Resultant Force: Calculating Forces with Trigonometry for a 800N and 600N Object

[Jud]

I'm sure this problem is pretty trivial to most but apologies in advance, I'm just starting out sorry.

Homework Statement

I'm not sure how to go about working this out due to the 800N

The Attempt at a Solution

F1 (800N) x-pos = 800 tan (90)?? , y-pos = ?
F2 (600N) x-pos = 600 cos(30) , y-pos = -600 sin(30)

Homework Statement

 

[UltrafastPED]

You can draw the parallelogram of forces, and the diagonal length (from your starting point) is the net force.

Or you can complete their conversion to vectors; F1 = 800 j, where j is the unit vector in the y-direction. Your F2 looks correct ... so now add them together!

 

[Jud]

You can draw the parallelogram of forces, and the diagonal length (from your starting point) is the net force.

Or you can complete their conversion to vectors; F1 = 800 j, where j is the unit vector in the y-direction. Your F2 looks correct ... so now add them together!

so,

F1 (800N) x-pos = 0 , y-pos = 800
F2 (600N) x-pos = 600 cos(30) , y-pos = -600 sin (30)

sqrt (519.615)^2 + (500)^2 = 721.1 N

is that correct?

 

[UltrafastPED]

It's certainly in the correct range ... if you are unsure you should recheck your calculations.

 

[Jud]

It's certainly in the correct range ... if you are unsure you should recheck your calculations.

Ok Thank you for the help, greatly appreciated.