Newphysicsguy
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Find \vec{a}+\vec{b} (Look at the picture)
The book shows the solutions as :
a. 4.1m Along the +x axis.
I however do not know where I went wrong in my calculations.
\vec{a}+\vec{b}=\begin{bmatrix}<br /> a_x+b_x\\<br /> a_y+b_y <br /> <br /> \end{bmatrix}=\begin{bmatrix}<br /> |\vec{a}|\cos(30)+|\vec{b}|\cos(45)\\<br /> |\vec{a}|\sin(30)-|\vec{b}|\sin(45) <br /> <br /> \end{bmatrix}=\begin{bmatrix}<br /> \frac{3\sqrt{3}}{2}+\sqrt{2}\\<br /> 1.5-\sqrt{2}<br /> <br /> \end{bmatrix}=\begin{bmatrix}<br /> 4.01\\<br /> .085<br /> \end{bmatrix}NOTE: If I posted this is the wrong location I am sorry. This is not homework, I am doing it on my own time.
EDIT: thank you for moving my post to the correct location.
The book shows the solutions as :
a. 4.1m Along the +x axis.
I however do not know where I went wrong in my calculations.
\vec{a}+\vec{b}=\begin{bmatrix}<br /> a_x+b_x\\<br /> a_y+b_y <br /> <br /> \end{bmatrix}=\begin{bmatrix}<br /> |\vec{a}|\cos(30)+|\vec{b}|\cos(45)\\<br /> |\vec{a}|\sin(30)-|\vec{b}|\sin(45) <br /> <br /> \end{bmatrix}=\begin{bmatrix}<br /> \frac{3\sqrt{3}}{2}+\sqrt{2}\\<br /> 1.5-\sqrt{2}<br /> <br /> \end{bmatrix}=\begin{bmatrix}<br /> 4.01\\<br /> .085<br /> \end{bmatrix}NOTE: If I posted this is the wrong location I am sorry. This is not homework, I am doing it on my own time.
EDIT: thank you for moving my post to the correct location.
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