Respuesta :
Answer:
[tex] R= \sqrt{F^2_{Rx} +F^2_{Ry}}= \sqrt{(-52.286)^2 +(60.39)^2} = 79.880 m[/tex]
And the angle would be:
[tex]\theta= tan^{-1} (\frac{F_{Ry}}{F_{Rx}}) =tan^{-1} (\frac{60.39}{-52.286})= -49.11 [/tex]
So the final position would be 79.880m and with an angle of [tex]\theta= -49.11 [/tex] Southwest
Explanation:
For this case we can use the method of components to solve this problem. And we have the following info given.
[tex] u_1 = 45m \theta_1 = 90[/tex]
[tex] u_2 = 45m \theta_2 = 160[/tex]
[tex] u_3 = 10m \theta_3 = 180[/tex]
Now we can find the components for each vector like this:
[tex] u_{x_1} = 45m *cos 90 =0 , u_{y_1}= 45m *sin 90 =45m[/tex]
[tex] u_{x_2} = 45m *cos 160 =-42.286 , u_{y_2}= 45m *sin 160 =15.39m[/tex]
[tex] u_{x_3} = 10m *cos 180 =-10 , u_{y_3}= 10m *sin 180 =0m[/tex]
Now we can find the net vectors on x and y:
[tex] F_{Rx}= 0 -42.286 -10= -52.286 m[/tex]
[tex] F_{Ry}= 45 +15.39 +0 m = 60.39 m[/tex]
And the resultant position would be:
[tex] R= \sqrt{F^2_{Rx} +F^2_{Ry}}= \sqrt{(-52.286)^2 +(60.39)^2} = 79.880 m[/tex]
And the angle would be:
[tex]\theta= tan^{-1} (\frac{F_{Ry}}{F_{Rx}}) =tan^{-1} (\frac{60.39}{-52.286})= -49.11 [/tex]
So the final position would be 79.880m and with an angle of [tex]\theta= -49.11 [/tex] Southwest
