A 1400 kg steel beam is supported by two ropes. Rope 1 has a direction 20 degrees to the left of the normal line. Rope two has a 30 degree direction, to the right of the normal line. (the two ropes are centered on the steel beam, at a 50 degree angle to each other.) What is the tension in rope 1? What is the tension in rope 2?
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Sum forces in the vertical direction and horizontal direction for two equations.
eq1. 1400kg * 9.81 = T1*cos(20) + T2cos(30)
eq2. 0 = T1*sin(20) – T2sin(30)
Solve the set of equations.
T1 = 6132N, T2 = 8964N
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T1*Sin(20) – T2*Sin(30) = 0
then…
T1*Sin(20) = T2*Sin(30)
then i did….
T1 = [(T2*Sin(30))/(Sin(20))]
using…
T1*Cos(20) + T2*Cos(30) = m*g.
Replace T1, with what i found earlier and…
[(T2*Sin(30))/(Sin(20))]*Cos(20) + T2*Cos(30) = m*g.
Pull out T2…..
T2([(Sin(30))/(Sin(20))]*[Cos(20)] + Cos(30)) = 13720 N
T2 = 13720 N / {([(Sin(30))/(Sin(20))]*[Cos(20)] + Cos(30))}
T2 = 6125.64 N
Then…. Plug in answer for T2 in the equation above to solve for T1
T1 = 8955.08 N
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