称为滑动摩擦;当一个表面滑过另一个表面时,摩擦力就属于滑动;如果你曾经试图将一个重物体滑过粗糙的表面,你可能会意识到,保持一个物体滑动比开始物体滑动更容易得多。当滑动对象停留在没有运动的表面时,摩擦力被称为静态摩擦力,比滑动摩擦略强一些。 相互对立的表面既具有静态摩擦系数,又具有滑动摩擦系数,两种值不同。例如,冰上冰层滑动摩擦系数为0.03,而冰上冰层的静动摩擦系数为0.10-三倍以上。

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  How can we pinpoint the exact amount of force needed to overcome static friction? Use the simulation below to find out:
  ::我们怎样才能确定克服静态摩擦所需的确切兵力?用下面的模拟来找出:

   

   

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  Examples
  ::实例

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  Example 1  
  ::例1

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  A box weighing 2000. N is sliding across a cement floor. The force pushing the box is 500. N, and the coefficient of sliding friction between the box and the floor is 0.20. What is the of the box?
  ::N 正在滑过水泥地板上。 推着盒子的力是500. N, 盒子和地板之间的滑动摩擦系数是0. 20。 盒子里是什么?

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  In this case, the box is sliding along the ground, so the normal force for the box is equal to its weight. Using the normal force and the coefficient of friction , we can find the frictional force. We can also find the mass of the box from its weight since we know the . Then we can find the net force and the acceleration.
  ::在这种情况下,盒子正在沿着地面滑动,所以盒子的正常力与其重量相等。用正常力和摩擦系数,我们可以找到摩擦力。我们也可以从它的重量中找到盒子的重量,因为我们知道。然后我们就可以找到净力和加速力。

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  $\begin{array}{r}{F}_F=\mu F_N=(0.20)(2000.\text{N})=400.\text{N}\end{array}$
  ::FFN=0.20(2000.N)=400.N

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  $\begin{array}{r}\text{mass of box}=\frac{\text{weight}}{g}=\frac{2000.\text{N}}{9.8\text{m}/{\text{s}}^2}=204\text{kg}\end{array}$
  ::N9.8米/s2=204千克

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  $\begin{array}{r}{F}_{net}=\text{pushing force}-\text{frictional force}=500.\text{N}-400.\text{N}=100.\text{N}\end{array}$
  ::N-400 N=100 N=100 N

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  $\begin{array}{r}a=\frac{F_{net}}{m}=\frac{100.\text{N}}{204\text{kg}}=0.49{\text{m/s}}^2\end{array}$
  ::a=Fnetm=100.N204公斤=0.49米/秒2

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  Example 2  
  ::例2

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  Two boxes are connected by a rope running over a , as shown in the figure below. The coefficient of sliding friction between box A and the table is 0.20. (Ignore the masses of the rope and the pulley and any friction in the pulley.) The mass of box A is 5.0 kg and the mass of box B is 2.0 kg. The entire system (both boxes) will move together with the same acceleration and . Find the acceleration of the system.
  ::如下图所示,两个框用一根绳子连接在一个框上。框A与表格之间的滑动摩擦系数为0.20。 (说明绳子和滑轮的质量以及滑轮中的任何摩擦。 )框A的质量为5.0千克,框B的质量为2.0千克。整个系统(两个框)都将与相同的加速度一起移动。查找系统的加速度。

  

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  The force acting to move the system is the weight of box B, and the force resisting the movement is the force of friction between the table and box A. The mass of the system is the sum of the masses of both boxes. The acceleration of the system can be found by dividing the net force by the total mass.
  ::移动系统的力量是B箱的重量,抵抗运动的力量是表格与A盒之间的摩擦力。系统的质量是两个盒子质量的总和。通过将净力量除以总质量,可以发现系统的加速。

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  $\begin{array}{r}{F}_N(\text{box A})=mg=(5.0\text{kg})(9.8{\text{m/s}}^2)=49\text{N}\end{array}$
  ::FN(框A)=mg=(5.0公斤)(9.8米/秒)=49海里

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  $\begin{array}{r}{F}_{\text{friction}}=\mu F_N=(0.20)(49\text{N})=9.8\text{N}\end{array}$
  ::FFM=(0.20(49 N)=9.8 N)

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  $\begin{array}{r}\text{Weight of box B}=mg=(2.0\text{kg})(9.8{\text{m/s}}^2)=19.6\text{N}\end{array}$
  ::B=mg=(2.0公斤)(9.8米/秒)=19.6牛顿

