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urk.nono
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Hello! I'm a student from Norway, and I've been struggling with a static equilibrium for a few hours now. The terms I use might not be "right", but I'll try to explain the problem to the best of my ability.
Find the forces Fa and Fb on this figure:
http://img44.imageshack.us/img44/4505/321k.jpg
[tex]\Sigma F_X = 0[/tex]
[tex]\Sigma F_Y = 0[/tex]
[tex]\Sigma \tau = 0[/tex] (sum torque = 0)
I solved it by drawing the forces on the figure:
http://img41.imageshack.us/img41/7513/321graph.jpg
After much fiddling with the "relevant equations" I came across the following:
20kN *3,5m - Fb * sin(30) * 5,5m = 0
=> Fb = 37,21kN
OR
20kN * 3,5m - Fb * cos(70) * 5,5m = 0
=> Fb = 37,21kN
Fa * cos(11,75) * 2m - 20m * 3,5m = 0
=> Fa = 35,74kN
I know from my measurements and by checking the solution in the book, it's pretty damn close. By which formula, or by what method I got no clue whatsoever.
I'd appreciate if someone could help me out by pointing me in the right direction
Thanks in advance
Urk
Homework Statement
Find the forces Fa and Fb on this figure:
http://img44.imageshack.us/img44/4505/321k.jpg
Homework Equations
[tex]\Sigma F_X = 0[/tex]
[tex]\Sigma F_Y = 0[/tex]
[tex]\Sigma \tau = 0[/tex] (sum torque = 0)
The Attempt at a Solution
I solved it by drawing the forces on the figure:
http://img41.imageshack.us/img41/7513/321graph.jpg
After much fiddling with the "relevant equations" I came across the following:
20kN *3,5m - Fb * sin(30) * 5,5m = 0
=> Fb = 37,21kN
OR
20kN * 3,5m - Fb * cos(70) * 5,5m = 0
=> Fb = 37,21kN
Fa * cos(11,75) * 2m - 20m * 3,5m = 0
=> Fa = 35,74kN
I know from my measurements and by checking the solution in the book, it's pretty damn close. By which formula, or by what method I got no clue whatsoever.
I'd appreciate if someone could help me out by pointing me in the right direction
Thanks in advance
Urk
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