A bag of cement of weight Fg hangs from three wires as .....??
A bag of cement of weight Fg hangs from three wires as shown in the
attached diagram (link). Two of the wires make angles theta1 and theta2
with the horizontal. If the system were in equilibrium, show that the
tension in the left hand wire is: T1 = (Fg.costheta2) /
sin(theta 1 + theta2)
Thanks !!
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Because the system is in equilibrium & there is no
acceleration, we can deduce the following:
T3 is balancing the weight. Therefore, T3 = Fg ----> (1)
Horizontal components of T1 & T2 are balancing each other. Therefore,
T1costheta1 = T2costheta2 ----> (2) And, finally vertical components of T1 & T2 are balancing T3. Therefore, T1sintheta1 + T2sintheta2 = T3 ----> (3) Now we have three equations, to get what is required, we just have to manipulate those equations & eliminate T2 & T3, because these terms don't appear in the required equation. T1costheta1 = T2costheta2 => T2 = T1costheta1 / costheta2 Putting this value in eq (3), T1sintheta1 + (T1costheta1 / costheta2)sintheta2 = Fg => T1 [ sintheta1costheta2 + costheta1sintheta2 ] = Fgcostheta2 => T1[sin(theta1 + theta2)] = Fgcostheta2 (Because, sin(a + b) = sinacosb + cosasinb) => T1 = (Fgcostheta2) / sin(theta1 + theta2) QED
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