Suggested languages for you:

Americas

Europe

Q9PE

Expert-verifiedFound in: Page 121

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Show that the sum of the vectors discussed in Example 3.2 gives the result shown in Figure 3.24.**

The location of the dock is \(52.94\;{\rm{m}}\), \({89.9^ \circ }\) north of west.

**Vectors are** **physical quantities that have magnitude and direction.**

The two vectors cannot be added by using simple algebraic rules. They can be added by using the triangle law of vector addition.

The vectors \({\rm{A}}\) and \({\rm{B}}\) are represented as,

Representation of vectors \({\rm{A}}\) and \({\rm{B}}\)

- The magnitude of the vector \({\rm{A}}\)is, \(A = 27.5\;{\rm{m}}\).
- The magnitude of the vector \({\rm{B}}\)is, \(B = 30\;{\rm{m}}\).

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

\({A_x} = {\rm{A}}\cos \left( {{{66}^ \circ }} \right)\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{A_x} &= \left( {27.5\;{\rm{m}}} \right) \times \cos \left( {{{66}^ \circ }} \right)\\ &= 11.185\;{\rm{m}}\end{aligned}\)

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

\({B_x} = {\rm{B}}\cos \left( {{{112}^ \circ }} \right)\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{B_x} &= \left( {30\;{\rm{m}}} \right) \times \cos \left( {{{112}^ \circ }} \right)\\ &= - 11.238\;{\rm{m}}\end{aligned}\)

The resultant of horizontal components vectors \({\rm{A}}\) and \({\rm{B}}\) is,

\({R_x} = {A_x} + {B_x}\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{R_x} &= \left( {11.185\;{\rm{m}}} \right) + \left( { - 11.238\;{\rm{m}}} \right)\\ &= - 0.053\;{\rm{m}}\end{aligned}\)

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

\({A_y} = A\sin \left( {{{66}^ \circ }} \right)\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{A_y} &= \left( {27.5\;{\rm{m}}} \right) \times \sin \left( {{{66}^ \circ }} \right)\\ &= 25.123\;{\rm{m}}\end{aligned}\)

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

\({B_y} = B\sin \left( {{{112}^ \circ }} \right)\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{B_y} &= \left( {30\;{\rm{m}}} \right) \times \sin \left( {{{112}^ \circ }} \right)\\ &= 27.816\;{\rm{m}}\end{aligned}\)

The resultant of vertical components vectors \({\rm{A}}\) and \({\rm{B}}\) is,

\({R_y} = {A_y} + {B_y}\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{R_y} &= \left( {25.123\;{\rm{m}}} \right) + \left( {27.816\;{\rm{m}}} \right)\\ &= 52.939\;{\rm{m}}\end{aligned}\)

The magnitude of the resultant vector is,

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

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}R &= \sqrt {{{\left( { - 0.053\;{\rm{m}}} \right)}^2} + {{\left( {52.939\;{\rm{m}}} \right)}^2}} \\ \approx 52.94\;{\rm{m}}\end{aligned}\)

The direction of the resultant vector is,

\(\theta = {\tan ^{ - 1}}\left( {\frac{{{R_y}}}{{{R_x}}}} \right)\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}\theta &= {\tan ^{ - 1}}\left( {\frac{{52.939\;{\rm{m}}}}{{ - 0.053\;{\rm{m}}}}} \right)\\ &= - {89.9^ \circ }\end{aligned}\)

Hence, the location of the dock is \(52.94\;{\rm{m}}\), \({89.9^ \circ }\) north of west.

94% of StudySmarter users get better grades.

Sign up for free