51.47 Newtons 60.94 to W of N
Explanation
Step 1
Free body diagram
Step 2
Magnitude
Let
[tex]\begin{gathered} Vx=-45\text{ N} \\ V_y=25\text{ N} \\ V=\text{ unknonw} \end{gathered}[/tex]so
[tex]\begin{gathered} V=\sqrt[]{V^2_x+V^2_y} \\ \text{replace} \\ V=\sqrt[]{-45^2+25^2} \\ V=\sqrt[]{2650} \\ V=51.47 \end{gathered}[/tex]Step 3
direction
we need to use the tan function
[tex]\begin{gathered} \tan \text{ }\alpha=\frac{opposit\text{e sid}e}{\text{adjacent side}} \\ \text{replace} \\ \tan x=\frac{45}{25} \\ x=\tan ^{-1}(\frac{45}{25}) \\ x=60.94 \end{gathered}[/tex]therefore,
the answer is
51.47 Newtons 60.94 to W of N
