# Couple of a Triangle Question

1. Oct 24, 2012

### tbarker5

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

In the figure F1= 27.9KN, F2= 39.8KN, F3= 61.8KN
What distance d is required for the net couple moment on the triangle to be 1140 counterclockwise?
Express your answer to three significant figures and include the appropriate units.

2. Relevant equations

I'm thinking the only really relevant equation is Mc=rXf or Mc=|r||F|sin(theta)

3. The attempt at a solution

So I wrote each moment in terms of d using trig. I got:
Mf1= 27900d Nm
Mf2= 19900d Nm
Mf3= 61800cos(30)d Nm
Then I set the sum of the three moments (with direction taken into respect) equal to 1140 Nm.
1140 = 11900d + 61800cos(30)d - 27900d
d~= 0.03038 m or 3.04*10^-2 m.

Unfortunately this was wrong and I have no clue where I went wrong. Any help would be greatly appreciated.

2. Oct 25, 2012

### CWatters

Where did 11900 come from? I know F2-F1=11900 but can't see how you got F2*D-F1*D from the drawing if indeed that's how you got to 11900.

Can you tell us which point you choose as origin, which direction you assumed as +ve and write the equation in terms of F1, F2 and F3 rather than substituting the values.

for example your equation above appears to be...

Total = (F2 - F1)d + F3Cos(30)d - F1d

Is that right?

3. Oct 25, 2012

### tbarker5

I didn't choose any point as an origin. The question asked about the couple on the entire triangle. And sorry here's my work for the equations:

It says that the total moment couple has to be 1140 counterclockwise, so I assumed counterclockwise as the positive direction for all the moments.
Then this is the part I wasn't too sure of. I figured that F2 and F3 would cause counter clockwise rotation while F1 would cause clockwise rotation.
Therefore 1140=MF2+MF3-MF1
subbing in: 1140=19900d+61800cos(30)d-27900d
1140=45520d
d=0.025m
WOW. I think I made an arithmetic error. just got it with 0.025m. I feel like an idiot. Thanks for the help though man