Answer
a) $\frac{\sqrt{3}}{2}\mathbf{i}+\frac{1}{2}\mathbf{j}$
b) $\mathbf{F}=\underline{-175\sqrt{3}\mathbf{i}-175\mathbf{j}}$.
c) The magnitude of the vector is $350$.
Work Step by Step
(a).
The boat is inclined on the ramp at an angle of ${{30}^{\circ }}$.
Hence, in the upward direction, the unit vector can be resolved having a unit magnitude.
Therefore, the unit vector $\mathbf{u}$ is
$\begin{align}
& \mathbf{u}=\cos {{30}^{\circ }}\mathbf{i}+\sin {{30}^{\circ }}\mathbf{j} \\
& =\frac{\sqrt{3}}{2}\mathbf{i}+\frac{1}{2}\mathbf{j}
\end{align}$
Hence, the unit vector is $\frac{\sqrt{3}}{2}\mathbf{i}+\frac{1}{2}\mathbf{j}$.
(b).
It is known that the vector projection of F onto u is
$\text{pro}{{\text{j}}_{\mathbf{u}}}\mathbf{F}=\frac{\mathbf{F}\cdot \mathbf{u}}{{{\left\| \mathbf{u} \right\|}^{2}}}\mathbf{u}$ …… (3)
Substituting equations (1) and (2) in equation (3), we get
$\begin{align}
& \text{pro}{{\text{j}}_{\mathbf{u}}}\mathbf{F}=\frac{\left( 0,-700 \right)\left( \frac{\sqrt{3}}{2},\frac{1}{2} \right)}{{{\left\| \mathbf{u} \right\|}^{2}}}\left( \frac{\sqrt{3}}{2}\mathbf{i}+\frac{1}{2}\mathbf{j} \right) \\
& =\left( 0\times \frac{\sqrt{3}}{2},-700\times \frac{1}{2} \right)\left( \frac{\sqrt{3}}{2}\mathbf{i}+\frac{1}{2}\mathbf{j} \right) \\
& =-175\sqrt{3}\mathbf{i}-175\mathbf{j}
\end{align}$
Hence, the vector projection of F onto a unit vector is $-175\sqrt{3}\mathbf{i}-175\mathbf{j}$.
(c).
It is known that the magnitude of a vector is
$\left\| \mathbf{A} \right\|=\sqrt{{{\left( \text{component of }\mathbf{i} \right)}^{2}}+{{\left( \text{component of }\mathbf{j} \right)}^{2}}}$
Hence, from (1), we get
$\begin{align}
& \left\| \text{pro}{{\text{j}}_{\mathbf{u}}}\mathbf{F} \right\|=\sqrt{{{\left( -175\sqrt{3} \right)}^{2}}+{{\left( -175 \right)}^{2}}} \\
& =350
\end{align}$
This means that a force of $350$ pounds is required to keep the boat from rolling down the ramp.
