Question

An old wooden telephone pole with a single cross member was blown off center during a recent storm. To temporarily stabilize the telephone pole, a repair crew has attached a cable from the end of the cross member to an eye-ring stake anchored in the ground. (Figure 3)

If the origin O of a coordinate system is placed where the pole meets the ground, the point where the cross member is attached to the pole is A(?1.00,1.35,11.5)m and the point where the cable is attached to the end of the cross member is B(?1.00,?1.15,13.0)m.

The cable's direction is {1.00i?1.00j?1.00k} and it has a tension of 775N .

An old wooden telephone pole with a single cross member was blown off center during a recent storm. To temporarily stabilize the telephone pole, a repair crew has attached a cable from the end of the cross member to an eye-ring stake anchored in the ground. (Figure 3) If the origin O of a coordinate system is placed where the pole meets the ground, the point where the cross member is attached to the pole is A(?1.00,1.35,11.5)m and the point where the cable is attached to the end of the cross member is B(?1.00,?1.15,13.0)m. The cable's direction is {1.00i?1.00j?1.00k} and it has a tension of 775N. What is Ma, the moment that tends to twist the telephone pole due to the tension in the cable applied at point B?

What is Ma, the moment that tends to twist the telephone pole due to the tension in the cable applied at point B?

Answer 1

Concepts and reason

Free body diagram is the representation of all forces acting on the body.

Vector form of representation of location of a point from a reference point is called position vector.

Magnitude of moment can be calculated by multiplying the magnitude of force with the perpendicular distance between the point where the force is applied and the point where the moment to be found.

Moment of a force about a point is given by cross product of position vector of any point on the line of action of force from the point about which moment has to be calculated and force vector.

Fundamentals

Procedure to write position vector of point, $A = \left( {{x_A},\;{y_A}\;,{z_A}\;} \right)$ with reference to origin is as follows:

${{\bf{r}}_{{\bf{OA}}}} = \left( {{x_A}{\bf{i}} + {y_A}{\bf{j}} + {z_A}{\bf{k}}} \right)$

Here, coordinates of point A along x, y, and z axis are ${x_A},\;{y_A}\;,{\rm{ and }}{z_A}$ respectively and unit vectors along x, y, and z axis are ${\bf{i}}\;,{\bf{j}},\;{\rm{and}}\;{\bf{k}}$ respectively.

Procedure to write position vector of point, $A = \left( {{x_A},\;{y_A}\;,{z_A}\;} \right)$ with reference to point $B = \left( {{x_B},\;{y_B}\;,{z_B}\;} \right)$ is as follows:

$\begin{array}{c}\\{{\bf{r}}_{{\bf{AB}}}} = {{\bf{r}}_{{\bf{OB}}}} - {{\bf{r}}_{{\bf{OA}}}}\\\\{{\bf{r}}_{{\bf{AB}}}} = \left( {{x_B}{\bf{i}} + {y_B}{\bf{j}} + {z_B}{\bf{k}}} \right) - \left( {{x_A}{\bf{i}} + {y_A}{\bf{j}} + {z_A}{\bf{k}}} \right)\\\\ = \left( {{x_B} - {x_A}} \right){\bf{i}} + \left( {{y_B} - {y_A}} \right){\bf{j}} + \left( {{z_B} - {z_A}} \right){\bf{k}}\\\end{array}$

Here, coordinates of point B along x, y, and z axis are ${x_B},\;{y_B}\;,{\rm{ and }}{z_B}$ respectively

Equation for calculation of magnitude of vector, ${\bf{r}} = \left( {{r_x}{\bf{i}} + {r_y}{\bf{j}} + {r_z}{\bf{k}}} \right)$ is as follows:

$\left| {\bf{r}} \right| = \sqrt {{{\left( {{r_x}} \right)}^2} + {{\left( {{r_y}} \right)}^2} + {{\left( {{r_z}} \right)}^2}}$

Here, ${r_x}$ is the x-component of the position vector (coefficient of i component of position vector), ${r_y}$ is the y-component of the position vector (coefficient of j component of position vector), and ${r_z}$ is the z-component of the position vector (coefficient of k component of position vector).

Calculation of unit vector for the direction along which the force applied is prominent.

Equation used for calculation of unit vector.

${\bf{u}} = \frac{{\bf{r}}}{{\left| {\bf{r}} \right|}}$

Equation used for force vector.

${\bf{F}} = F.{\bf{u}}$

Moment about a point can be calculated in two ways.

First way:

It can be calculated by multiplying the magnitude of force with the perpendicular distance between the point of application of force and point about which moment is to be calculated.

Second way:

Use vector notation of position vector and force. The cross product of and distance vector and this force vector gives the moment in Cartesian vector form.

Basic equations for the calculation of internal forces are as follows:

Equation for calculation of moment about point O due force applied at point B.

