Question

Figure 1) 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.

Answer 1

In the vertical direction, the forces are balanced:
Fnet = 0
T1sin?1 + T2sin?2 = mg

In the horizontal direction, the forces are also balanced:
Fnet = 0
T1cos?1 = T2cos?2
T2 = T1(cos?1)/(cos?2)

Using this result in the first equation:
T1sin?1 + T2sin?2 = mg
T1sin?1 + T1(cos?1)(sin?2)/(cos?2) = mg
T1sin?1 + T1(cos?1)(tan?2) = mg
T1(sin?1 + (cos?2)(tan?2)) = mg

T1 = mg / (sin?1 + (cos?2)(tan?2))

Answer 2

In the vertical direction, the forces are balanced:
Fnet = 0
T1sin?1 + T2sin?2 = mg

In the horizontal direction, the forces are also balanced:
Fnet = 0
T1cos?1 = T2cos?2
T2 = T1(cos?1)/(cos?2)

Using this result in the first equation:
T1sin?1 + T2sin?2 = mg
T1sin?1 + T1(cos?1)(sin?2)/(cos?2) = mg
T1sin?1 + T1(cos?1)(tan?2) = mg
T1(sin?1 + (cos?2)(tan?2)) = mg

T1 = mg / (sin?1 + (cos?2)(tan?2))

Answer 3

In the vertical direction, the forces are balanced:
Fnet = 0
T1sinθ1 + T2sinθ2 = mg

In the horizontal direction, the forces are also balanced:
Fnet = 0
T1cosθ1 = T2cosθ2
T2 = T1(cosθ1)/(cosθ2)

Using this result in the first equation:
T1sinθ1 + T2sinθ2 = mg
T1sinθ1 + T1(cosθ1)(sinθ2)/(cosθ2) = mg
T1sinθ1 + T1(cosθ1)(tanθ2) = mg
T1(sinθ1 + (cosθ2)(tanθ2)) = mg

T1 = mg / (sinθ1 + (cosθ2)(tanθ2))