# 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)

Relevance

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

• 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

• 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.