Edit to “Solution”
en/2.1.66.md
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| ### Statement | |||
| $2.1.66.$ [Insert the problem statement] | |||
| @@ -4,7 +4,7 @@Statement | |||
| ### Solution | |||
| − |  | ||
| If the rider uses minimum speed possible, the friction from the surface is maximum possible, i.e. $f=\mu N$. As the vertical component of the total force is zero, we have | |||
| \[f\sin\alpha-N\cos\alpha-mg=0\qquad\Rightarrow N(\mu\sin\alpha-\cos\alpha)=mg.\] | |||
| The horizontal component of the total force serves as the centripetal force, so | |||
| \[f\cos\alpha+N\sin\alpha=\frac{mv^2}{R\sin\alpha}\qquad\Rightarrow N(\mu\cos\alpha+\sin\alpha)=\frac{mv^2}{R\sin\alpha}.\] | |||
| Eliminating $N$ from both equations and solving for $v$, we have | |||
| \[v=\sqrt{\frac{gR\sin\alpha(\mu\cos\alpha+\sin\alpha)}{\mu\sin\alpha-\cos\alpha}}=\sqrt{\frac{gR\sin\alpha(\mu+\tan\alpha)}{\mu\tan\alpha-1}},\] | |||
| which is possible when $\tan\alpha>1/\mu$. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||
| unchanged lines 15 | |||
| ### Statement | ### Statement | ||
| $2.1.66.$ [Insert the problem statement] | $2.1.66.$ [Insert the problem statement] | ||
| @@ -4,7 +4,7 @@Statement | |||
| ### Solution | ### Solution | ||
|  | ||
| If the rider uses minimum speed possible, the friction from the surface is maximum possible, i.e. $f=\mu N$. As the vertical component of the total force is zero, we have | If the rider uses minimum speed possible, the friction from the surface is maximum possible, i.e. $f=\mu N$. As the vertical component of the total force is zero, we have | ||
| \[f\sin\alpha-N\cos\alpha-mg=0\qquad\Rightarrow N(\mu\sin\alpha-\cos\alpha)=mg.\] | \[f\sin\alpha-N\cos\alpha-mg=0\qquad\Rightarrow N(\mu\sin\alpha-\cos\alpha)=mg.\] | ||
| The horizontal component of the total force serves as the centripetal force, so | The horizontal component of the total force serves as the centripetal force, so | ||
| \[f\cos\alpha+N\sin\alpha=\frac{mv^2}{R\sin\alpha}\qquad\Rightarrow N(\mu\cos\alpha+\sin\alpha)=\frac{mv^2}{R\sin\alpha}.\] | \[f\cos\alpha+N\sin\alpha=\frac{mv^2}{R\sin\alpha}\qquad\Rightarrow N(\mu\cos\alpha+\sin\alpha)=\frac{mv^2}{R\sin\alpha}.\] | ||
| Eliminating $N$ from both equations and solving for $v$, we have | Eliminating $N$ from both equations and solving for $v$, we have | ||
| \[v=\sqrt{\frac{gR\sin\alpha(\mu\cos\alpha+\sin\alpha)}{\mu\sin\alpha-\cos\alpha}}=\sqrt{\frac{gR\sin\alpha(\mu+\tan\alpha)}{\mu\tan\alpha-1}},\] | \[v=\sqrt{\frac{gR\sin\alpha(\mu\cos\alpha+\sin\alpha)}{\mu\sin\alpha-\cos\alpha}}=\sqrt{\frac{gR\sin\alpha(\mu+\tan\alpha)}{\mu\tan\alpha-1}},\] | ||
| which is possible when $\tan\alpha>1/\mu$. | which is possible when $\tan\alpha>1/\mu$. | ||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | [Insert a concise answer or boxed result] | ||
| unchanged lines 15 | |||