Question

Can u please help me solve. I am reviewing for a final

Answer 1

henSolution:

Given that;

An object is suspended by two cables attached at a single point.

The diagrammatic representation is shown below

For part A;

Writing the vector in component form

$\begin{gathered} \vec{v_1}=110\sin50\degree(\vec{i})+110\cos50\degree(\vec{j}) \\ \vec{v_1}=70.7\vec{i}+84.3\vec{j} \\ \vec{v_2}=60\cos160\degree(\vec{i})+60\cos160\degree(\vec{j}) \\ \vec{v_2}=-56.4\vec{i}+20.5\vec{j} \end{gathered}$

Hence, the answer is

$\begin{gathered} v_1=70.7\vec{\imaginaryI}+84.3\vec{j} \\ v_2=-56.4\vec{\imaginaryI}+20.5\vec{j} \end{gathered}$

For part B;

Finding the dot product of the vectors

$\begin{gathered} \vec{v_1}\cdot\vec{v_2}=(70.7\vec{\mathrm{i}}+84.3\vec{j})\cdot(-56.4\vec{\mathrm{i}}+20.5\vec{j}) \\ \vec{v_1}\cdot\vec{v_2}=-3987.48+1728.15=-2259.33 \\ \vec{v_1}\cdot\vec{v_2}=-2259.33 \end{gathered}$

Hence, the dot product is

$\vec{v_1}\vec{\cdot v_2}=-2259.33$

For part C

For the angles between vectors, using the dot product

$\vec{v_1}\cdot\vec{v_2}=\lvert{\lvert\vec{{v_1}}\rvert}\rvert\lvert{\lvert\vec{{v_2}}\rvert}\rvert\cos\theta$

$\begin{gathered} (110\cdot60)\cos\theta=-2259.33 \\ \cos\theta=\frac{-2259.33}{6600} \\ \cos\theta=-0.34232 \\ \theta=\cos^{-1}(-0.34232) \\ \theta=110\degree\text{ \lparen nearest degree\rparen} \end{gathered}$

Hence, the angle between them is 110° (nearest degree)