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  $\begin{array}{r}{F}_{net}=19.6\text{N}-9.8\text{N}=9.8\text{N}\end{array}$
  ::Fnet=19.6 N-9.8 N=9.8 N

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  $\begin{array}{r}a=\frac{F_{net}}{\text{mass}}=\frac{9.8\text{N}}{7.0\text{kg}}=1.4{\text{m/s}}^2\end{array}$
  ::a = Fnetmas=9.8 N7.0千克=1.4 m/s2

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  In the simulation below, you can adjust the type of shoe a runner wears to change the coefficient of static friction between the runner’s shoe and the track surface. Play around and observe how a greater static friction coefficient means the sprinter can push off with more force without the danger of slipping.
  ::在下面的模拟中,您可以调整跑者穿戴的鞋型,以改变跑者鞋和轨道表面之间的静态摩擦系数。 玩耍并观察更大的静态摩擦系数如何意味着短跑者可以以更大的力力力推开而不会滑落的危险。

   

   

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  Further Reading
  ::继续阅读

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  Summary
  ::摘要

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  • Friction is caused by bodies sliding over rough surfaces.
    ::摩擦是由身体滑过粗糙表面造成的。
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  • The degree of surface roughness is indicated by the coefficient of friction, $\begin{array}{r}\mu \end{array}$ .
    ::表面粗糙度以摩擦系数(微克)表示。
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  • The force of friction is calculated by multiplying the coefficient of friction by the normal force.
    ::摩擦力的计算方法是将摩擦系数乘以正常力量。
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  • The frictional force always opposes motion.
    ::摩擦力量总是反对运动
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  • The net force is found by subtracting the frictional force from the applied force.
    ::净力是通过从所施武力中减去摩擦力而发现的。

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  Review
  ::回顾

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  1. A 52 N sled is pulled across a cement sidewalk at constant speed. A horizontal force of 36 N is exerted. What is the coefficient of sliding friction between the sidewalk and the metal runners of the sled?
     ::52 N 雪橇以恒定速度拉过水泥人行道。 水平力为36 N。 人行道和雪橇金属跑者之间的滑动摩擦系数是多少?
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  2. If the coefficient of sliding friction between a 25 kg crate and the floor is 0.45, how much force is required to move the crate at a constant velocity across the floor?
     ::如果25公斤的箱箱与地板之间的滑动摩擦系数是0.45,那么以恒定速度将箱子移动到地板上需要多少力?
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  3. A smooth wooden 40.0 N block is placed on a smooth wooden table. A force of 14.0 N is required to keep the block moving at constant velocity.
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     1. What is the coefficient of sliding friction between the block and the table top?
        ::区块和表顶之间的滑动摩擦系数是多少?
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     2. If a 20.0 N brick is placed on top of the wooden block, what force will be required to keep the block and brick moving at constant velocity?
        ::如果在木制砖块顶部放置了20.0N砖块,需要用什么力量才能使砖块和砖块以恒定速度移动?
     
     ::平滑的木制40.0N块放在平滑的木制桌上。要保持块以不变的速度移动,需要14.0N的力量。块与表顶之间的滑动摩擦系数是多少?如果木制块上方放置了20.0N砖,那么要保持块和砖以恒定的速度移动,需要何种力量?

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  Explore More
  ::探索更多

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  Use the  resource below to answer the following questions.
  ::使用以下资源回答下列问题。

   

   

   

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  1. 1 lbf (pound-force) = 4.44 N. Given this information, how many newtons of force did it take to rip the two phonebooks apart?
     ::1磅(重力)=4.44牛顿。 根据这一信息,用多少牛顿的力量将两部电话簿撕开?
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  2. Why do you think there is so much friction between the two phonebooks?
     ::为什么你认为 这两本电话簿之间有这么多摩擦?

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  Resources
  ::资源

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  When you fall thousands of feet from the sky, it seems like something strange is happening with the laws of physics. Turns out, everything relies on a simple force called drag. Though it may seem skydivers are in free fall, air resistance allows skydivers to speed up, slow down, and even change direction. 
  ::当你从天空坠落数千英尺时, 物理定律似乎正在发生一些奇怪的事情。 事实证明,一切都依赖于一种叫拖力的简单力量。 尽管它似乎在自由坠落,但空气阻力却让天体加速,减慢,甚至改变方向。