The net force is 279 N at 53.7 degrees north of east
Explanation:
The two forces in the problem are:
[tex]F_y= 225 N[/tex], in the north direction
[tex]F_x = 165 N[/tex], in the east direction
Where we have taken the east direction as positive x-direction and the north direction as positive y-direction.
We notice that the two forces are perpendicular to each other: so, they form the sides of a right triangle, of which the hypothenuse is the magnitude of their resultant.
Therefore, the magnitude of the net force can be found by using Pythagorean's theorem:
[tex]F=\sqrt{F_x^2+F_y^2}=\sqrt{(165)^2+(225)^2}=279 N[/tex]
And the direction can be found as follows:
[tex]\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{225}{165})=53.7^{\circ}[/tex]
And the angle is measured as north from east.
Learn more about vector addition here:
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