Solving Problems with Newton's Laws - Hanging Chandelier

by tiggrulz13
Tags: chandelier, laws, newton, solving
 P: 12 1. The problem statement, all variables and given/known data A chandelier with mass m is attached to the ceiling of a large concert hall by two cables. Because the ceiling is covered with intricate architectural decorations (not indicated in the figure, which uses a humbler depiction), the workers who hung the chandelier couldn't attach the cables to the ceiling directly above the chandelier. Instead, they attached the cables to the ceiling near the walls. Cable 1 has tension T1 and makes an angle of θ1 with the ceiling. Cable 2 has tension T2 and makes an angle of θ2 with the ceiling. Find an expression for T1, the tension in cable 1, that does not depend on T2. Express your answer in terms of some or all of the variables m, θ1, and θ2, as well as the magnitude of the acceleration due to gravity g. 2. Relevant equations 3. The attempt at a solution So far I have found the x and y components: ΣFx = 0 = -T1cosθ1 + T2cosθ2 ΣFy = 0 = T1sinθ1 + T2sinθ2 - mg Now I need to eliminate T2 from this pair of equations and solve for T1, but I can't figure out how to do this.
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P: 5,346
 Quote by tiggrulz13 1. The problem statement, all variables and given/known data A chandelier with mass m is attached to the ceiling of a large concert hall by two cables. Because the ceiling is covered with intricate architectural decorations (not indicated in the figure, which uses a humbler depiction), the workers who hung the chandelier couldn't attach the cables to the ceiling directly above the chandelier. Instead, they attached the cables to the ceiling near the walls. Cable 1 has tension T1 and makes an angle of θ1 with the ceiling. Cable 2 has tension T2 and makes an angle of θ2 with the ceiling. Find an expression for T1, the tension in cable 1, that does not depend on T2. Express your answer in terms of some or all of the variables m, θ1, and θ2, as well as the magnitude of the acceleration due to gravity g. 2. Relevant equations 3. The attempt at a solution So far I have found the x and y components: ΣFx = 0 = -T1cosθ1 + T2cosθ2 ΣFy = 0 = T1sinθ1 + T2sinθ2 - mg Now I need to eliminate T2 from this pair of equations and solve for T1, but I can't figure out how to do this.
2 equations - 4 unknowns T1 and T2, θ1, θ2?

You can't get rid of T2?

Try harder. You know directly that T2 = T1*(Cosθ1/Cosθ2)

Surely something will occur to you.
 P: 12 I tried that, but I kept getting an answer where T1 was equal to something that had T1 in it. Such as: T1 = (mg - T1cosθ1tanθ2) / sinθ1 I got that when I plugged T2 = T1 (cosθ1)/(cosθ2) into the y component.
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P: 5,346

Solving Problems with Newton's Laws - Hanging Chandelier

 Quote by tiggrulz13 I tried that, but I kept getting an answer where T1 was equal to something that had T1 in it. Such as: T1 = (mg - T1cosθ1tanθ2) / sinθ1 I got that when I plugged T2 = T1 (cosθ1)/(cosθ2) into the y component.
And this is a problem because ...?
 P: 12 Is it allowed to have T1 in the answer? I didn't think it was because it said in terms of some or all of the variables m, θ1, and θ2, as well as the magnitude of the acceleration due to gravity g. It never mentions T1.
 P: 3 T1 is not suppose t be in the answer. i have the same question for my assignment. the answer ends up being : mg/sin(theta1)+tan(theta2)*cos(theta1) i too substituted T2 = T1 (cosθ1)/(cosθ2) for T2 in the second equation and got 0=T1sin(theta1)+T1(cos(theta1)*sin(theta2)/cos(theta2))-mg after moving the mg to the other side i made the sin(theta2)/cos(theta2)= tan(theta2), and then factored out the two. mg=T1(sin(theta1)+tan(theta2)cos(theta1) devided both sides by sin(theta1)+tan(theta2)cos(theta1) and got mg/sin(theta1)+tan(theta2)*cos(theta1)

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