### Select your language

Suggested languages for you:

Americas

Europe

Q9PE

Expert-verified
Found in: Page 121

### College Physics (Urone)

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.

See the step by step solution

## Step 1: Vectors

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.

## Step 2: Representation of vectors

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

Representation of vectors $${\rm{A}}$$ and $${\rm{B}}$$

## Step 3: Given data:

• The magnitude of the vector $${\rm{A}}$$is, $$A = 27.5\;{\rm{m}}$$.
• The magnitude of the vector $${\rm{B}}$$is, $$B = 30\;{\rm{m}}$$.

## Step 4: Horizontal component of vectors

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}

## Step 5: Vertical component of vectors

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}

## Step 6: Resultant vector

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.