Have block B on table weight=711 N, coefficient of static friction is 0.25. Have cord attatched horizontally to B, goes to knot. Another cord is attatched to knot at 30 deg. from horizontal and is attatched to wall. Another cord from knot is vertical, and is attatched to block A. What is the maximum weight of A that will allow the system to remain stationary?
2006-10-16 06:12:35 · 3 answers · asked by joe s 1 in Science & Mathematics ➔ Physics
30 above
2006-10-16 07:01:37 · update #1
http://en.wikipedia.org/wiki/Coefficient_of_friction
2006-10-16 17:35:47 · answer #1 · answered by Surya M. 3 · 0⤊ 0⤋
I can not draw the diagram, but it seem that solution looks like this:
F1 = 711 N (weight of block B)
F2 = (711 N)(0.25) = 177.75 N (friction force needed to move block B in direction to the knot)
In the knot, there would be an equilibrium of forces:
wA = F2 + F3........(Eq.1)
where : wA = máximum weight of block A
F2 = the friction force above
F3 = the force exerted by the cord attached to the wall
F3 = 177 N / cos 30° =177 / 0.866 = 204.95 N (cord is attached 30 degress respect to horizontal)
Using Eq.1:
wA = 177.5 N + 204.95 N = 382.45 N
So, the maximun weight allowed for block A in order to get stationery the system is 382.4 N
Good luck!
2006-10-17 04:18:39 · answer #2 · answered by CHESSLARUS 7 · 0⤊ 0⤋
Is the second cord attached 30 degrees above, or below the horizon?
2006-10-16 06:42:39 · answer #3 · answered by VTNomad 4 · 0⤊ 0⤋