The steel I-beam in the drawing has a weight of 9.00 kN and is being lifted at a constant velocity. What is the tension in each cable attached to its ends? (Assume 𝛼 = 71.1°.)
To determine the tension in each cable attached to the steel I-beam, we can analyze the forces acting on the beam in equilibrium. Since the beam is being lifted at a constant velocity, we can assume that the net force acting on the beam is zero.
Let's break down the forces acting on the I-beam:
1. Weight of the beam: The weight of the I-beam is given as 9.00 kN. Since weight is a force acting downwards, we can represent it as a downward force of magnitude 9.00 kN.
2. Tension in the cables: There are two cables attached to the ends of the I-beam. Each cable applies a force on the beam to lift it. Since the beam is in equilibrium, the tension in each cable must be equal.
Now let's analyze the forces using vector components:
1. Vertical forces:
- The weight of the I-beam acts vertically downward.
- The tension in the cables acts vertically upward.
2. Horizontal forces:
- The tension in the cables acts horizontally inward.
We need to find the tension in each cable. To do this, we can analyze the forces acting vertically.
The tension in each cable can be determined using trigonometry. We'll use the angle 𝛼 = 71.1° and the vertical component of the tension.
Vertical component of the tension = Tension * sin(𝛼)
Since the I-beam is at equilibrium, the vertical component of the tension must be equal to the weight of the beam.
Tension * sin(𝛼) = Weight
Now we can solve for the tension:
Tension = Weight / sin(𝛼)
Substituting the given values:
Weight = 9.00 kN
𝛼 = 71.1°
Tension = 9.00 kN / sin(71.1°)
Using a calculator, we can find the value of sin(71.1°) ≈ 0.945.
Tension = 9.00 kN / 0.945
Tension ≈ 9.52 kN
Therefore, the tension in each cable attached to the ends of the I-beam is approximately 9.52 kN.