An equilateral triangle 35.0 m on a side has a m1 = 5.00-kg mass at one corner, a m2 = 70.00-kg mass at another corner, and a m3 = 125.00 kg mass at the third corner. Find the magnitude and direction of the net force acting on the 5.00-kg mass.

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An equilateral triangle 35.0 m on a side has a m1 = 5.00-kg mass at one corner, a m2 = 70.00-kg mass at another corner, and a m3 = 125.00 kg mass at the third corner. Find the magnitude and direction of the net force acting on the 5.00-kg mass.

Physics
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Here is what I did: F=Gm1m2/r^2 F(1,2)=((6.67e-11)*5*70)/35^2 = 1.9e-11N Fy(1,2) = F(1,2)sin(60) = 1.9e-11*sin(60)=1.65e-11N Fx(1,2) = F(1,2)cos(60) =1.9e-11*cos60) = 9.53e-12 N Found F the same way at F(1,2) t0 get 3.4e-11N for F(1,3) Fy(1,3) = F(1,3)sin(120) =3.4e-11N*0.87 = 2.94e-11 N Fx(1,3) = F(1,3)cos(120) = 3.4e-11N*(-0.5) = -1.7e-11 N So, the net force on the y-axis is Fy = Fy(1,2)+Fy(1,3) = 4.6e-11 N and on the x-axis Fx = Fx(1,2)+Fx(1,3) = -7.47e-12 N Then to get FtotalI do sqrt(Fy/fx) For the angle I would do arctan(Fy/Fx) to get -80.77 degrees but its not correct. Please help me with this extremely long problem. I dont know where I messed up.
|dw:1433713175772:dw| You are in the good direction solving this, just make some considerations... if you set the masses as in the picture, the calculation for F(1,2) is exactly as you made it. for F(1,3) the angle would be 0º, not 120º, so you can fix your calculation for that case. finally to get the total magnitude of the force you should use: \[F=\sqrt{F_x^2+F_y^2}\] and for the angle: \[\theta = Arctan(\frac{F_y}{F_x})\]
So that would make Fx(1.3)=3.4e-11 N and Fy(1.3)=0N I did arctan (1.65e-11/((3.4e-11 + 9.53e-12))) to get 20.8 degrees but its not correct. Am I missing something?

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everything you did looks ok to me, let me check the numbers!
i get 20.66º, its very close to your answer... do you have an answer for this? or you just get right/wrong when you submit?
also, do you have a picture for the problem, because if we put the 5Kg mass in other vertice the direction may be different, of course the magnitude of F would be the same.
No it does not give the correct answers. But here is the picture
1 Attachment
oh! so m1=5Kg is in the upper vertex, we need to recalculate! For F(1,2) the angle should be -60º, it will have a negative y-component, since it points to m2, down and to the right For F(1,3) the angle is -120º, and will have both negative components, since it points to m3 down and to the left |dw:1433720613749:dw| it should work now...
I get the same answers as before except the Fx and Fy totals are different signs which doesnt change the magnitude of force. Thank you for all your time btw
do you get a different value for the angle now? glad to be able to help you :D
I got 80.77 which is an answer I already got earlier and it wasnt correct :/ this problem might just be a lost cause for me lol
here are my results for the total force: \[F_x=-1.29\times 10^{-11}N\] \[F_y=-1.12\times 10^{-11}N\] this makes sense since m3 is the bigger one, and it is placed in the lower left vertex, so the net force points approx to it. for the angle i get 40.97º using: \[\theta = Arctan{(\frac{F_y}{F_x})}=40.97º\] now, we must be careful here, since inverse trigonometric functions often yield results in the 1st or 2nd quadrant, but if we take a look at the resulting vector for the force we can see that the angle should be one of the 3rd quadrant, so i guess what we get here is the angle measured from the negative x direction, so if we want to give the result measured from the positive x direction, as is usually done, we need to either add 180º, obtaining a positive angle, in the 3rd quadrant or give the equivalent negative value. i like the first option here, so the answer would be: \[\theta = 180º + 40.97º = 220.97º\approx 221º\] or \[\theta = -180º + 40.97º = -139.03º\approx -140º\] well, that is what i get, hope that even if it is not the correct answer at least you get an understanding of the problem and the thinking involved. You made an excellent job from the begining! :D

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