Question

A sign is to be hung from the end of a thin pole, and the polesupported by a single cable. Your design firm brainstorms the sixscenarios shown below. In scenarios A, B, and D, the cable isattached halfway between the midpoint and end of the pole. In C,the cable is attached to the mid-point of the pole. In E and F, thecable is attached to the end of the pole.


A sign is to be hung from the end of a thin pole, and the polesupported by a single cable. Your design firm brainstorms the sixscenarios shown below. In scenarios A, B, and D, the cable isattached halfway between the midpoint and end of the pole. In C,the cable is attached to the mid-point of the pole. In E and F, thecable is attached to the end of the pole. Rank the design scenarios (A throughF) on the basis of the tension in the supportingcable. Rank from largest tosmallest. To rank items as equivalent, overlap them.

Rank the design scenarios (A throughF) on the basis of the tension in the supportingcable.

Rank from largest tosmallest. To rank items as equivalent, overlap them.

Answer 1

Concepts and reason

The concept used to solve this problem is tension in the cable that exerts a torque about the pole.

Initially, write an equation for the tension in the cable that exerts the torque about the pole. Later, calculate the weight of the sign that exerts a torque. Finally, rearrange for the tension equation by equating the tension in the cable that exerts the torque about the pole to the by the weight of the sign that exerts a torque and rank based on the tension.

Fundamentals

The expression for the tension in the cable that exerts the torque about the pole is as follows:

$\tau = rT\sin \theta$

Here, r is the perpendicular distance, T is the tension in the string, and $\theta$ is the angle between the distance and the tension vector.

The expression for the weight of the sign that exerts a torque is as follows:

$\tau ' = Lmg$

Here, L is the length, m is the mass, and g is the acceleration due to gravity.

When the system is under equilibrium, the expression is as follows:

$\tau = \tau '$

Substitute $rT\sin \theta$ for $\tau$ and $Lmg$ for $\tau '$ in the equation $\tau = \tau '$ and rearrange for the expression of T.

$\begin{array}{c}\\rT\sin \theta = Lmg\\\\T = \frac{{Lmg}}{{r\sin \theta }}\\\end{array}$

Therefore, the expression for the tension in the cable is as follows:

$T = \frac{{Lmg}}{{r\sin \theta }}$

Substitute 8.00 m for L, $\frac{3}{4}$ for r, and $60^\circ$ for $\theta$ in the equation $T = \frac{{Lmg}}{{r\sin \theta }}$.

$\begin{array}{c}\\{T_{\rm{A}}} = \frac{{\left( {8.00{\rm{ m}}} \right)mg}}{{\left( {\frac{3}{4}} \right)\sin 60^\circ }}\\\\ = \frac{{\left( {8.00{\rm{ m}}} \right)mg}}{{\left( {\frac{3}{4}} \right)\left( {\frac{{\sqrt 3 }}{2}} \right)}}\\\\ = \frac{{8mg}}{{3\sqrt 3 }}\\\end{array}$

Substitute 4.00 m for L, $\frac{3}{4}$ for r, and $90^\circ$ for $\theta$ in the equation $T = \frac{{Lmg}}{{r\sin \theta }}$.

$\begin{array}{c}\\{T_{\rm{B}}} = \frac{{\left( {4.00{\rm{ m}}} \right)mg}}{{\left( {\frac{3}{4}} \right)\sin 90^\circ }}\\\\ = \frac{{\left( {4.00{\rm{ m}}} \right)mg}}{{\left( {\frac{3}{4}} \right)\left( {1.00} \right)}}\\\\ = \frac{{4mg}}{3}\\\end{array}$

Substitute L for L, $\frac{1}{2}$ for r, and $90^\circ$ for $\theta$ in the equation $T = \frac{{Lmg}}{{r\sin \theta }}$.

$\begin{array}{c}\\{T_{\rm{C}}} = \frac{{Lmg}}{{\left( {\frac{1}{2}} \right)\sin 90^\circ }}\\\\ = \frac{{Lmg}}{{\left( {\frac{1}{2}} \right)\left( {1.00} \right)}}\\\\ = 2mg\\\end{array}$

Substitute $4\sqrt 2$ for L, $\frac{3}{4}$ for r, and $45^\circ$ for $\theta$ in the equation $T = \frac{{Lmg}}{{r\sin \theta }}$.

$\begin{array}{c}\\{T_{\rm{D}}} = \frac{{\left( {4\sqrt 2 } \right)mg}}{{\left( {\frac{3}{4}} \right)\sin 45^\circ }}\\\\ = \frac{{\left( {4\sqrt 2 } \right)mg}}{{\left( {\frac{3}{4}} \right)\left( {\frac{1}{{\sqrt 2 }}} \right)}}\\\\ = \frac{{4\sqrt 2 mg}}{3}\\\end{array}$

Substitute L for L, L for r, and $90^\circ$ for $\theta$ in the equation $T = \frac{{Lmg}}{{r\sin \theta }}$.

$\begin{array}{c}\\{T_{\rm{E}}} = \frac{{Lmg}}{{L\sin 90^\circ }}\\\\ = \frac{{Lmg}}{{L\left( {1.00} \right)}}\\\\ = mg\\\end{array}$

Substitute L for L, L for r, and $30^\circ$ for $\theta$ in the equation $T = \frac{{Lmg}}{{r\sin \theta }}$.

$\begin{array}{c}\\{T_{\rm{F}}} = \frac{{Lmg}}{{L\sin 30^\circ }}\\\\ = \frac{{Lmg}}{{L\left( {\frac{1}{2}} \right)}}\\\\ = 2mg\\\end{array}$

Ans:

The ranking of the tension form largest to smallest is ${T_{\rm{C}}} = {T_{\rm{F}}} > {T_{\rm{D}}} > {T_{\rm{A}}} > {T_{\rm{B}}} > {T_{\rm{E}}}$ .