A street lamp weighs 150 N. It is supported by two wires that form an angle of 140° with each other. The tensions in the wires are equal.
(a) What is the tension in each wire?
(b) If the angle between the wires is reduced to 110°, what is the tension in each wire?
To find the tension in each wire, we can use the principle of equilibrium. Since the lamp is stationary, the net force acting on it must be zero in both cases. Let's calculate the tensions in each wire.
(a) Given:
Weight of the lamp (W) = 150 N
Since the angles between the wires are equal, let's denote the tension in each wire as T.
In equilibrium, the vertical components of the tensions in the wires must balance the weight of the lamp.
Step 1: Resolving forces vertically
The vertical component of tension in one wire is T * sin(140° / 2) = T * sin(70°) = T * 0.9397
Similarly, the vertical component of tension in the other wire is also T * 0.9397.
Step 2: Equating vertical forces
The sum of the vertical components of tension in the wires must equal the weight of the lamp for equilibrium.
2 * T * 0.9397 = 150
Step 3: Solving for T
2 * T * 0.9397 = 150
T = 150 / (2 * 0.9397)
T ≈ 79.74 N
Therefore, the tension in each wire is approximately 79.74 N.
(b) To calculate the tension in each wire when the angle between them is reduced to 110°, we'll follow a similar process.
Step 1: Resolving forces vertically
The vertical component of tension in one wire is T * sin(110° / 2) = T * sin(55°) = T * 0.8192
Similarly, the vertical component of tension in the other wire is also T * 0.8192.
Step 2: Equating vertical forces
2 * T * 0.8192 = 150
Step 3: Solving for T
2 * T * 0.8192 = 150
T = 150 / (2 * 0.8192)
T ≈ 91.83 N
Therefore, when the angle between the wires is reduced to 110°, the tension in each wire is approximately 91.83 N.