The magnitude and direction of the resultant force are 10.0N and 101.36⁰
The sum of force along the force horizontal is expressed as:
[tex]\sum F_x=-15.0N+15cos30\\\sum F_x = -15.0N + 12.99N\\\sum F_x = -2.01N[/tex]
The sum of force along the vertical is expressed as:
[tex]\sum F_y = +10N[/tex]
Calculate the magnitude of the force:
[tex]F=\sqrt{(\sum F_x)^2+(\sum F_y)^2} \\F=\sqrt{(-2.01)^2+(10)^2} \\F=\sqrt{104.0401}\\F= 10.2N[/tex]
Hence the magnitude of the resultant force is 10.2N
Find the direction
[tex]\theta = tan^{-1}\frac{y}{x}\\ \theta = tan^{-1}\frac{10}{-2.01}\\\theta =tan^{-1}(-4.9751)\\\theta =-78.63^0\\\theta = 180-78.63 = 101.36^0[/tex]
Hence the magnitude and direction of the resultant force are 10.0N and 101.36⁰
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