A street lamp weighs 150 N. It is supported by two wires that form an angle of 100° with each other. The tensions in the wires are equal.
(a) What is the tension in each wire supporting the street lamp?
(b) If the angle between the wires supporting the street lamp is reduced to 70°, what is the tension in each wire?
sugma
To find the tension in each wire supporting the street lamp, we can divide the weight of the lamp by the cosine of the angle formed by the wires.
(a)
The weight of the street lamp is given as 150 N. Since the tensions in the two wires are equal, let's call the tension in each wire T.
Using the cosine rule, we can calculate the tension in each wire using the formula:
T = weight / cos(angle)
T = 150 N / cos(100°)
Calculating the cosine of 100°:
cos(100°) ≈ -0.173648
T ≈ 150 N / (-0.173648)
T ≈ -863.04 N
Since tension cannot be negative, we can discard this result.
Therefore, there is no valid tension for the given angle of 100°.
(b)
Now, let's calculate the tension in each wire when the angle is reduced to 70°.
Using the same formula as before:
T = weight / cos(angle)
T = 150 N / cos(70°)
Calculating the cosine of 70°:
cos(70°) ≈ 0.34202
T ≈ 150 N / 0.34202
T ≈ 438.10 N
Therefore, the tension in each wire when the angle is reduced to 70° is approximately 438.10 N.
To solve this problem, we can start by drawing a free-body diagram of the street lamp. Let's label the tensions in the two wires as T1 and T2.
(a) To find the tension in each wire when the angle between them is 100°, we can use the concept of vector resolution. Since the tensions in the wires are equal, we can represent them as T.
In the vertical direction:
The weight of the street lamp is acting downward, so we have:
ΣFy = -Tsin(100°) - Tsin(100°) + 150N = 0
In the horizontal direction:
The sum of the horizontal forces is equal to zero since the lamp is not accelerating horizontally:
ΣFx = -Tcos(100°) + Tcos(100°) = 0
Simplifying the equations, we have:
-2Tsin(100°) + 150N = 0
Tsin(100°) = 75N
From the equation Tsin(100°) = 75N, we can solve for T:
T = 75N / sin(100°)
T ≈ 86.4N
Therefore, the tension in each wire supporting the street lamp is approximately 86.4N.
(b) Now, let's consider the case where the angle between the wires is reduced to 70°.
Using the same approach as before, we can write the equations:
In the vertical direction:
ΣFy = -Tsin(70°) - Tsin(70°) + 150N = 0
In the horizontal direction:
ΣFx = -Tcos(70°) + Tcos(70°) = 0
Simplifying the equations, we have:
-2Tsin(70°) + 150N = 0
Tsin(70°) = 75N
From the equation Tsin(70°) = 75N, we can solve for T:
T = 75N / sin(70°)
T ≈ 84.0N
Therefore, the tension in each wire when the angle between them is reduced to 70° is approximately 84.0N.
angle of wire from horizontal = 90-50 = 40 degrees
2 * T sin 40 = 150
T = 75/sin 40