A street lamp weighs 160 N. It is supportedequally by two wires that form an angle of 119� with each other.
What is the tension of each of these wires?
Answer in units of N
To find the tension in each wire supporting the street lamp, we can use trigonometry. Let's call the tension in one of the wires T.
The weight of the street lamp (160 N) is being supported equally by the two wires, so each wire carries half of the weight. Therefore, the weight supported by each wire is 160 N / 2 = 80 N.
Now, let's consider one of the wires. The weight (80 N) acts downward vertically. The tension (T) in the wire is acting at an angle of 119 degrees with respect to the horizontal.
To find the tension in the wire, we can resolve the weight into its horizontal and vertical components. The vertical component is given by W * sin(θ), where θ is the angle between the wire and the vertical direction. In this case, sin(119) gives us the vertical component of 80 N * sin(119).
The tension in the wire can then be found using the Pythagorean theorem. The tension T is equal to the square root of the sum of the squares of the horizontal and vertical components of the weight.
Using the equation T^2 = (80 N * sin(119))^2 + (80 N * cos(119))^2, we can calculate T.
T^2 = (80 N * sin(119))^2 + (80 N * cos(119))^2
T^2 = (80 N * -0.51504)^2 + (80 N * -0.85717)^2
T^2 = 3326.87616 N^2 + 4899.0912 N^2
T^2 = 8225.96736 N^2
Taking the square root of both sides, we get:
T = √(8225.96736 N^2) ≈ 90.75 N
Therefore, the tension in each wire is approximately 90.75 N.
To find the tension in each wire, we can start by breaking down the forces acting on the street lamp.
Let's assume that the tension in one wire is T1 and the tension in the other wire is T2.
Since the street lamp is in equilibrium (not accelerating), the sum of the vertical forces acting on it must be zero.
The vertical component of tension T1 can be found using the formula: T1 * sin(θ) = mg
where θ is the angle between the wires (119 degrees) and m is the mass of the street lamp (converted to weight: 160 N).
So, T1 * sin(119) = 160 N
Next, let's find the vertical component of tension T2. Since the wires are at equal angles, we can use the same formula: T2 * sin(θ) = mg.
Thus, T2 * sin(119) = 160 N
Now we can solve for T1 and T2.
T1 = 160 N / sin(119)
T1 ≈ 209.856 N (rounded to three decimal places)
T2 = 160 N / sin(119)
T2 ≈ 209.856 N (rounded to three decimal places)
Therefore, the tension in each wire, T1 and T2, is approximately 209.856 N.