applied mechanics question?
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)
- 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
[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
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
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:
Forgive my units.