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Question Number 99982 by Dwaipayan Shikari last updated on 24/Jun/20

Commented by mr W last updated on 24/Jun/20

a=(g/μ)=g tan θ ⇒θ=tan^(−1) (1/μ)=(π/2)−tan^(−1) μ

$${a}=\frac{{g}}{\mu}={g}\:\mathrm{tan}\:\theta \\ $$$$\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mu}=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \mu \\ $$

Commented by Dwaipayan Shikari last updated on 24/Jun/20

f=μma f=mg (a/g)=tanθ , a=(g/μ) ,tanθ=(1/μ) θ=tan^(−1) (1/μ) θ=tan^(−1) (2)

$${f}=\mu{ma} \\ $$$${f}={mg} \\ $$$$\frac{{a}}{{g}}={tan}\theta\:\:,\:\:\:{a}=\frac{{g}}{\mu}\:\:,{tan}\theta=\frac{\mathrm{1}}{\mu}\:\:\:\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mu}\:\:\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

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