Question

Express each of the forces in Cartesian vector form and determine the magnitude and coordinate direction angles of the resultant force.

Answer 1

Concepts and reason

The concepts required to solve this question are vector algebra, unit vector, position vector, resultant of two vectors and triangle law of vector addition.

Vector algebra: This is a bunch of concepts which describes the methodology of the different vector operations such as vector product and scalar product of vectors and other operations.

Unit vector: A unit vector is a vector along a direction which has the magnitude of $1$ . These vectors are generally used to denote a direction as ${\bf{\hat i}}$ is assigned x-direction, ${\bf{\hat j}}$ is assigned to y-direction and ${\bf{\hat k}}$ is assigned to z-direction in a general 3-dimensional field.

Position vector: Position vector shows the location of a point in a vector space in terms of the unit vectors in different directions.

Resultant of two vectors: The resultant of the two vectors is expressed as the vector sum of two vectors in Cartesian vector form.

Triangle Law of vector addition: When two vectors are considered as the adjacent side of a triangle, the resultant is given by the third side of the triangle. The orientation of the vectors is such that the head of first vector is at the tail of the second vector. The tail of the resultant vector is at the tail of the first vector and the head is at the head of the second vector.

Calculate the unit vector along the CA and use it to calculate the vector form of the force ${F_1}$ , calculate the unit vector along BA which would be used to calculate the vector form of ${F_2}$ . Use the expression for the resultant of two vectors to calculate the resultant of the provided forces. Use the expression for the angle between the resultant vector and x, y and z axes to calculate the respective coordinate direction angle of the resultant force.

Fundamentals

Position vector:

The expression of a position vector ${\bf{OA}}$ in three-dimension space is:

${\bf{OA}} = x{\bf{\hat i}} + y{\bf{\hat j}} + z{\bf{\hat k}}$

Here, $x$ , $y$ and $z$ are the coordinate points in three-dimension space with respect to origin O.

The position vector of a point A with respect to the fixed-point O is the vector ${\bf{OA}}$ .

Triangle law for vector addition:

Consider that the position vector of a point B with respect to the fixed-point O is the vector ${\bf{OB}}$ . The position vector of a point A with respect to the fixed-point O is the vector ${\bf{OA}}$ .

The expression for triangle law of vector addition is:

${\bf{OA}} + {\bf{AB = OB}}$

To solve for vector ${\bf{AB}}$ , rearrange the expression as:

${\bf{AB = OB}} - {\bf{OA}}$

Unit vector:

Unit vector for any vector ${\bf{AB}}$ is $\mathop {{\bf{AB}}}\limits^ \wedge$ which is calculated as,

$\mathop {{\bf{AB}}}\limits^ \wedge = \frac{{{\bf{AB}}}}{{\left| {{\bf{AB}}} \right|}}$

For a vector ${\bf{AB}} = a{\bf{\hat i}}{\rm{ + b}}{\bf{\hat j}} + c{\bf{\hat k}}$ , magnitude of this vector is calculated as,

$\left| {{\bf{AB}}} \right| = \sqrt {{a^2} + {b^2} + {c^2}}$

The expression of force in the direction of $\mathop {{\bf{AB}}}\limits^ \wedge$ as a Cartesian vector in three-dimensional space is,

${\bf{F}} = \left| {\bf{F}} \right|\mathop {{\bf{AB}}}\limits^ \wedge$

Resultant of two vector (in Cartesian vector form):

The expression of resultant of two vectors (in Cartesian vector form) for two vectors represented as, ${\bf{A}} = {a_1}{\bf{\hat i}} + {a_2}{\bf{\hat j}} + {a_3}{\bf{\hat k}}$ and ${\bf{B}} = {b_1}{\bf{\hat i}} + {b_2}{\bf{\hat j}} + {b_3}{\bf{\hat k}}$ , is:

