Posted by Anna on .
Two ropes pull on a ring. One exerts a 57 N force at 30.0°, the other a 57 N force at 60.0°
(a) What is the net force on the ring?
Magnitude
__?__N
Direction
__?__°
(b) What are the magnitude and direction of the force that would cause the ring to be in equilibrium?
__?__N
__?__°

Physics! 
Jennifer,
The first rope has an x component of force Fx1 and a y component of force Fy1:
Fx1 = 57*sin 30 = 28.5
Fy1 = 57*cos 30 = 49.3
The second rope has an x component of force Fx2 and a y component of force Fy2:
Fx2 = 57*sin 60 = 49.3
Fy2 = 57*cos 60 = 28.5
The net force in the x direction Fnetx and the net force in the y direction Fnet y are given by:
Fnetx = Fx1 + Fx2 = 28.5 + 49.3 = 77.8
Fnety = Fy1 + Fy2 = 49.3 + 28.5 = 77.8
The magnitude of this force is given by
(Fnetx^2 + Fnety^2)^.5 = 110 N
at an angle tan(theta) = Fnety/Fnetx = 77.8/77.8 = 45 degrees
b) A force that would cause the ring to be balanced would be equal in magnitude and opposite in direction, so 180 degrees from 45, or at 225 degrees, and magnitude 110N 
Physics! 
Elena,
F(x) =F1(x) +F2(x)
F(x)= 57•cos30°+57•cos60° =
F(y) =F1(y) +F2(y)
F(y) =57•sin30°+57•sin 60° =
F=sqrt{F(x)²+F(y)²}
tan θ=F(y)/F(x)
F(x) =F1(x) +F2(x) +F3(x) =0
57•cos30°+57•cos60° +F3(x) =0
F(y) =F1(y) +F2(y) +F3(y) =0
57•sin30°+57•sin 60° +F3(y) =0
F3=sqrt{F3(x)²+F3(y)²}
tan α=F3(y)/F3(x)