What Distance d is Required for Net Couple Moment on a Triangle?

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SUMMARY

The discussion focuses on calculating the required distance \( d \) for achieving a net couple moment of 1140 Nm counterclockwise on a triangle subjected to three forces: \( F1 = 27.9 \) kN, \( F2 = 39.8 \) kN, and \( F3 = 61.8 \) kN. The correct approach involves using the equation for moments, \( M_c = r \times F \), leading to the equation \( 1140 = (F2 - F1)d + F3 \cos(30)d - F1d \). The final calculation reveals that the correct value for \( d \) is approximately 0.025 m after correcting an arithmetic error.

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Homework Statement



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.
link to image http://imgur.com/cBnLJ

Homework Equations



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

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.
 
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1140 = 11900d + 61800cos(30)d - 27900d

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?
 
CWatters said:
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?

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