$\begin{array}{c}\\{\bf{R}} = {\bf{A}} + {\bf{B}}\\\\ = \left( {{a_1}{\bf{\hat i}} + {a_2}{\bf{\hat j}} + {a_3}{\bf{\hat k}}} \right) + \left( {{b_1}{\bf{\hat i}} + {b_2}{\bf{\hat j}} + {b_3}{\bf{\hat k}}} \right)\\\\ = \left( {{a_1} + {b_1}} \right){\bf{\hat i}} + \left( {{a_2} + {b_2}} \right){\bf{\hat j}} + \left( {{a_3} + {b_3}} \right){\bf{\hat k}}\\\end{array}$

Here, ${\bf{R}}$ is the resultant of the vectors ${\bf{AB}}$ and ${\bf{BC}}$ .

Expression for ‘coordinate direction angle’ or ‘angle formed by a resultant vector’ with respective $x$ , $y$ and $z$ axis is written as,

Coordinate direction angle of the resultant vector with $x$ axis $\left( {{\theta _x}} \right)$ .

${\theta _x} = {\cos ^{ - 1}}\left( {\frac{{{{\bf{R}}_x}}}{{\left| {\bf{R}} \right|}}} \right)$

Coordinate direction angle of the resultant vector with $y$ axis $\left( {{\theta _y}} \right)$ .

${\theta _y} = {\cos ^{ - 1}}\left( {\frac{{{{\bf{R}}_y}}}{{\left| {\bf{R}} \right|}}} \right)$

Coordinate direction angle of the resultant vector with $z$ axis $\left( {{\theta _z}} \right)$ .

${\theta _z} = {\cos ^{ - 1}}\left( {\frac{{{{\bf{R}}_z}}}{{\left| {\bf{R}} \right|}}} \right)$

Here, ${{\bf{R}}_x}$ , ${{\bf{R}}_y}$ and ${{\bf{R}}_z}$ are the $x$ , $y$ and $z$ components of resultant vector ${\bf{R}}$ .

Draw the diagram provided in the question.

Here, ${z_c}$ is the distance between point C and the x axis.

Calculate the length ${z_c}$ .

$\begin{array}{l}\\\frac{{12}}{5} = \frac{{{z_c}}}{{2.5{\rm{ ft}}}}\\\\{z_c} = 6{\rm{ ft}}\\\end{array}$

Determine the position vector of the point C with respect to the origin at O.

${\bf{C}} = - 2.5{\bf{\hat i}} + 0{\bf{\hat j}} + 6{\bf{\hat k}}$

Determine the position vector of the point A with respect to the origin at O.

${\bf{A}} = 0{\bf{\hat i}} + 4{\bf{\hat j}} + 0{\bf{\hat k}}$

Determine the position vector along CA (using triangle law property).

${\bf{CA}} = {\bf{C}} - {\bf{A}}$

Substitute $\left( { - 2.5{\bf{\hat i}} + {\rm{0}}{\bf{\hat j}} + {\rm{6}}{\bf{\hat k}}} \right){\rm{ ft}}$ for ${\bf{C}}$ and $\left( {0{\bf{\hat i}} + {\rm{4}}{\bf{\hat j}} + {\rm{0}}{\bf{\hat k}}} \right){\rm{ ft}}$ for ${\bf{A}}$ .

$\begin{array}{c}\\{\bf{CA}} = \left( { - 2.5{\bf{\hat i}} + {\rm{0}}{\bf{\hat j}} + {\rm{6}}{\bf{\hat k}}} \right){\rm{ ft}} - \left( {0{\bf{\hat i}} + {\rm{4}}{\bf{\hat j}} + {\rm{0}}{\bf{\hat k}}} \right){\rm{ ft}}\\\\ = \left( { - 2.5{\bf{\hat i}} - {\rm{4}}{\bf{\hat j}} + {\rm{6}}{\bf{\hat k}}} \right){\rm{ ft}}\\\end{array}$

Calculate the magnitude of the position vector along the CA $\left( {\left| {{\bf{CA}}} \right|} \right)$ .

