Question

In Fig. 12-69, a package of mass mhangs 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$\mathit{\varphi }\mathbf{=}\mathbf{}\mathbf{40}\mathbf{°}$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

Short answer

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$\text{0.77mg}$.

Step-by-step solution

Listing the given quantities

$\varphi =40°$

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:

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

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

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

Free body diagram

(a) Calculations of the angle of cord 2 with the horizontal

$\begin{array}{l}{\sum }_{}^{}{\text{F}}_{\text{x}}=0\\ {\text{T}}_{\text{1}}\text{cos40}°-{\text{T}}_{\text{2}}\text{cosθ}=0\\ {\text{T}}_{\text{1}}\text{cos40}°={\text{T}}_{\text{2}}\text{cosθ}\\ {\text{T}}_{\text{1}}=\frac{{\text{T}}_{\text{2}}\text{cosθ}}{\text{cos40}°}\end{array}$

$\begin{array}{l}{\sum }_{}^{}{\text{F}}_{\text{y}}=0\\ {\text{T}}_{\text{1}}{\text{sin40°+T}}_{\text{2}}\text{sinθ}-\text{mg}=0\\ \frac{{\text{T}}_{\text{2}}\text{cosθ}}{\text{cos40}°}\times {\text{sin40°+T}}_{\text{2}}\text{sinθ}=\text{mg}\\ {\text{T}}_{\text{2}}(\text{sinθ}+\text{cosθtan40}°)=\text{mg}\\ {\text{T}}_{\text{2}}=\frac{\text{mg}}{(\text{sinθ}+\text{cosθtan40}°)}\end{array}$

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

$\begin{array}{l}\frac{d}{d\theta }\left(T_2\right)=\frac{d}{d\theta }\left[\frac{\text{mg}}{(\text{sinθ}+\text{cosθtan40}°)}\right]\\ 0=\frac{-\text{mg}(\text{tan40}°\times -\text{sinθ}+\text{cosθ})}{{\left((\text{tan40}°\text{×cosθ})\text{+sinθ}\right)}^{\text{2}}}\\ (\text{tan40}°\times -\text{sinθ}+\text{cosθ})=0\\ \text{tan40}°=\cot \theta \\ \theta ={\cot }^{-1}(\text{tan40}°)\\ \theta =50°\end{array}$

(b) Calculations of the minimum tension in the cord in terms of mg                

As we know,

$\begin{array}{c}{\text{T}}_{\text{2}}\text{=}\frac{\text{mg}}{(\text{tan40}°\text{×cosθ})\text{+sinθ}}\\ \text{=}\frac{\text{mg}}{(\text{tan40}°\text{×cos50}°)\text{+sin50}°}\\ \text{=0.77mg}\end{array}$

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