A light fixture that weighs 82.2 N is hung symmetrically by three cables that make an angle of 67.1 degrees with the horizontal. Determine the tension (in Newtons) in one of the cables.
To find the tension in one of the cables, we can break down the weight of the light fixture into horizontal and vertical components.
1. Start by finding the vertical component of the weight:
Vertical component = Weight * sin(angle)
Vertical component = 82.2 N * sin(67.1°)
Vertical component ≈ 73.8 N
2. Since there are three cables, each cable will share an equal amount of the vertical component of the weight. Therefore, the tension in one of the cables will be:
Tension = Vertical component / 3
Tension ≈ 73.8 N / 3
Tension ≈ 24.6 N
Therefore, the tension in one of the cables is approximately 24.6 Newtons.
To determine the tension in one of the cables, we can use the concept of equilibrium. In equilibrium, the sum of all forces acting on an object is zero. So, in this case, the sum of the forces in the vertical direction (up and down) should be equal to zero.
Let's analyze the forces acting on the light fixture:
1. The weight of the light fixture acts vertically downward with a force of 82.2 N.
2. The tension in each cable acts diagonally upward and inward at an angle of 67.1 degrees with the horizontal. Since there are three cables, we denote the tension in one cable as T.
Considering the vertical forces, we have:
Downward force (weight) - Upward forces (tensions) = 0
82.2 N - 3T sin(67.1 degrees) = 0
To find the tension, we isolate T by moving the terms around:
3T sin(67.1 degrees) = 82.2 N
T sin(67.1 degrees) = 82.2 N / 3
T = (82.2 N / 3) / sin(67.1 degrees)
Using a scientific calculator, we can calculate the value of T:
T = (82.2 N / 3) / sin(67.1 degrees)
T ≈ 40 N
Therefore, the tension in one of the cables is approximately 40 Newtons.
draw a diagram
[m * g * (1/3) ] / (tension) = sin(67.1º)