2006-09-19 10:53:23 · 5 answers · asked by Pattie F 1 in Science & Mathematics ➔ Physics
If a body is in equilibrium under the action of three coplanar forces, what must be true of the lines of action of the three forces?
2006-09-19 11:19:30 · update #1
I don't know your definition of 'lines of action' but the vector sum of the forces is zero.
2006-09-19 17:14:06 · answer #1 · answered by Frank N 7 · 0⤊ 0⤋
F = ma = SUM(f) = 0; it truly is it. The sum of all forces performing interior the plane on the body is 0. So there's a cyber web rigidity F = 0 = ma and, subsequently, the body isn't accelerating. that's in equilibrium. word, a = 0 ability the body's speed is consistent, it does no longer unavoidably advise the body isn't transferring, even with the reality that that ought to correctly be as v = 0 is a continuing as well.
2016-11-28 02:37:08 · answer #2 · answered by Anonymous · 0⤊ 0⤋
The forces cancel each other out.
If the forces are the same: they make 120 & 120 & 120 degrees.
2006-09-19 11:34:42 · answer #3 · answered by · 5 · 0⤊ 0⤋
seems like part of your question got chopped off
2006-09-19 11:16:07 · answer #4 · answered by sojsail 7 · 0⤊ 0⤋
their resolved actions must be equal!
2006-09-19 11:28:43 · answer #5 · answered by Anonymous · 0⤊ 0⤋