$\begin{array}{c}\\\left| {{\bf{CA}}} \right| = \sqrt {{{\left( { - 2.5{\rm{ ft}}} \right)}^2} + {{\left( { - {\rm{4 ft}}} \right)}^2} + {{\left( {{\rm{6 ft}}} \right)}^2}} \\\\ = \sqrt {{\rm{6}}{\rm{.25 f}}{{\rm{t}}^2} + {\rm{16 f}}{{\rm{t}}^2} + {\rm{36 f}}{{\rm{t}}^2}} \\\\ = \sqrt {{\rm{58}}{\rm{.25 f}}{{\rm{t}}^2}} \\\\ = 7.63{\rm{ ft}}\\\end{array}$

Write the expression for the unit vector along CA.

${{\bf{u}}_{CA}} = \frac{{{\bf{CA}}}}{{\left| {{\bf{CA}}} \right|}}$

Here, ${{\bf{u}}_{CA}}$ is unit vector along CA.

Substitute $\left( { - 2.5{\bf{\hat i}} - {\rm{4}}{\bf{\hat j}} + {\rm{6}}{\bf{\hat k}}} \right){\rm{ ft}}$ for ${\bf{CA}}$ and $7.63{\rm{ ft}}$ for $\left| {{\bf{CA}}} \right|$ .

$\begin{array}{c}\\{{\bf{u}}_{{\bf{CA}}}} = \frac{{\left( { - 2.5{\bf{\hat i}} - {\rm{4}}{\bf{\hat j}} + {\rm{6}}{\bf{\hat k}}} \right){\rm{ ft}}}}{{7.63{\rm{ ft}}}}\\\\ = - 0.328{\bf{\hat i}} - 0.524{\bf{\hat j}} + 0.786{\bf{\hat k}}\\\end{array}$

Write the expression for the vector form of the force ${F_1}$ .

${{\bf{F}}_{\bf{1}}} = {F_1}{{\bf{u}}_{{\rm{CA}}}}$

Substitute $80{\rm{ lb}}$ for ${F_1}$ and $- 0.328{\bf{\hat i}} - 0.524{\bf{\hat j}} + 0.786{\bf{\hat k}}$ for ${{\bf{u}}_{{\rm{CA}}}}$ .

$\begin{array}{c}\\{{\bf{F}}_{\bf{1}}} = 80{\rm{ lb}}\left( { - 0.328{\bf{\hat i}} - 0.524{\bf{\hat j}} + 0.786{\bf{\hat k}}} \right)\\\\ = \left( { - 26.24{\bf{\hat i}} - 41.92{\bf{\hat j}} + 62.88{\bf{\hat k}}} \right){\rm{ lb}}\\\end{array}$

Calculate the position vector of B.

${\bf{B}} = \left( {2{\bf{\hat i}} + {\rm{0}}{\bf{\hat j}} - {\rm{6}}{\bf{\hat k}}} \right){\rm{ ft}}$

Write the expression for the position vector along BA.

${\bf{BA}} = {\bf{B}} - {\bf{A}}$

Substitute $\left( {2{\bf{\hat i}} + {\rm{0}}{\bf{\hat j}} - {\rm{6}}{\bf{\hat k}}} \right){\rm{ ft}}$ for ${\bf{B}}$ and $\left( {0{\bf{\hat i}} + {\rm{4}}{\bf{\hat j}} + {\rm{0}}{\bf{\hat k}}} \right){\rm{ ft}}$ for A.

$\begin{array}{c}\\{\bf{BA}} = \left( {2{\bf{\hat i}} + {\rm{0}}{\bf{\hat j}} - {\rm{6}}{\bf{\hat k}}} \right){\rm{ ft}} - \left( {0{\bf{\hat i}} + {\rm{4}}{\bf{\hat j}} + {\rm{0}}{\bf{\hat k}}} \right){\rm{ ft}}\\\\ = \left( {2{\bf{\hat i}} - {\rm{4}}{\bf{\hat j}} - {\rm{6}}{\bf{\hat k}}} \right){\rm{ ft}}\\\end{array}$

Calculate the magnitude of vector BA.

