# Help with force vectors problem

1. Dec 1, 2007

### Bensky

Help with force vectors problem :(

1. The problem statement, all variables and given/known data

The scale in Fig. P12 is being pulled on via three ropes. (The force on each rope is F = 230 N.) What net force does the scale read?

^Fig. P12

2. Relevant equations
Frx = f1x + f2x + f3x
Fry = F1y + f2y + f3y
a^2+b^2=c^2
F = ma (not needed to solve though)

3. The attempt at a solution
N = newtons

F1x = -(cos30$$\circ$$ * 230)
F1x = -199.185 N

F1y = -(sin(30$$\circ$$ * 230)
F1y = -115 N

F2y = -sin(0$$\circ$$ * 230)
F2y = -230N

F3x = cos30$$\circ$$ * 230)
F3x = 199.185 N

F3y = -(sin30$$\circ$$ * 230)
F3y = -115 N

Frx = -199.185 + 0 + 199.185
Frx = 0

Fry = -115 + -230 + -115
Fry = -460 N
---------------------------
a^2 + b^2 = c^2
0^2 + (-460)^2 = c^2
c = 460 N
(done)

Ok, so the problem is with F2y. I think I might have screwed up since it's facing down - I was thinking about the unit circle and realized that its at a 270 degree angle, not a 0 degree angle. I have one last chance to get this problem right, so I thought I would ask here. Is the angle where I screwed up or did I mess up somewhere else? >_<

Any help is appriciated, thanks.

-Bensky

Last edited: Dec 1, 2007
2. Dec 1, 2007

### cristo

Staff Emeritus
Your problem is in your resolving of the forces. Presuming you are using a usual y=vertical coordinate system, then the y component of each of the two forces at 30 degree angles is 230cos(30). You should be able to notice this from that fact that cos(0)=1 which implies that 230cos(0)=230, which we would expect for the vertical component of a vertical vector.

Before doing this problem, you should be able to look at it and simplify the work you need to do. Since the situation is symmetrical, clearly the horizontal components of the vectors will vanish, thus we are left with 230+(the vertical part of each of the forces at an angle).

Last edited: Dec 1, 2007