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There are 3 ropes connected to a small, very light ring. Two of the ropes are anchored to walls at right angles w/ rope 1 pulling along the negative x-axis and rope 2 pulling along the positive y-axis, and the third rope pulls down to the right at an angle of 28 degrees below the x-axis, and there is a force of 92N on rope 3. What are T1 and T2, the magnitudes of tension forces in the first two ropes?


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Added Thu, 18 Jun '15
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For the ring to be in static equilibrium, we must have, 

F(net)_x = (T_1)_x + (T_2)_x + (T_3)_x = 0
F(net)_y = (T_1)_y + (T_2)_y + (T_3)_y = 0

The components of the force are,
(T_1)_x = -T_1
(T_2)_x = 0
(T_3)_x = T_3*cos(28)
(T_1)_y = 0
(T_2)_y = T_2
(T_3)_y = -T_3 * sin(28)

Using the equation for F(net)_x we can solve for T_1 
-T_1 + 0 + T*cos(28) = 0

T_1 = 92*cos(28)
       = 81.23 N

Using the equation for F(net)_y we can solve for T_2
0+T_2 - T_3*sin(28) = 0 

T_2 = 92*sin(28) 
       =  43.2N
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Added Thu, 18 Jun '15
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