Question

(a) Repeat the problem two problems prior, but for the second leg you walk \(20.0{\rm{ m}}\) in a direction \(40.0^\circ \) north of east (which is equivalent to subtracting \({\rm{B}}\) from \({\rm{A}}\) —that is, to finding \({\rm{R'}} = {\rm{A}} - {\rm{B}}\)).

(b) Repeat the problem two problems prior, but now you first walk \(20.0{\rm{ m}}\) in a direction \(40.0^\circ \) south of west and then \(12.0{\rm{ m}}\) in a direction \(20.0^\circ \) east of south (which is equivalent to subtracting \({\rm{A}}\) from \({\rm{B}}\) —that is, to finding \({\rm{R''}} = {\rm{B}} - {\rm{A}} = - {\rm{R'}}\)). Show that this is the case.

Short answer

(a) The magnitude of the resultant vector is \(26.62{\rm{ m}}\) and is directed towards \(65.1^\circ \) the north of the east.

(b) The magnitude of the resultant vector is \(26.62{\rm{ m}}\) and is directed towards \(65.1^\circ \) south of west.

Step-by-step solution

Triangle rule of vector addition

When two vectors are taken along two sides of a triangle, the third side of the triangle is always in reverse order. It represents the magnitude and direction of the resultant vector, according to the triangle rule of vector addition.

Given data

• Magnitude of vector \({\rm{A}}\) \(A = 12\,{\rm{m}}\).
• Direction of vector \({\rm{A}}\) \({\theta _A} = 20^\circ \).
• Magnitude of vector \({\rm{B}}\) \(B = 20.0\,{\rm{m}}\).
• Direction of vector \({\rm{B}}\) \({\theta _B} = 40.0^\circ \).

(a) Vector Representation

The vectors are represented as,

Fig: Representation of vector

Horizontal component of the resultant vector

The horizontal component of the vector \({\rm{A}}\) is,

\({A_x} = - A\sin {\theta _A}\)

Here \(A\) is the magnitude of a vector \({\rm{A}}\)and \({\theta _A}\) is the angle between \(y\)-the axis and the vector \({\rm{A}}\).

Substitute \(12{\rm{ m}}\) for \(A\) and \(20^\circ \) for \({\theta _A}\),

\(\begin{aligned}{}{A_x} = - \left( {12{\rm{ m}}} \right) \times \sin \left( {20^\circ } \right)\\ = - 4.1{\rm{ m}}\end{aligned}\)

The horizontal component of the vector \({\rm{B}}\) is,

\({B_x} = B\cos {\theta _B}\)

Here \(B\) is the magnitude of the vector \({\rm{B}}\), and \({\theta _B}\) is made by the vector \({\rm{B}}\) with the horizontal.

Substitute \(20{\rm{ m}}\) for \(B\) and \(40^\circ \) for \({\theta _B}\),

\(\begin{aligned}{}{B_x} = \left( {20{\rm{ m}}} \right) \times \cos \left( {40^\circ } \right)\\ = 15.32{\rm{ m}}\end{aligned}\)

The resultant horizontal component of the vector \({\rm{A}}\) and \({\rm{B}}\) is,

\({R'_x} = {A_x} + {B_x}\)

Substitute \( - 4.1{\rm{ m}}\) for \({A_x}\) and \(15.32{\rm{ m}}\) for \({B_x}\),

\(\begin{aligned}{}{{R'}_x} = \left( { - 4.1{\rm{ m}}} \right) + \left( {15.32{\rm{ m}}} \right)\\ = 11.22{\rm{ m}}\end{aligned}\)

Vertical component of the resultant vector

The vertical component of the vector \({\rm{A}}\) is,

\({A_y} = A\cos {\theta _A}\)

Here \(A\) is the magnitude of the vector \({\rm{A}}\)and \({\theta _A}\) is the angle between \(y\)the -axis and the vector \({\rm{A}}\).

Substitute \(12{\rm{ m}}\) for \(A\) and \(20^\circ \) for \({\theta _A}\),

\(\begin{aligned}{}{A_y} = \left( {12{\rm{ m}}} \right) \times \cos \left( {20^\circ } \right)\\ = 11.28{\rm{ m}}\end{aligned}\)

The vertical component of the vector \(B\) is,

\({B_y} = B\sin {\theta _B}\)

Here \(B\) is the magnitude of a vector \({\rm{B}}\), and \({\theta _B}\) is the made by the vector \({\rm{B}}\) with the horizontal.

Substitute \(20{\rm{ m}}\) for \(B\) and \(40^\circ \) for \({\theta _B}\),

\(\begin{aligned}{}{B_y} = \left( {20{\rm{ m}}} \right) \times \sin \left( {40^\circ } \right)\\ = 12.86{\rm{ m}}\end{aligned}\)

The resultant vertical component of the vector \({\rm{A}}\) and \({\rm{B}}\) is,

\({R'_y} = {A_y} + {B_y}\)

Substitute \({\rm{11}}{\rm{.28 m}}\) for \({A_y}\) and \(12.86{\rm{ m}}\) for \({B_y}\),

\(\begin{aligned}{}{{R'}_y} = \left( {11.28{\rm{ m}}} \right) + \left( {12.86{\rm{ m}}} \right)\\ = 24.14{\rm{ m}}\end{aligned}\)

Resultant

The resultant of vector \({\rm{A}}\) and \({\rm{B}}\) is,

\[{R}'=\sqrt{{{R_x}'^2}+{{R_y}'^2}}\]

Substitute \(11.22{\rm{ m}}\) for \({R'_x}\) and \(24.14{\rm{ m}}\) for \({R'_y}\),

\(\begin{aligned}{}R' = \sqrt {{{\left( {11.22{\rm{ m}}} \right)}^2} + {{\left( {24.14{\rm{ m}}} \right)}^2}} \\ = 26.62{\rm{ m}}\end{aligned}\)

The direction of the resultant vector is,

\(\theta = {\tan ^{ - 1}}\left( {\frac{{{{R'}_y}}}{{{{R'}_x}}}} \right)\)

Substitute \(11.22{\rm{ m}}\) for \({R'_x}\) and \(24.14{\rm{ m}}\) for \({R'_y}\),

\(\begin{aligned}{}\alpha = {\tan ^{ - 1}}\left( {\frac{{24.14{\rm{ m}}}}{{11.22{\rm{ m}}}}} \right)\\ = 65.1^\circ \end{aligned}\)

Hence, the magnitude of the resultant vector is \(26.62{\rm{ m}}\) and is directed towards \(65.1^\circ \) north of east.

