Q12E

Find the components of \({v_{tot}}\) along a set of perpendicular axes rotated \(30^\circ \) counterclockwise relative to those in Figure 3.55.

Figure: The two velocities \({{\rm{v}}_{\rm{A}}}\) and \({{\rm{v}}_{\rm{B}}}\) add to give a total \({{\rm{v}}_{{\rm{tot}}}}\)

Verified

Answer

The \(x\) component of \({v_{tot}}\) is \(6.35\;{\rm{m/s}}\) and \(y\) component of \({v_{tot}}\) is \(2.19\;{\rm{m/s}}\).

1Step 1: Components of vectors

When the components of a vector quantity are combined together, they generate the same effect as a single vector.

Given data:

2Step 2: Angle of total velocity vector with horizontal

The angle of between \(x\)-axis and \({{\rm{v}}_{{\rm{tot}}}}\) is

\(\begin{array}{c}\alpha = \left( {26.5^\circ } \right) + \left( {22.5^\circ } \right)\\ = 49^\circ \end{array}\)

When the perpendicular axes are rotated by \(30^\circ \) in counterclockwise, the angle formed between \(x\)-axis and \({{\rm{v}}_{{\rm{tot}}}}\) is

\(\theta = \alpha - 30^\circ \)

Substitute \(49^\circ \) for \(\alpha \).

\(\begin{array}{c}\theta = 49^\circ - 30^\circ \\ = 19^\circ \end{array}\)

3Step 3: Horizontal component of total velocity vector

The horizontal component of \({{\rm{v}}_{{\rm{tot}}}}\) is

\({v_{to{t_x}}} = {v_{tot}}\cos \theta \)

Substitute \(6.72\;{\rm{m/s}}\) for \({v_{tot}}\) and \(19^\circ \) for \(\theta \).

\(\begin{array}{c}{v_{to{t_x}}} = \left( {6.72\;{\rm{m/s}}} \right) \times \cos \left( {19^\circ } \right)\\ = 6.35\;{\rm{m/s}}\end{array}\)

Hence, the \(x\) component of \({v_{tot}}\) is \(6.35\;{\rm{m/s}}\).

4Step 4: Vertical component of total velocity vector

The vertical component of \({{\rm{v}}_{{\rm{tot}}}}\) is

\({v_{to{t_y}}} = {v_{tot}}\sin \theta \)

Substitute \(6.72\;{\rm{m/s}}\) for \({v_{tot}}\) and \(19^\circ \) for \(\theta \).

\(\begin{array}{c}{v_{to{t_y}}} = \left( {6.72\;{\rm{m/s}}} \right) \times \sin \left( {19^\circ } \right)\\ = 2.19\;{\rm{m/s}}\end{array}\)

Hence, the \(y\) component of \({v_{tot}}\) is \(2.19\;{\rm{m/s}}\).