Question

As shown, a pipe is anchored to a wall at point A. During the pipe's installation, several forces are applied to the pipe at different locations. If F1 = 80.0N , F2 = 83.0N , F3 = 92.0N , F4 = 74.0N , d1 = 0.151m , d2 = 0.198m , and d3 = 0.272m , what is MRA, the net moment about point A due to these forces? Assume that moments acting counterclockwise about point A are positive whereas moments acting clockwise are negative.
$1108610_001.jpg$

Answer 1

Concepts and reason

Moment:

The magnitude of the moment can be determined by the product of force and the perpendicular distance to the force.

General sign conventions for moment:

Moment is considered as negative in clockwise direction and positive in counter clockwise direction.

Fundamentals

Moment acting is calculated by the product of the force magnitude and the perpendicular distance between the point of application of force and the point where moment is considered.

${\rm{Moment}}\left( M \right){\rm{ = Force}}\left( F \right) \times {\rm{Perpendicular distance}}\left( d \right)$

Total moment acting about a point is the summation of force moment caused by each individual force.

$\sum M = {F_1}{d_1} + {F_2}{d_2} + {F_3}{d_3}.......... + {F_n}{d_n}$

Here, suffices represent the number of forces acting.

The free body diagram is shown in figure (1),

Take the expression for summation of all force moments.

$\sum M = {F_1}{d_1} + {F_2}{d_2} + {F_3}{d_3}.......... + {F_n}{d_n}$

From the figure (1), equation reduces to

$\sum {{M_A}} = {F_1}{d_2} - {F_2}{d_2} + {F_3}\left( 0 \right) - {F_4}\left( {{d_3} - {d_1}} \right)$ …… (1)

From equation (1), the expression is written as,

$\sum {{M_A}} = {F_1}{d_2} - {F_2}{d_2} + {F_3}\left( 0 \right) - {F_4}\left( {{d_3} - {d_1}} \right)$

Substitute $80{\rm{ N}}$ for ${F_1}$ , 0.198 m for ${d_2}$ , $83{\rm{ N}}$ for ${F_2}$ , ${\rm{92 N}}$ for ${F_3}$ , ${\rm{74 N}}$ for ${F_4}$ , 0.272 m for ${d_3}$ and 0.151 m for ${d_1}.$

$\begin{array}{l}\\\sum {{M_A}} = \left( {80 \times 0.198} \right) - \left( {83 \times 0.198} \right) + \left( {92 \times 0} \right) - \left( {74 \times \left( {0.272 - 0.151} \right)} \right)\\\\\sum {{M_A}} = - 9.548{\rm{ N}} \cdot {\rm{m}}\\\end{array}$

Ans:

Magnitude of moment about point A $\left( {\sum {{M_A}} } \right)$ is $- {\bf{9}}{\bf{.548 N}} \cdot {\bf{m}}$ .