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Q60P

Expert-verifiedFound in: Page 351

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**In Fig. 12-69, a package of mass m hangs from a short cord that is tied to the wall via cord 1 and to the ceiling via cord 2. Cord 1 is at angle${\mathit{\varphi}}{\mathbf{=}}{\mathbf{}}{\mathbf{40}}{\mathbf{\xb0}}$ **

a) The angle of cord 2 with the horizontal i.e.$\text{\theta}=50\xb0$ **.**

b) The minimum tension in cord 2 in terms of mg is the$\text{0.77mg}$.

** **

$\varphi =40\xb0$

**From the Free Body Diagram and condition for static equilibrium of the system, we can find the angle and the tension in cord 2.**

**Equations:**

${{\mathbf{\sum}}}_{}^{}{{\mathit{F}}}_{{\mathbf{x}}}{\mathbf{=}}{\mathbf{0}}$

${{\mathbf{\sum}}}_{}^{}{{\mathit{F}}}_{{\mathbf{y}}}{\mathbf{=}}{\mathbf{0}}$

${\mathbf{\text{Weight}}}{\mathbf{=}}{\mathbf{\text{mg}}}$

** **

$\begin{array}{l}{\sum}_{}^{}{\text{F}}_{\text{x}}=0\\ {\text{T}}_{\text{1}}\text{cos 40}\xb0-{\text{T}}_{\text{2}}\text{cos \theta}=0\\ {\text{T}}_{\text{1}}\text{cos 40}\xb0={\text{T}}_{\text{2}}\text{cos \theta}\\ {\text{T}}_{\text{1}}=\frac{{\text{T}}_{\text{2}}\text{cos \theta}}{\text{cos 40}\xb0}\end{array}$

$\begin{array}{l}{\sum}_{}^{}{\text{F}}_{\text{y}}=0\\ {\text{T}}_{\text{1}}{\text{sin 40\xb0 +T}}_{\text{2}}\text{sin\theta}-\text{mg}=0\\ \frac{{\text{T}}_{\text{2}}\text{cos \theta}}{\text{cos 40}\xb0}\times {\text{sin 40\xb0+T}}_{\text{2}}\text{sin\theta}=\text{mg}\\ {\text{T}}_{\text{2}}(\text{sin\theta}+\text{cos \theta tan40}\xb0)=\text{mg}\\ {\text{T}}_{\text{2}}=\frac{\text{mg}}{(\text{sin\theta}+\text{cos\theta tan40}\xb0)}\end{array}$

As the value in the ${\text{T}}_{\text{2}}$should be minimum, we take derivative at both sides, and we get,

$\begin{array}{l}\frac{d}{d\theta}\left({T}_{2}\right)=\frac{d}{d\theta}\left[\frac{\text{mg}}{(\text{sin\theta}+\text{cos\theta tan40}\xb0)}\right]\\ 0=\frac{-\text{mg}(\text{tan 40}\xb0\text{}\times -\text{sin \theta}+\text{cos \theta})}{{\left((\text{tan 40}\xb0\text{\xd7cos \theta})\text{+sin\theta}\right)}^{\text{2}}}\\ (\text{tan 40}\xb0\text{}\times -\text{sin \theta}+\text{cos \theta})=0\\ \text{tan 40}\xb0\text{}=\mathrm{cot}\theta \\ \theta ={\mathrm{cot}}^{-1}(\text{tan 40}\xb0\text{})\\ \theta =50\xb0\end{array}$

As we know,

$\begin{array}{c}{\text{T}}_{\text{2}}\text{=}\frac{\text{mg}}{(\text{tan 40}\xb0\text{\xd7cos \theta})\text{+sin\theta}}\\ \text{=}\frac{\text{mg}}{(\text{tan 40}\xb0\text{\xd7cos 50}\xb0\text{})\text{+sin 50}\xb0\text{}}\\ \text{=0.77mg}\end{array}$

The minimum tension in cord 2 in terms of mg is the$\text{0.77mg}$.

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