Question

Forces of 70 N at 50 degrees, and 20 N at an angle of 220 degrees, measured counter-clockwise from the positive x-axis, act on an object. What is the magnitude of the resultant force (in Newtons)? What is the angle of the resultant force with respect to x-axis? (positive values only)

Answer 1

A force _{$\small \overrightarrow{F_{1}}$} =70 N at an angle of 50^o is acting on the object counterclock wise from +x - axis. So, x and y components of _{$\small \overrightarrow{F_{1}}$} will be;

  $\small \overrightarrow{F_{1_{_{_{_{x}}}}}}=70cos(50^{o})=45\: N$

  $\small \overrightarrow{F_{1_{_{_{y}}}}}=70sin(50^{o})=53.6\: N$

The force _{$\small \overrightarrow{F_{2}}$} =20 N at an angle of 220^o is acting on the object counterclock wise from +x - axis. So, x and y components of _{$\small \overrightarrow{F_{2}}$} will be;

  $\small \overrightarrow{F_{2_{_{_{x}}}}}=20cos(220^{o})=-15.3\: N$

  $\small \overrightarrow{F_{2_{_{_{y}}}}}=20sin(220^{o})=-12.9\: N$

Now by adding the x and y components of _{$\small \overrightarrow{F_{1}}$} and _{$\small \overrightarrow{F_{2}}$} , x and y components of resultant force $\small \overrightarrow{R}$ can be calculated;

$\small \overrightarrow{R_{x}}=\overrightarrow{F_{1_{_{_{x}}}}}+\overrightarrow{F_{2_{_{_{x}}}}}=45+(-15.3)=29.7\: N$

$\small \overrightarrow{R_{y}}=\overrightarrow{F_{1_{_{_{y}}}}}+\overrightarrow{F_{2_{_{_{y}}}}}=53.6+(-12.9)=40.7\: N$

Therefore the magnitude of resultant force $\small \overrightarrow{R}$ can be calculated as;

  $\small \overrightarrow{R}=\sqrt{\left ( \overrightarrow{R_{x}} \right )^{2}+\left ( \overrightarrow{R_{y}} \right )^{2}}=\sqrt{(29.7)^{2}+(40.7)^{2}}$

  $\small \Rightarrow \overrightarrow{R}=50.38\: N$

The angle of resultant force with respect to x - axis will be;

$\small \theta =tan^{-1}\left (\frac{\overrightarrow{R_{y}}}{\overrightarrow{R_{x}}} \right )=tan^{-1}\left (\frac{40.7}{29.7} \right )=53.88^{o}$