$\begin{array}{c}\\\left| {{\bf{BA}}} \right| = \sqrt {{{\left( {2{\rm{ ft}}} \right)}^2} + {{\left( { - {\rm{4 ft}}} \right)}^2} + {{\left( { - {\rm{6 ft}}} \right)}^2}} \\\\ = \sqrt {{\rm{4 f}}{{\rm{t}}^2} + {\rm{16 f}}{{\rm{t}}^2} + {\rm{36 f}}{{\rm{t}}^2}} \\\\ = \sqrt {{\rm{56 f}}{{\rm{t}}^2}} \\\\ = 7.48{\rm{ ft}}\\\end{array}$

Calculate the unit vector along BA.

${{\bf{u}}_{BA}} = \frac{{{\bf{BA}}}}{{\left| {{\bf{BA}}} \right|}}$

Here, ${{\bf{u}}_{BA}}$ is unit vector along the BA.

Substitute $\left( {2{\bf{\hat i}} - {\rm{4}}{\bf{\hat j}} - {\rm{6}}{\bf{\hat k}}} \right){\rm{ ft}}$ for BA and $7.48{\rm{ ft}}$ for $\left| {{\bf{BA}}} \right|$ .

$\begin{array}{c}\\{{\bf{u}}_{BA}} = \frac{{\left( {2{\bf{\hat i}} - {\rm{4}}{\bf{\hat j}} - {\rm{6}}{\bf{\hat k}}} \right){\rm{ ft}}}}{{7.48{\rm{ ft}}}}\\\\ = 0.267{\bf{\hat i}} - 0.535{\bf{\hat j}} - 0.802{\bf{\hat k}}\\\end{array}$

Write the expression for the vector form of the force ${F_2}$ .

${{\bf{F}}_{\bf{2}}} = {F_2}{{\bf{u}}_{BA}}$

Substitute $50{\rm{ lb}}$ for ${F_2}$ and $0.267{\bf{\hat i}} - 0.535{\bf{\hat j}} - 0.802{\bf{\hat k}}$ for ${{\bf{u}}_{BA}}$

$\begin{array}{c}\\{{\bf{F}}_{\bf{2}}} = 50{\rm{ lb}}\left( {0.267{\bf{\hat i}} - 0.535{\bf{\hat j}} - 0.802{\bf{\hat k}}} \right)\\\\ = \left( {13.35{\bf{\hat i}} - 26.75{\bf{\hat j}} - 40.10{\bf{\hat k}}} \right){\rm{ lb}}\\\end{array}$

Write the expression for the resultant force.

${\bf{F}} = {{\bf{F}}_{\bf{1}}} + {{\bf{F}}_{\bf{2}}}$

Substitute $\left( { - 26.24{\bf{\hat i}} - 41.92{\bf{\hat j}} + 62.88{\bf{\hat k}}} \right){\rm{ lb}}$ for ${{\bf{F}}_{\bf{1}}}$ and $\left( {13.35{\bf{\hat i}} - 26.75{\bf{\hat j}} - 40.10{\bf{\hat k}}} \right){\rm{ lb}}$ for ${{\bf{F}}_{\bf{2}}}$ .

$\begin{array}{c}\\{\bf{F}} = \left( { - 26.16{\bf{\hat i}} - 41.92{\bf{\hat j}} + 62.88{\bf{\hat k}}} \right){\rm{ lb}} + \left( {13.35{\bf{\hat i}} - 26.75{\bf{\hat j}} - 40.10{\bf{\hat k}}} \right){\rm{ lb}}\\\\ = \left( { - 12.89{\bf{\hat i}} - 68.67{\bf{\hat j}} + 22.78{\bf{\hat k}}} \right){\rm{ lb}}\\\end{array}$

Calculate the magnitude of the resultant force.

