F1 = 1150 N, F2 = 875 N and F3 = 1450 N are applied at the same point. The angle between F1 and F2 is 90 degrees and the angle between F2 and F3 is 120 degrees. F2 is between F1 and F3. Find the resultant.

the answer is 190 N and 126.3 degrees.

Question

F1 = 1150 N, F2 = 875 N and F3 = 1450 N are applied at the same point. The angle between F1 and F2 is 90 degrees and the angle between F2 and F3 is 120 degrees. F2 is between F1 and F3. Find the resultant.

the answer is 190 N and 126.3 degrees.

summersnow8 · Answer

|dw:1437436008915:dw|

summersnow8 · Answer

This is my work: 
Fx components: 1150 N + (1450 cos 210) = -105.7368 
Fy compnents: 875 N + (1450 sin 210) = 150 

$$R= \sqrt{105.7368^{2}+ 105^{2}} = 183.52 N $$

irishboy123 · Answer

looks good

summersnow8 · Answer

@IrishBoy123, the back of the book says the answer is 190 N and 126.3 degrees. I am not getting that answer

irishboy123 · Answer

sounds like your diagram is wrong so we could try some reverse engineering....

summersnow8 · Answer

can you help me draw a new one then? or just go off the text?

irishboy123 · Answer

i drew it differently and got your answer again!!  let me see if there is another way

you clearly know your stuff so it is annoying that you have to guess the diagram....

summersnow8 · Answer

okay

irishboy123 · Answer

|dw:1437492586634:dw|

i wrote a little script to find the angle $ \theta$ as per my diagram that gives a net force of 190N

The output:
HIT!! angle = 225.21 F = 190.0 net angle= -39.53
HIT!! angle = 240.25 F = 189.9 net angle= -34.99

so it looks to me like you get a net 190N at $ \theta $ = 225.2 or 240.25 and the angle of that 190N is -39 or -35.......

IOW i think the book is wrong, if that helps at all....

[the script is attached if you are interested]