${{\bf{M}}_O} = {{\bf{r}}_{OB}} \times {\bf{F}}$

Here, ${{\bf{r}}_{OB}}$ is the position of B with respect to O, ${{\bf{M}}_O}$ is the moment, and ${\bf{F}}$ is the force.

Calculate the moment about the axis using following relation:

$M = {{\bf{u}}_{OA}} \cdot \left( {{{\bf{r}}_{OB}} \times {\bf{F}}} \right)$

Here, the unit vector along pole axis.

General rule for cross product of the vectors is as follows:

$\begin{array}{l}\\{\bf{i}} \times {\bf{i}} = 0\\\\{\bf{i}} \times {\bf{j}} = {\bf{k}}\\\\{\bf{i}} \times {\bf{k}} = - {\bf{j}}\\\\{\bf{j}} \times {\bf{j}} = 0\\\\{\bf{j}} \times {\bf{i}} = - {\bf{k}}\\\\{\bf{j}} \times {\bf{k}} = {\bf{i}}\\\\{\bf{k}} \times {\bf{k}} = 0\\\\{\bf{k}} \times {\bf{i}} = {\bf{j}}\\\\{\bf{k}} \times {\bf{j}} = - {\bf{i}}\\\end{array}$

General sign convention for moment: The moment is considered positive in counter-clockwise direction and negative in clockwise direction.

General sign convention for axis: Distance along the axis is positive and opposite to the axis is negative.

Write the coordinates of point A, B.

$\begin{array}{l}\\O = \left( {0,0,0} \right)\\\\A = \left( { - 1,1.35,11.5} \right)\\\\B = \left( { - 1, - 1.15,13} \right)\\\end{array}$

Write the position vector of point A with respect to origin.

${{\bf{r}}_{OA}} = \left( { - {\bf{i}} + 1.35{\bf{j}} + 11.5{\bf{k}}} \right)\,{\rm{m}}$

Write the position vector of point B with respect to origin.

${{\bf{r}}_{OB}} = \left( { - {\bf{i}} - 1.15{\bf{j}} + 13{\bf{k}}} \right)\,{\rm{m}}$

Find the magnitude of ${{\bf{r}}_{OA}}$ .

$\begin{array}{c}\\{r_{OA}} = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( {1.35} \right)}^2} + {{11.5}^2}} \\\\ = 11.622\,{\rm{m}}\\\end{array}$

Calculate the unit vector of OA as shown.

$\begin{array}{c}\\{{\bf{u}}_{OA}} = \frac{{{{\bf{r}}_{OA}}}}{{{r_{OA}}}}\\\\ = \frac{{ - {\bf{i}} + 1.35{\bf{j}} + 11.5{\bf{k}}}}{{11.622}}\\\\ = - 0.086{\bf{i}} + 0.11615{\bf{j}} + 0.9895{\bf{k}}\\\end{array}$

Calculate the unit vector for cable direction.

$\begin{array}{c}\\{{\bf{u}}_{cable}} = \frac{{1{\bf{i}} - 1{\bf{j}} - 1{\bf{k}}}}{{\sqrt {{1^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2}} }}\\\\ = \frac{1}{{\sqrt 3 }}\left[ {{\bf{i}} - {\bf{j}} - {\bf{k}}} \right]\\\end{array}$

Write the tension vector along cable.

$\begin{array}{c}\\{\bf{T}} = T\left( {{{\bf{u}}_{cable}}} \right)\\\\ = 775\left( {\frac{1}{{\sqrt 3 }}\left[ {{\bf{i}} - {\bf{j}} - {\bf{k}}} \right]} \right)\\\\ = \left[ {447.44{\bf{i}} - 447.44{\bf{j}} - 447.44{\bf{k}}} \right]\,{\rm{N}}\\\end{array}$

Calculate the twisting moment due to the tension in the cable.

$\begin{array}{c}\\{M_a} = {{\bf{u}}_{OA}} \cdot \left( {{{\bf{r}}_{OB}} \times {\bf{T}}} \right)\\\\ = \left[ { - 0.086{\bf{i}} + 0.11615{\bf{j}} + 0.9895{\bf{k}}} \right] \cdot \left( {\left[ { - {\bf{i}} - 1.15{\bf{j}} + 13{\bf{k}}} \right] \times \left[ {447.44{\bf{i}} - 447.44{\bf{j}} - 447.44{\bf{k}}} \right]} \right)\\\\ = \left[ { - 0.086{\bf{i}} + 0.11615{\bf{j}} + 0.9895{\bf{k}}} \right] \cdot \left( {6331.28{\bf{i}} + 5369.28{\bf{j}} + 961.996{\bf{k}}} \right)\\\\ = \left( { - 0.086 \times 6331.28 + 0.11615 \times 5369.28 + 0.9895 \times 961.996} \right)\\\end{array}$

${M_a} = 1031.046\,{\rm{N - m}}$

Ans:

Therefore, the twisting moment due to the tension in the cable is $1031.046\,{\rm{N - m}}$ .