Question

Determine the magnitude of resultant force and angle with positive\(x\)-axis.The magnitude of first vector as \(20{\rm{lb}}\) and an angle with positive \(x\)-axis as \({45^^\circ }\) and the magnitude of second vector as \(16{\rm{lb}}\) and angle with positive \(x\)-axis as\( - 30^\circ \).

Short answer

The magnitude of resultant force and angle with positive \(x\)-axis are \(28.7{\rm{lb}}\) and \(12.4^\circ \) respectively.

Step-by-step solution

Formula used

Write the expression for vector in terms of horizontal and vertical components.

\({\bf{F}} = |{\bf{F}}|\cos \theta {\bf{i}} + |{\bf{F}}|\sin \theta {\bf{j}} \ldots \ldots \ldots (1)\)

Here,

\(|{\bf{F}}|\)is magnitude of vector\({\bf{F}}\).

Write the expression for resultant vector of two vectors\({{\bf{F}}_1}\)and\({{\bf{F}}_2}({\bf{F}})\)

\({\bf{F}} = {{\bf{F}}_1} + {{\bf{F}}_2}{\rm{ }} \ldots \ldots \ldots {\rm{(2) }}\)

Write the expression for magnitude of vector\({F_1}{\bf{i}} + {F_2}{\bf{j}}\)

\(|{\bf{F}}| = \sqrt {F_1^2 + F_2^2} \ldots \ldots \ldots {\rm{ (3) }}\)

Write the expression for\(\tan \theta \)of a vector\({\bf{F}}\)

\(\tan \theta = \frac{{{F_2}}}{{{F_1}}}{\rm{ }} \ldots \ldots \ldots {\rm{(4) }}\)

Determine the magnitude of resultant force and angle with positive \(x\)-axis

As given, the magnitude of first vector as \(20{\rm{lb}}\) and an angle with positive \(x\)-axis as \({45^^\circ }\) and the magnitude of second vector as \(16{\rm{lb}}\) and angle with positive \(x\)-axis as \( - {30^^\circ }\)shown in figure 1.

Substitute \(20{\rm{lb}}\) for\(|{\bf{F}}|\), and \({45^^\circ }\) for \(\theta \) in equation (1),

\(\begin{aligned}{l}{{\bf{F}}_1} = (20{\rm{lb}})\cos \left( {{{45}^^\circ }} \right){\bf{i}} + (20{\rm{lb}})\sin \left( {{{45}^^\circ }} \right){\bf{j}}\\{{\bf{F}}_1} = (20{\rm{lb}})\left( {\frac{1}{{\sqrt 2 }}} \right){\bf{i}} + (20{\rm{lb}})\left( {\frac{1}{{\sqrt 2 }}} \right){\bf{j}}\\{{\bf{F}}_1} = 10\sqrt 2 {\bf{i}} + 10\sqrt 2 {\bf{j}}{\rm{lb}}\end{aligned}\)

Substitute \(16{\rm{lb}}\) for\(|{\bf{F}}|\), and \( - {30^^\circ }\) for \(\theta \) in equation (1),

\(\begin{aligned}{l}{{\bf{F}}_2} = (16{\rm{lb}})\cos \left( { - {{30}^^\circ }} \right){\bf{i}} + (16{\rm{lb}})\sin \left( { - {{30}^^\circ }} \right){\bf{j}}\\{{\bf{F}}_2} = (16{\rm{lb}})\left( {\frac{{\sqrt 3 }}{2}} \right){\bf{i}} + (16{\rm{lb}})\left( { - \frac{1}{2}} \right){\bf{j}}\\{{\bf{F}}_2} = 8\sqrt 3 {\bf{i}} - 8{\bf{j}}{\rm{lb}}\end{aligned}\)

Substitute \(10\sqrt 2 {\bf{i}} + 10\sqrt 2 {\bf{j}}\) lb for \({{\bf{F}}_1}\) and \(8\sqrt 3 {\bf{i}} + 8{\bf{j}}{\rm{lb}}\) for \({{\bf{F}}_2}\) in equation (2),

\(\begin{aligned}{l}{\bf{F}} = (10\sqrt 2 {\bf{i}} + 10\sqrt 2 {\bf{j}}){\rm{lb}} + (8\sqrt 3 {\bf{i}} - 8{\bf{j}}){\rm{lb}}\\{\bf{F}} = (10\sqrt 2 + 8\sqrt 3 ){\bf{i}} + (10\sqrt 2 - 8){\bf{jl}}b\end{aligned}\)

Substitute \(10\sqrt 2 + 8\sqrt 3 \) for \({F_1}\) and \(10\sqrt 2 - 8\) for \({F_2}\) in equation (3),

\(\begin{aligned}{l}|{\bf{F}}| = \sqrt {{{(10\sqrt 2 + 8\sqrt 3 )}^2} + {{(10\sqrt 2 - 8)}^2}} {\rm{lb}}\\|{\bf{F}}| = \sqrt {{{(27.998)}^2} + {{(6.142)}^2}} {\rm{lb}}\\|{\bf{F}}| = \sqrt {783.888 + 37.724} {\rm{lb}}\\|{\bf{F}}| = \sqrt {821.612} {\rm{lb}}\\|{\bf{F}}| = 28.66{\rm{lb}}\\|{\bf{F}}| \simeq 28.7{\rm{lb}}\end{aligned}\)

Re-arrange the equation (4)

\(\theta = {\tan ^{ - 1}}\left( {\frac{{{F_2}}}{{{F_1}}}} \right)\)

Substitute \(10\sqrt 2 + 8\sqrt 3 \) for \({F_1}\) and \(10\sqrt 2 - 8\) for \({F_2}\),

\(\begin{aligned}{l}\theta = {\tan ^{ - 1}}\left( {\frac{{10\sqrt 2 - 8}}{{10\sqrt 2 + 8\sqrt 3 }}} \right)\\\theta = {\tan ^{ - 1}}\left( {\frac{{6.142}}{{27.998}}} \right)\\\theta = {\tan ^{ - 1}}(0.21937)\quad \\\theta = {12.37^^\circ }\theta \\\theta \simeq {12.4^^\circ }\end{aligned}\)

Thus, the magnitude of resultant force and angle with positive \(x\)-axis are \(28.7{\rm{lb}}\) and \(12.4^\circ \) respectively.