Question

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.

Answer 1

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.

Answer 2

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:
\[\theta = Arctan(\frac{F_y}{F_x})\]