Applied mechanics question?

Question

http://img85.imageshack.us/img85/9896/appliedmechanicsquestiosr2.jpg 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...

Answer 1

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

Answer 2

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

Answer 3

I drew a diagram and wrote the calculations in the picture, hope this helps:

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

Forgive my units.