# applied mechanics question?

http://img85.imageshack.us/img85/9896/appliedmecha...

I have this quesiton for one of my tutorials in university its in my applied mechanics course. Despite being rushed through a very quick explanation i still dont get it. I know your meant to get out 2 equations or something but i cant seem to do it. Could someone have an attempt. btw those are the answers down the bottom (in kN)

### 3 Answers

- smciLv 71 decade agoFavorite Answer
Just take horizontal and vertical components, that's all there is to it, or in fact any 2D equilibrium-of-forces question

(ok maybe sometimes we prefer to take radial and tangential components, but they're always somehow orthogonal):

W=14kN in -y-direction

N acts @ 20deg left of vertical +y-direction

Thus:

[Equilibrium of forces in y-dirn]

W=14k = N cos 20

N = 14k/cos 20 ... gives you N

[Equilibrium of forces in x-dirn]

T = N sin 20

= 14k tan 20 ... gives you T

Solutions:

T=5.1kN in -x-direction

N=14.9kN @ 20deg left of vertical +y-direction

- cp_exit_105Lv 41 decade ago
The sum of the forces in the x direction =0.

The sum of the forces in the y direction =0.

let i and j be the unit vectors for x and y

W = -14kN j

T = Ti

N = -cos(70)Ni + sin(70)Nj

so:

sin(70)N -14kN = 0

N = 14898

-cos(70)N + T = 0

T = 5095

- Chris HLv 41 decade ago
I drew a diagram and wrote the calculations in the picture, hope this helps:

http://img108.imageshack.us/img108/8328/test1xk0.j...

Forgive my units.