$\begin{array}{c}\\\left| {\bf{F}} \right| = \sqrt {{{\left( { - 12.89{\rm{ lb}}} \right)}^2} + {{\left( { - 68.67{\rm{ lb}}} \right)}^2} + {{\left( {22.78{\rm{ lb}}} \right)}^2}} \\\\ = \sqrt {5400.6494} {\rm{ lb}}\\\\ = 73.489{\rm{ lb}}\\\\ \approx 73.49{\rm{ lb}}\\\end{array}$

Write the expression for the coordinate direction angle of the resultant force with respect to x-axis.

$\cos {\theta _x} = \frac{{{{\bf{F}}_x}}}{{\left| {\bf{F}} \right|}}$

Here, ${\theta _x}$ is angle between the resultant force and x-axis and ${{\bf{F}}_x}$ is component of the resultant force in the x-direction (coefficient of ${\bf{\hat i}}$ ).

Substitute $73.47{\rm{ lb}}$ for $\left| {\bf{F}} \right|$ and $- 12.89{\rm{ lb}}$ for ${{\bf{F}}_x}$ .

$\begin{array}{l}\\\cos {\theta _x} = \frac{{ - 12.81{\rm{ lb}}}}{{73.47{\rm{ lb}}}}\\\\\cos {\theta _x} = - 0.1754\\\\{\theta _x} = {\cos ^{ - 1}}\left( { - 0.1754} \right)\\\\ = 100.10^\circ \\\end{array}$

Write the expression for the resultant of the force with respect to y-axis.

$\cos {\theta _y} = \frac{{{{\bf{F}}_y}}}{{\left| {\bf{F}} \right|}}$

Here, ${\theta _y}$ is angle between the force and y-axis and ${{\bf{F}}_y}$ is component of the resultant force in the y-direction (coefficient of ${\bf{\hat j}}$ ).

Substitute $73.47{\rm{ lb}}$ for $\left| {\bf{F}} \right|$ and $- 68.67{\rm{ lb}}$ for ${{\bf{F}}_y}$ .

$\begin{array}{l}\\\cos {\theta _y} = \frac{{ - 68.67{\rm{ lb}}}}{{73.47{\rm{ lb}}}}\\\\{\theta _y} = {\cos ^{ - 1}}\left( { - 0.935} \right)\\\\ = 159.23^\circ \\\end{array}$

Write the expression for the resultant of the force with respect to z-axis.

$\cos {\theta _z} = \frac{{{{\bf{F}}_z}}}{{\left| {\bf{F}} \right|}}$

Here, ${\theta _z}$ is angle between the force and z-axis and ${{\bf{F}}_z}$ is component of the resultant force in the z-direction (coefficient of ${\bf{\hat k}}$ ).

Substitute $73.47{\rm{ lb}}$ for $\left| {\bf{F}} \right|$ and ${\rm{22}}{\rm{.78 lb}}$ for ${{\bf{F}}_z}$ .

$\begin{array}{l}\\\cos {\theta _z} = \frac{{22.78{\rm{ lb}}}}{{73.47{\rm{ lb}}}}\\\\{\theta _z} = {\cos ^{ - 1}}\left( { - 0.310} \right)\\\\ = 71.94^\circ \\\end{array}$

Ans:

The force vector ${{\bf{F}}_{\bf{1}}}$ in Cartesian vector form is $\left( { - 26.24{\bf{\hat i}} - 41.92{\bf{\hat j}} + 62.88{\bf{\hat k}}} \right){\rm{ lb}}$ .

The force vector ${{\bf{F}}_2}$ in Cartesian vector form is $\left( {13.35{\bf{\hat i}} - 26.75{\bf{\hat j}} - 40.10{\bf{\hat k}}} \right){\rm{ lb}}$ .

The magnitude of resultant force is $73.49{\rm{ lb}}$ .

The coordinate direction angle of the resultant force with the $x$ -axis is $100.10^\circ$ .

The coordinate direction angle of the resultant force with $y$ -axis is $159.23^\circ$ .

The coordinate direction angle of the resultant force with $z$ axis is $71.94^\circ$ .