Step 1 We are given the force and distance as shown in figure.
We are asked to find the magnitude of the projected component of the force along the pipe AO.
Step 2 Write the position vector of point A.
\[{\vec r_A} = \left( {0\hat i + 4\hat j + 6\hat k} \right)\;{\rm{m}}\]
Write the position vector of point B.
\[{\vec r_B} = \left( {4\hat i + 5\hat j + 0\hat k} \right)\;{\rm{m}}\]
Write the position vector of the line AB$\left( {{{\vec r}_{AB}}} \right)$.
\[{\vec r_{AB}} = {\vec r_B} - {\vec r_A}\]
Step 3 To find the position vector of the line AB$\left( {{{\vec r}_{AB}}} \right)$, substitute the position vectors of point A and B in the above expression, we get:
\[\begin{array}{c} {{\vec r}_{AB}} = \left( {4\hat i + 5\hat j + 0\hat k} \right)\;{\rm{m}} - \left( {0\hat i + 4\hat j + 6\hat k} \right)\;{\rm{m}}\\ {{\vec r}_{AB}} = \left( {4\hat i - 0\hat i} \right)\;{\rm{m}} + \left( {5\hat j - 4\hat j} \right)\;{\rm{m}} + \left( {0\hat k - 6\hat k} \right)\;{\rm{m}}\\ {{\vec r}_{AB}} = \left( {4\hat i + \hat j - 6\hat k} \right)\;{\rm{m}} \end{array}\]
To find the magnitude of the vector along line AB$\left( {\left| {{{\vec r}_{AB}}} \right|} \right)$, we need to solve the above equation as follows:
\[\begin{array}{c} \left| {{{\vec r}_{AB}}} \right| = \sqrt {{{\left( 4 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( { - 6} \right)}^2}} \\ \left| {{{\vec r}_{AB}}} \right| = \sqrt {53} \\ \left| {{{\vec r}_{AB}}} \right| = 7.28\;{\rm{m}} \end{array}\]
Step 4 The unit vector along the line AB can be given as follows:
\[{\vec a_{AB}} = \frac{{{{\vec r}_{AB}}}}{{\left| {{{\vec r}_{AB}}} \right|}}\]
To find the unit vector along the line AB, substitute the values in the above expression, we get:
\[\begin{array}{c} {{\vec a}_{AB}} = \frac{{\left( {4\hat i + \hat j - 6\hat k} \right)\;{\rm{m}}}}{{7.28\;{\rm{m}}}}\\ {{\vec a}_{AB}} = 0.549\hat i + 0.137\hat j - 0.824\hat k \end{array}\]
Step 5 The force vector $\left( {\vec F} \right)$ can be given as follows:
\[\vec F = F \cdot {\vec a_{AB}}\]
Here, F is the magnitude of force.
To find the force vector $\left( {\vec F} \right)$, substitute the values in the above expression, we get:
\[\begin{array}{c} \vec F = \left( {400\;{\rm{N}}} \right) \cdot \left( {0.549\hat i + 0.137\hat j - 0.824\hat k} \right)\\ \vec F = \left( {219.6\hat i + 54.8\hat j - 329.6\hat k} \right)\;{\rm{N}} \end{array}\]
Step 6 Write the position vector of point O.
\[{\vec r_O} = \left( {0\hat i + 0\hat j + 0\hat k} \right)\;{\rm{m}}\]
Write the position vector along the pipe AO$\left( {{{\vec r}_{AO}}} \right)$.
\[{\vec r_{AO}} = {\vec r_O} - {\vec r_A}\]
Step 7 To find the position vector of pipe AO$\left( {{{\vec r}_{AO}}} \right)$, substitute the position of vector of point O and A in the above expression, we get:
\[\begin{array}{c} {{\vec r}_{AO}} = \left( {0\hat i + 0\hat j + 0\hat k} \right)\;{\rm{m}} - \left( {0\hat i + 4\hat j + 6\hat k} \right)\;{\rm{m}}\\ {{\vec r}_{AO}} = \left( {0\hat i - 0\hat i} \right)\;{\rm{m}} + \left( {0\hat j - 4\hat j} \right)\;{\rm{m}} + \left( {0\hat k - 6\hat k} \right)\;{\rm{m}}\\ {{\vec r}_{AO}} = \left( {0\hat i - 4\hat j - 6\hat k} \right)\;{\rm{m}} \end{array}\]
To find the magnitude of the vector along pipe AO$\left( {\left| {{{\vec r}_{AO}}} \right|} \right)$, we need to solve the above equation as follows:
\[\begin{array}{c} \left| {{{\vec r}_{AO}}} \right| = \sqrt {{{\left( 0 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2}} \\ \left| {{{\vec r}_{AO}}} \right| = \sqrt {52} \\ \left| {{{\vec r}_{AO}}} \right| = 7.21\;{\rm{m}} \end{array}\]
Step 8 The unit vector along the pipe AO can be given as follows:
\[{\vec a_{AO}} = \frac{{{{\vec r}_{AO}}}}{{\left| {{{\vec r}_{AO}}} \right|}}\]
To find the unit vector along the pipe AO, substitute the values in the above expression, we get:
\[\begin{array}{c} {{\vec a}_{AO}} = \frac{{\left( {0\hat i - 4\hat j - 6\hat k} \right)\;{\rm{m}}}}{{7.21\;{\rm{m}}}}\\ {{\vec a}_{AO}} = 0\hat i - 0.555\hat j - 0.832\hat k \end{array}\]
Step 9 The magnitude of the force along the pipe AO can be given as follows:
\[{F_{AO}} = \vec F \cdot {\vec a_{AO}}\]
Here, $\vec F$is the force vector.
To find the magnitude of the force, substitute the values in the above expression, we get:
\[\begin{array}{c} {F_{AO}} = \left( {219.6\hat i + 54.8\hat j - 329.6\hat k} \right)\;{\rm{N}} \cdot \left( {0\hat i - 0.555\hat j - 0.832\hat k} \right)\\ {F_{AO}} = \left( {219.6\hat i \times 0\hat i} \right)\;{\rm{N}} + \left( {54.8\hat j \times \left( { - 0.555\hat j} \right)} \right)\;{\rm{N}} + \left( { - 329.6\hat k \times - 0.832\hat k} \right)\;{\rm{N}}\\ {F_{AO}} = 0\;{\rm{N}} - 30.414\;{\rm{N}} + 274.22\;{\rm{N}}\\ {F_{AO}} = 243.863\;\;{\rm{N}} \approx 244\;{\rm{N}}\\ {F_{AO}} = 244\;{\rm{N}} \end{array}\]
