Use components to find Direction of the resultant of the three pulls?

[valeriex0x]

Workmen are trying to free an SUV stuck in the mud. To extricate the vehicle, they use three horizontal ropes, producing the force vectors shown in the figure. (985N @ 31 Degrees Quad 1, 788 N 32 degrees Quad 2, and 411N 53 degrees quad 3.



How do I find the direction of the resultant of the three pulls? Enter your answer as the angle counted from +x axis counterclockwise direction.



I have already found the three x components of the three pulls. I have already found the y components of each of the three pulls. And I have already found the magnitude of the resultant of the three pulls. All of my answers to those parts are correct. I just don't understand what I am supposed to do to find the direction of the resultant of the three pulls. Please help!

 

[grzz]

Please explain what '32 degrees Quad 2' and '53 degrees Quad 3' mean?

 

[valeriex0x]

V1 points north east, V2 points north west, and V3 points Southwest. Angle theta (31, 32, 53) is between the vector and the x axis. carteesion plane

 

[valeriex0x]

y components:
985N sin(31)= 507
788N cos(32)= 668
411N sin(53)=-328

x components:
985N cos(31)= 844
788N sin(32)= -417
411N cos(53)= -247

tan^-1 (507/844)= 30.98
tan^-1 (668/-417)= -58.02
tan^-1 (-328/-247)= 53.01

 

[valeriex0x]

Sum Fx= 179.4
Sum Fy= 847

total Fsum: rad (179)^2 + (847)^2 = 865 magnitude of the resultant of the three pulls

 

[grzz]

Suppose the resultant R is in the 4th quad i.e. NW. Then the angle A between the resultant R and the resultant of all x-components, X, is given by cosA = X/R.

 

[valeriex0x]

So are you saying I should use cos^-1 (x/865)? I'm not sure I'm following what x would be! :/

 

[grzz]

y components:
985N sin(31)= 507
788N cos(32)= 668
411N sin(53)=-328

x components:
985N cos(31)= 844
788N sin(32)= -417
411N cos(53)= -247

tan^-1 (507/844)= 30.98
tan^-1 (668/-417)= -58.02
tan^-1 (-328/-247)= 53.01

If the angle 32 degrees is betweem 788N and the negative x-axis, then the y-comp is given by 788sin32 and not by 788cos32.

 

[valeriex0x]

wait! cos^-1 (179/865) = 77.77? degreees!?

 

[valeriex0x]

yayy! the answer was 77 degrees. Thanks for the guidance. regarding the sin cos stuff with the components all those answers were correct so i might have explained the quads incorrectly. thanks for urrrrr help!

 

[grzz]

So are you saying I should use cos^-1 (x/865)? I'm not sure I'm following what x would be! :/

I said X , the resultant of all x-components of the three given forces.