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January 30, 2015

January 30, 2015

Posted by **Anna** on Thursday, October 25, 2012 at 2:19pm.

(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**, Thursday, October 25, 2012 at 3:33pmThe 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**, Thursday, October 25, 2012 at 3:37pmF(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)

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