Question

Answer: M = 51.8 j - 193.2 k N.m

Determine the moment associated with the pair of 400 N forces applied to the T-shaped structure.

Answer 1

Concepts and reason

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.

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

Fundamentals

Resolve each force into its components to completely specify the force vector. In three dimensional problems, a force can be written as

${\bf{F}} = {F_x}{\bf{i}} + {F_y}{\bf{j}} + {F_z}{\bf{k}}$

Here, the component of force in x-direction is ${F_x}$ , the component of force along y-direction is ${F_y}$ , and the component of force in z-direction is ${F_z}$ .

Position vector between two points,

$\begin{array}{l}\\{{\bf{r}}_A} = {x_1}{\bf{i}} + {y_1}{\bf{j}} + {z_1}{\bf{k}}\\\\{{\bf{r}}_B} = {x_2}{\bf{i}} + {y_2}{\bf{j}} + {z_2}{\bf{k}}\\\end{array}$

${{\bf{r}}_{AB}} = \left( {{x_2} - {x_1}} \right){\bf{i}} + \left( {{y_2} - {y_1}} \right){\bf{j}} + \left( {{z_2} - {z_1}} \right){\bf{k}}$

Moment of force acting at some point C about a point A can be calculated using the formula,

${{\bf{M}}_A} = {\bf{r}} \times {\bf{F}}$

General rule for cross product of the vectors,

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

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

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

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 the points.

$\begin{array}{l}\\A\left( { - 0.25,0,0.45} \right)\\\\B\left( {0.25,0,0.45} \right)\\\end{array}$

Calculate the position vector of the cable AB.

$\begin{array}{c}\\{\bf{r}} = A - B\\\\ = \left( { - 0.25 - 0.25} \right){\bf{i}} + \left( {0 - 0} \right){\bf{j}} + \left( {0.45 - 0.45} \right){\bf{k}}\\\\ = \left( { - 0.5{\bf{i}}} \right)\,{\rm{m}}\\\end{array}$

Calculate the force vector.

$\begin{array}{c}\\{\bf{F}} = 400\left( {\cos 15^\circ {\bf{j}} + \sin 15^\circ {\bf{k}}} \right)\\\\ = 386.37{\bf{j}} + 103.53{\bf{k}}\\\end{array}$

Calculate the moment of force using the following expression

${\bf{M}} = {\bf{r}} \times {\bf{F}}$

Substitute $- 0.5\,{\bf{i}}$ for ${\bf{r}}$ and $386.37{\bf{j}} + 103.53{\bf{k}}$ for ${\bf{F}}$ .

$\begin{array}{c}\\{\bf{M}} = \left( { - 0.5\,{\bf{i}}} \right) \times \left( {386.37{\bf{j}} + 103.53{\bf{k}}} \right)\\\\ = \left( {51.76\,{\bf{j}} - 193.18\,{\bf{k}}} \right)\,{\rm{N}} \cdot {\rm{m}}\\\end{array}$

Ans:

Thus, the moment associated is $\left( {51.76\,{\bf{j}} - 193.18\,{\bf{k}}} \right)\,{\rm{N}} \cdot {\rm{m}}$