Answer:
[tex]F_r = 200N[/tex]
Explanation:
Given
Let the two forces be
[tex]F_1 = 130N[/tex]
[tex]F_2 = 110N[/tex]
and
[tex]\tan(\theta) = \frac{12}{5}[/tex]
Required
Determine the resultant force
Resultant force (Fr) is calculated using:
[tex]F_r^2 = F_1^2 + F_2^2 + 2F_1F_2\cos(\theta)[/tex]
This means that we need to first calculate [tex]\cos(\theta)[/tex]
Given that:
[tex]\tan(\theta) = \frac{12}{5}[/tex]
In trigonometry:
[tex]\tan(\theta) = \frac{Opposite}{Adjacent}[/tex]
By comparing the above formula to [tex]\tan(\theta) = \frac{12}{5}[/tex]
[tex]Opposite = 12[/tex]
[tex]Adjacent = 5[/tex]
The hypotenuse is calculated as thus:
[tex]Hypotenuse^2 = Opposite^2 + Adjacent^2[/tex]
[tex]Hypotenuse^2 = 12^2 + 5^2[/tex]
[tex]Hypotenuse^2 = 144 + 25[/tex]
[tex]Hypotenuse^2 = 169[/tex]
[tex]Hypotenuse = \sqrt{169[/tex]
[tex]Hypotenuse = 13[/tex]
[tex]\cos(\theta)[/tex] is then calculated using:
[tex]\cos(\theta)= \frac{Adjacent}{Hypotenuse}[/tex]
[tex]\cos(\theta)= \frac{5}{13}[/tex]
Substitute values for [tex]F_1[/tex], [tex]F_2[/tex] and [tex]cos(\theta)[/tex] in
[tex]F_r^2 = F_1^2 + F_2^2 + 2F_1F_2\cos(\theta)[/tex]
[tex]F_r^2 = 130^2 + 110^2 + 2*130*110*\frac{5}{13}[/tex]
[tex]F_r^2 = 16900 + 12100 + 11000[/tex]
[tex]F_r^2 = 40000[/tex]
Take square roots of both sides
[tex]F_r = \sqrt{40000[/tex]
[tex]F_r = 200N[/tex]
Hence, the resultant force is 200N