(b) Vector Representation

The vectors are represented as,

Representation of vector

Horizontal component of the resultant vector

The horizontal component of the vector \({\rm{A}}\) is,

\({A_x} = A\sin {\theta _A}\)

Here \(A\) is the magnitude of the vector \({\rm{A}}\)and \({\theta _A}\) is the angle between \(y\)-axis and the vector \({\rm{A}}\).

Substitute \(12{\rm{ m}}\) for \(A\) and \(20^\circ \) for \({\theta _A}\),

\(\begin{aligned}{}{A_x} = \left( {12{\rm{ m}}} \right) \times \sin \left( {20^\circ } \right)\\ = 4.1{\rm{ m}}\end{aligned}\)

The horizontal component of the vector \({\rm{B}}\) is,

\({B_x} = - B\cos {\theta _B}\)

Here \(B\) is the magnitude of the vector \({\rm{B}}\) and \({\theta _B}\) is the angle made by the vector \({\rm{B}}\) with the horizontal.

Substitute \(20{\rm{ m}}\) for \(B\) and \(40^\circ \) for \({\theta _B}\),

\(\begin{aligned}{}{B_x} = - \left( {20{\rm{ m}}} \right) \times \cos \left( {40^\circ } \right)\\ = - 15.32{\rm{ m}}\end{aligned}\)

The resultant horizontal component of the vector \({\rm{A}}\) and \({\rm{B}}\) is,

\({R''_x} = {A_x} + {B_x}\)

Substitute \(4.1{\rm{ m}}\) for \({A_x}\) and \( - 15.32{\rm{ m}}\) for \({B_x}\),

\(\begin{aligned}{}{{R''}_x} = \left( {4.1{\rm{ m}}} \right) + \left( { - 15.32{\rm{ m}}} \right)\\ = - 11.22{\rm{ m}}\end{aligned}\)

Vertical component of the resultant vector

The vertical component of the vector \({\rm{A}}\) is,

\({A_y} = - A\cos {\theta _A}\)

Here \(A\) is the magnitude of the vector \({\rm{A}}\)and \({\theta _A}\) is the angle between \(y\)-axis and the vector \({\rm{A}}\).

Substitute \(12{\rm{ m}}\) for \({\rm{A}}\) and \(20^\circ \) for \({\theta _A}\),

\(\begin{aligned}{}{A_y} = - \left( {12{\rm{ m}}} \right) \times \cos \left( {20^\circ } \right)\\ = - 11.28{\rm{ m}}\end{aligned}\)

The vertical component of the vector \({\rm{B}}\) is,

\({B_y} = - B\sin {\theta _B}\)

Here \(B\) is the magnitude of the vector \({\rm{B}}\), and \({\theta _B}\) is the made by the vector \({\rm{B}}\) with the horizontal.\({\rm{B}}\)

Substitute \(20{\rm{ m}}\) for \(B\) and \(40^\circ \) for \({\theta _B}\),

\(\begin{aligned}{}{B_y} = - \left( {20{\rm{ m}}} \right) \times \sin \left( {40^\circ } \right)\\ = - 12.86{\rm{ m}}\end{aligned}\)

The resultant vertical component of the vector \({\rm{A}}\) and is,

\({R''_y} = {A_y} + {B_y}\)

Substitute \( - {\rm{11}}{\rm{.28 m}}\) for \({A_y}\) and \( - 12.86{\rm{ m}}\) for \({B_y}\),

\(\begin{aligned}{}{{R''}_y} = \left( { - 11.28{\rm{ m}}} \right) + \left( { - 12.86{\rm{ m}}} \right)\\ = - 24.14{\rm{ m}}\end{aligned}\)

Resultant

The resultant of vector \({\rm{A}}\) and \({\rm{B}}\) is,

\({R}'=\sqrt{{{R_x}'^2}+{{R_y}'^2}}\)

Substitute \( - 11.22{\rm{ m}}\) for \({R''_x}\) and \( - 24.14{\rm{ m}}\) for \({R''_y}\),

\(\begin{aligned}{}R'' = \sqrt {{{\left( {11.22{\rm{ m}}} \right)}^2} + {{\left( {24.14{\rm{ m}}} \right)}^2}} \\ = 26.62{\rm{ m}}\end{aligned}\)

The direction of the resultant vector is,

\(\beta = {\tan ^{ - 1}}\left( {\frac{{{{R''}_y}}}{{{{R''}_x}}}} \right)\)

Substitute \( - 11.22{\rm{ m}}\) for \({R''_x}\) and \( - 24.14{\rm{ m}}\) for \({R''_y}\),

\(\begin{aligned}{}\beta = {\tan ^{ - 1}}\left( {\frac{{ - 24.14{\rm{ m}}}}{{ - 11.22{\rm{ m}}}}} \right)\\ = 65.1^\circ \end{aligned}\)

Hence, the magnitude of the resultant vector is \(26.62{\rm{ m}}\) and is directed towards \(65.1^\circ \) south of west.