Express forces in Cartesian vector form.
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Question “Express forces in Cartesian vector form.”
Express each of the forces in Cartesian vector form and determine the magnitude and coordinate direction angles of the resultant force.
Answer
Vector algebra, unit vector, position, resultant from two vectors, and triangle law for vector addition are the concepts needed to answer this question.
Vector algebra This is a collection of concepts that describes the method of different vector operations, such as vector product or scalar products of vectors.
Unit vector: A vector that runs along a direction with the magnitude of 1 is called a unit vector.
Position vector:Position Vector shows the position of a point within a vector space, in terms of unit vectors in various directions.
Resultant from two vectors The vector sum of two vectors, in Cartesian vector form, is the resultant of these two vectors.
Triangle Law on vector addition: Two vectors can be considered the adjacent sides of a triangle. The resultant is the third side. The vectors are oriented so that the head and tail of the first vector are at the same time. The tail of a resultant vector is located at the tail of a first vector, while the head is at its head.
Calculate the unit vector along with the CA to calculate the vector form for the force {F_1}
Position vector:
The expression of a position vector \bfOA \bfOA = x\bf\hat i + y\bf\hat j + z\bf\hat k
Here, x
The position vector of a point A with respect to the fixed-point O is the vector \bfOA
Triangle law to vector addition:
Consider that the position vector of a point B with respect to the fixed-point O is the vector \bfOB
The following is the expression for triangle law vector addition:
\bfOA + \bfAB = OBTo solve for vector \bfAB \bfAB = OB - \bfOA
Unit vector:
Unit vector for any vector \bfAB \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}} \left| {{\bf{AB}}} \right| = \sqrt {{a^2} + {b^2} + {c^2}}
The expression of force towards \mathop {{\bf{AB}}}\limits^ \wedge {\bf{F}} = \left| {\bf{F}} \right|\mathop {{\bf{AB}}}\limits^ \wedge
The resultant of two vector in Cartesian vector form:
The expression of the resultant of two vectors in Cartesian vector form for two vectors is {\bf{A}} = {a_1}{\bf{\hat i}} + {a_2}{\bf{\hat j}} + {a_3}{\bf{\hat k}} \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}}
Expression for “coordinate direction angle” or “angle formed by a resultant Vector”. Use x
CodeMathTag19: Coordinated direction angle for the resultant vector
{\theta _x} = {\cos ^{ - 1}}\left( {\frac{{{{\bf{R}}_x}}}{{\left| {\bf{R}} \right|}}} \right)CodeMathTag22: Coordinated direction angle for the resultant vector
{\theta _y} = {\cos ^{ - 1}}\left( {\frac{{{{\bf{R}}_y}}}{{\left| {\bf{R}} \right|}}} \right)CodeMathTag24: Coordinated direction angle for the resultant vector
{\theta _z} = {\cos ^{ - 1}}\left( {\frac{{{{\bf{R}}_z}}}{{\left| {\bf{R}} \right|}}} \right)Here, {{\bf{R}}_x}
Draw the diagram in the question.
Here, {z_c}
Find the length {z_c} \begin{array}{l}\\\frac{{12}}{5} = \frac{{{z_c}}}{{2.5{\rm{ ft}}}}\\\\{z_c} = 6{\rm{ ft}}\\\end{array}
Find the position vector for the point C relative to the origin at O.
{\bf{C}} = - 2.5{\bf{\hat i}} + 0{\bf{\hat j}} + 6{\bf{\hat k}}Find the position vector for the point A relative to the origin at O.
{\bf{A}} = 0{\bf{\hat i}} + 4{\bf{\hat j}} + 0{\bf{\hat k}}Use triangle law property to determine the position vector along .
{\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}} \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 and direction vectors 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 to describe the unit vector along CA.
{{\bf{u}}_{CA}} = \frac{{{\bf{CA}}}}{{\left| {{\bf{CA}}} \right|}}Here, {{\bf{u}}_{CA}}
Substitute \left( { - 2.5{\bf{\hat i}} - {\rm{4}}{\bf{\hat j}} + {\rm{6}}{\bf{\hat k}}} \right){\rm{ ft}} \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 to represent the vector form for force {F_1} {{\bf{F}}_{\bf{1}}} = {F_1}{{\bf{u}}_{{\rm{CA}}}}
Substitute 80{\rm{ lb}} \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 for B.
{\bf{B}} = \left( {2{\bf{\hat i}} + {\rm{0}}{\bf{\hat j}} - {\rm{6}}{\bf{\hat k}}} \right){\rm{ ft}}Describe the expression that describes 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}} \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 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 BB.
{{\bf{u}}_{BA}} = \frac{{{\bf{BA}}}}{{\left| {{\bf{BA}}} \right|}}Here, {{\bf{u}}_{BA}}
Substitute \left( {2{\bf{\hat i}} - {\rm{4}}{\bf{\hat j}} - {\rm{6}}{\bf{\hat k}}} \right){\rm{ ft}} \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 to represent the vector form for force {F_2} {{\bf{F}}_{\bf{2}}} = {F_2}{{\bf{u}}_{BA}}
Substitute 50{\rm{ lb}} \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 to describe 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}} \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 and direction of the 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}The expression to calculate the coordinate angle angle of the force resulting from x-axis.
\cos {\theta _x} = \frac{{{{\bf{F}}_x}}}{{\left| {\bf{F}} \right|}}Here, {\theta _x}
Substitute 73.47{\rm{ lb}} \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}
The expression to calculate the resultant of force in relation to yaxis.
\cos {\theta _y} = \frac{{{{\bf{F}}_y}}}{{\left| {\bf{F}} \right|}}Here, {\theta _y}
Substitute 73.47{\rm{ lb}} \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}
The expression to calculate the resultant of force in relation to z-axis.
\cos {\theta _z} = \frac{{{{\bf{F}}_z}}}{{\left| {\bf{F}} \right|}}Here, {\theta _z}
Substitute 73.47{\rm{ lb}}
\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}}}
The force vector {{\bf{F}}_2}
The 73.49{\rm{ lb}} is the magnitude of the resultant force
The x indicates the coordinate direction angle for the resultant force
CodeMathTag22 – The coordinate direction angle for the resultant force
CodeMathTag24 – The coordinate direction angle for the resultant force
Conclusion
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