# Physics!

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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! - ,

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! - ,

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)