Find an expression for T_1, the tension in cable 1, that does not depend on T_2?
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 T_1 and makes an angle of theta_1 with the ceiling. Cable 2 has tension T_2 and makes an angle of theta_2 with the ceiling. http://session.masteringphysics.com/problemAsset/1... Express your answer in terms of some or all of the variables m, theta_1, and theta_2, as well as the magnitude of the acceleration due to gravity g.See Answer 10 Add Answers
You want to use Newton's second law in the x and y direction and set each equation equal to zero since the chandelier isn't accelerating.
FORCES IN THE X:
-T1*cos(θ1) + T2*cos(θ2) = 0
If you look at the figure, you'll notice that the x components of the tensions point in opposite direction.
T1*cos(θ1) = T2*cos(θ2)
Since you want to find an expression for T1 in terms of T2, solve for T2.
==(1)==> T2 = T1*cos(θ1) / cos(θ2)
FORCES IN THE Y:
T1*sin(θ1) + T2*sin(θ2) - mg = 0
==(2)==> T1*sin(θ1) + T2*sin(θ2) = mg
Since you want to find an expression for T1 in terms of T2, substitute equation (1) into equation (2) for T2:
T1*sin(θ1) + T1*cos(θ1)*sin(θ2)/cos(θ2) = mg
But sin(θ2)/cos(θ2) = tan(θ2)
T1*sin(θ1) + T1*cos(θ1)*tan(θ2) = mg
T1[sin(θ1) + cos(θ1)*tan(θ2)] = mg
EXPRESSION FOR T1 INDEPENDENT OF T2:
==> T1 = mg / [sin(θ1) + cos(θ1)*tan(θ2)]