Engineering Mechanics: Statics and Dynamics, 14th Edition

Engineering Mechanics: Statics and Dynamics, 14th Edition

Authors: Russell C. Hibbeler

ISBN-13: 978-0133915426

See our solution for Question 34P from Chapter 4 from Hibbeler's Engineering Mechanics.

Problem 34P

Chapter:
Problem:
Determine the moment of the force of F about point A.

Step-by-Step Solution

Step 1

We are given the force acting along the rope is $F = 600\;{\rm{N}}$.


We are asked the moment of the force of $F = 600\;{\rm{N}}$ about point A.


 
Step 2

The following is the diagram of the system:

Images


We have the vector representation of point A is $A = \left( {0i + 0j + 4k} \right){\rm{ m}}$.

We have the vector representation of point B is $B = \left( {4\sin 45^\circ i + 0j + 4\cos 45^\circ k} \right){\rm{ m}}$.

We have the vector representation of point C is $C = \left( {6i + 6j + 0k} \right){\rm{ m}}$.


 
Step 3

The position vector of component AB is,

\[{R_{AB}} = B - A\]
 
Step 4

Substitute the values in the above expression.

\begin{array}{l} {R_{AB}} = \left( {4\sin 45^\circ i + 0j + 4\cos 45^\circ k} \right){\rm{ m}} - \left( {0i + 0j + 4k} \right){\rm{ m}}\\ {R_{AB}} = \left( {2.8i + 0j - 1.2k} \right)\;{\rm{m}} \end{array}
 
Step 5

The position vector of component BC is,

\[{R_{BC}} = C - B\]
 
Step 6

Substitute the values in the above expression.

\begin{array}{l} {R_{BC}} = \left( {6i + 6j + 0k} \right){\rm{ m}} - \left( {4\sin 45^\circ i + 0j + 4\cos 45^\circ k} \right){\rm{ m}}\\ {R_{BC}} = \left( {3.2i + 6j - 2.8k} \right){\rm{ m}} \end{array}
 
Step 7

The Cartesian vector form of force is,

\[F = F \cdot \frac{{{R_{BC}}}}{{\left| {{R_{BC}}} \right|}}\]
 
Step 8

Substitute the values in the above expression.

\begin{array}{l} F = \left( {600\;{\rm{N}}} \right) \cdot \frac{{\left( {3.2i + 6j - 2.8k} \right){\rm{ m}}}}{{\sqrt {{{\left( {3.2} \right)}^2} + {{\left( 6 \right)}^2} + {{\left( { - 2.8} \right)}^2}} \;{\rm{m}}}}\\ F = \left( {261.1i + 489.6j - 228.5k} \right){\rm{ N}} \end{array}
 
Step 9

The moment of the force of $F = 600\;{\rm{N}}$ about point A is,

\[M = F \times {R_{AB}}\]
 
Step 10

Substitute the values in the above expression.

\[\begin{array}{l} M = \left( {261.1i + 489.6j - 228.5k} \right){\rm{ N}} \times \left( {2.8i + 0j - 1.2k} \right)\;{\rm{m}}\\ M = \left| {\begin{array}{*{20}{c}} i&j&k\\ {261.1}&{489.6}&{ - 228.5}\\ {2.8}&0&{ - 1.2} \end{array}} \right|\\ M = \left\{ \begin{array}{l} i\left[ {\left\{ {489.6 \times \left( { - 1.2} \right)} \right\} - \left\{ {\left( { - 228.5} \right) \times 0} \right\}} \right]\\ - j\left[ {\left\{ {261.1 \times \left( { - 1.2} \right)} \right\} - \left\{ {\left( { - 228.5} \right) \times 2.8} \right\}} \right]\\ + k\left[ {\left\{ {\left( {261.1 \times 0} \right)} \right\} - \left\{ {\left( {489.6 \times 2.8} \right)} \right\}} \right] \end{array} \right\}\\ M = \left( { - 587.5i - 326.5j - 1370.9k} \right)\;{\rm{N}} \cdot {\rm{m}} \end{array}\]