Chapter 9, Problem 67GP

(a)

To determine

To Find: The tension in the cable at lowest point.

Answer to Problem 67GP

  $0.29mg$

Explanation of Solution

Given:

The weight of the flexible steel cable, $mg$

Formula used:

Let the tension at the lowest point, $F_{T2}$

The tension at the points of attachment, $F_{T1}$

  

From the equilibrium condition in x-direction:

  $\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{T2}-F_{T1}\cos 60°=0\mathrm{.....}(1)\end{array}$

From the equilibrium condition in y-direction:

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{T1}\sin 60°-\frac{1}{2}mg=0\mathrm{.....}(2)\end{array}$

Calculation:

From equation (2):

  $\begin{array}{l}{F}_{T1}\sin 60°-\frac{1}{2}mg=0\\ F_{T1}=\frac{mg}{2\sin 60°}=0.58mg\end{array}$

Substitute the value of $F_{T1}$ in equation (1):

  $\begin{array}{l}{F}_{T2}-(0.58mg)\cos 60°=0\\ F_{T2}=0.29mg\end{array}$

Conclusion:

The value of tension at the lowest point is $0.29mg$ .

(b)

To determine

To find: The tension in the cable at point of attachment.

Answer to Problem 67GP

  $0.58mg$

Explanation of Solution

Given:

The weight of the flexible steel cable, $mg$

Formula used:

Let the tension at the lowest point, $F_{T2}$

The tension at the points of attachment, $F_{T1}$

  

From the equilibrium condition in x-direction:

  $\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{T2}-F_{T1}\cos 60°=0\mathrm{.....}(1)\end{array}$

From the equilibrium condition in y-direction:

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{T1}\sin 60°-\frac{1}{2}mg=0\mathrm{....}(2)\end{array}$

Calculation:

From equation (2):

  $\begin{array}{l}{F}_{T1}\sin 60°-\frac{1}{2}mg=0\\ F_{T1}=\frac{mg}{2\sin 60°}=0.58mg\end{array}$

Conclusion:

The value of tension at the point of attachment is $0.58mg$ .

(c)

To determine

To calculate: The direction of the tension force in both cases.

Explanation of Solution

Given:

The weight of the flexible steel cable, $mg$

Let the tension at the lowest point, $F_{T2}$

The tension at the points of attachment, $F_{T1}$

The direction of tension is shown in the diagram below.

  

Conclusion:

The direction of tension force at the lowest point is along horizontal & at the point of attachment make an angle 60° with the x-direction.