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Expert-verified Found in: Page 351 ### Fundamentals Of Physics

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${\mathbit{\varphi }}{\mathbf{=}}{\mathbf{}}{\mathbf{40}}{\mathbf{°}}$ with the horizontal; cord 2 is at angle${\mathbf{\text{θ}}}$. (a) For what value of ${\mathbf{\text{θ}}}$ is the tension in cord 2 minimized? (b) In terms of mg, what is the minimum tension in cord 2 a) The angle of cord 2 with the horizontal i.e.$\text{θ}=50°$ .

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

See the step by step solution

## Step 1: Listing the given quantities

$\varphi =40°$

## Step 2: Understanding the concept of force and tension

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:

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

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

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

## Step 3: Free body diagram ## Step 4: (a) Calculations of the angle of cord 2 with the horizontal

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

$\begin{array}{l}{\sum }_{}^{}{\text{F}}_{\text{y}}=0\\ {\text{T}}_{\text{1}}{\text{sin 40° +T}}_{\text{2}}\text{sinθ}-\text{mg}=0\\ \frac{{\text{T}}_{\text{2}}\text{cos θ}}{\text{cos 40}°}×{\text{sin 40°+T}}_{\text{2}}\text{sinθ}=\text{mg}\\ {\text{T}}_{\text{2}}\left(\text{sinθ}+\text{cos θtan40}°\right)=\text{mg}\\ {\text{T}}_{\text{2}}=\frac{\text{mg}}{\left(\text{sinθ}+\text{cosθ tan40}°\right)}\end{array}$

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

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

## Step 5: (b) Calculations of the minimum tension in the cord in terms of mg

As we know,

$\begin{array}{c}{\text{T}}_{\text{2}}\text{=}\frac{\text{mg}}{\left(\text{tan 40}°\text{×cos θ}\right)\text{+sinθ}}\\ \text{=}\frac{\text{mg}}{\left(\text{tan 40}°\text{×cos 50}°\text{}\right)\text{+sin 50}°\text{}}\\ \text{=0.77mg}\end{array}$

The minimum tension in cord 2 in terms of mg is the$\text{0.77mg}$. ### Want to see more solutions like these? 