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# 斜面是什么?

## 我们在处理斜面时如何运用牛顿第二定律？

${a}_{\perp }=\frac{\mathrm{\Sigma }{F}_{\perp }}{m}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}{a}_{\parallel }=\frac{\mathrm{\Sigma }{F}_{\parallel }}{m}$

## 斜面上一个物体的正向力${F}_{N}$‍ 是什么?

${F}_{N}=mg\text{cos}\theta$

## 涉及到斜面的问题是什么样的？

### 例 1: 滑雪橇

${a}_{\parallel }=\frac{\mathrm{\Sigma }{F}_{\parallel }}{m}\phantom{\rule{1em}{0ex}}\text{(对平行方向使用牛顿第二定律)}$
${a}_{\parallel }=\frac{mg\text{sin}\theta -{F}_{k}}{m}\phantom{\rule{1em}{0ex}}\text{(代入平行力)}$
${a}_{\parallel }=\frac{mg\text{sin}\theta -{\mu }_{k}{F}_{N}}{m}\phantom{\rule{1em}{0ex}}\text{(代入摩擦力公式)}$
${a}_{\parallel }=\frac{\overline{)m}g\text{sin}\theta -{\mu }_{k}\left(\overline{)m}g\text{cos}\theta \right)}{\overline{)m}}\phantom{\rule{1em}{0ex}}\text{(消掉分子和分母中都有的质量)}$
${a}_{\parallel }=g\text{sin}\theta -{\mu }_{k}\left(g\text{cos}\theta \right)\phantom{\rule{1em}{0ex}}\text{(加速度与质量压根没关系！)}$
${a}_{\parallel }=\left(9.8\frac{\text{m}}{{\text{s}}^{2}}\right)\text{sin}{30}^{o}-\left(0.150\right)\left(9.8\frac{\text{m}}{{\text{s}}^{2}}\right)\text{cos}{30}^{o}\phantom{\rule{1em}{0ex}}\text{(代入数值)}$
${a}_{\parallel }=3.63\frac{\text{m}}{{\text{s}}^{2}}\phantom{\rule{1em}{0ex}}\text{(结果)}$

### 例 2: 陡峭车道

${a}_{\parallel }=\frac{\mathrm{\Sigma }{F}_{\parallel }}{m}\phantom{\rule{1em}{0ex}}\text{(对平行方向使用牛顿第二定律)}$
${a}_{\parallel }=\frac{mg\text{sin}\theta -{F}_{s}}{m}\phantom{\rule{1em}{0ex}}\text{(代入平行重力分量与静态摩擦力)}$
$0=\frac{mg\text{sin}\theta -{F}_{s}}{m}\phantom{\rule{1em}{0ex}}\text{(因为汽车没有溜车, 所以加速度为零)}$

$0=mg\text{sin}\theta -{\mu }_{s}{F}_{N}\phantom{\rule{1em}{0ex}}\text{(代入最大静态摩擦力公式}\right)$
$0=mg\text{sin}\theta -{\mu }_{s}\left(mg\text{cos}\theta \right)\phantom{\rule{1em}{0ex}}\text{(代入斜面上正向力的表达式}\right)$
$0=\text{sin}\theta -{\mu }_{s}\left(\text{cos}\theta \right)\phantom{\rule{1em}{0ex}}\text{(角度与质量没有关系！)}$
$\text{sin}\theta ={\mu }_{s}\left(\text{cos}\theta \right)\phantom{\rule{1em}{0ex}}\text{(解 sin}\theta \right)$
$\frac{\text{sin}\theta }{\text{cos}\theta }={\mu }_{s}\phantom{\rule{1em}{0ex}}\text{(两边同除cos}\theta \right)$
$\theta =ta{n}^{-1}\left({\mu }_{s}\right)\phantom{\rule{1em}{0ex}}\text{(去掉左侧的tan)}$
$\theta =ta{n}^{-1}\left(0.75\right)\phantom{\rule{1em}{0ex}}\text{(代入数值)}$
$\theta ={37}^{o}\phantom{\rule{1em}{0ex}}\text{(结果)}$