The steel I-beam in the drawing has a weight of 9.03 × 103 N and is being lifted at a constant velocity. What is the tension in each cable attached to its ends?

With both angles being 70 degrees

To find the tension in each cable attached to the ends of the steel I-beam, we need to utilize the concept of equilibrium. When the I-beam is being lifted at a constant velocity, the forces acting on it must be balanced.

First, let's draw a free-body diagram of the I-beam. We have two cables attached to its ends, so at each end, there is a tension force acting upwards. We will label these tensions as T1 and T2.

Next, we need to consider the weight of the I-beam. The weight acts downward and has a magnitude of 9.03 × 10^3 N. We can represent this force as W.

From the given information, we know that both angles are 70 degrees. We can split the weight force into two components: one parallel to the cables (downwards) and one perpendicular to the cables (horizontally).

Using trigonometry, we can determine that the component of the weight parallel to the cables is W * sin(70°), and the component perpendicular to the cables is W * cos(70°).

Since the I-beam is being lifted at a constant velocity, the net force on it must be zero. Therefore, the sum of the vertical forces (T1 and T2) must balance the weight of the I-beam. In other words, T1 + T2 = W * sin(70°).

Now, we also know that the horizontal components of the weight and the tensions in the cables must balance. The horizontal component of the weight is W * cos(70°). Since the I-beam is lifted at constant velocity, there is no acceleration in the horizontal direction, which means the horizontal forces are balanced. Therefore, T1 = T2.

Substituting T2 for T1 in the equation T1 + T2 = W * sin(70°), we get 2 * T1 = W * sin(70°), or T1 = (W * sin(70°)) / 2.

Finally, to find the tension in each cable, we can substitute the given weight of the I-beam, W = 9.03 × 10^3 N, into the equation.

T1 = (9.03 × 10^3 N * sin(70°)) / 2.
T1 ≈ 3.77 × 10^3 N (rounded to two decimal places).

Therefore, the tension in each cable attached to the ends of the steel I-beam is approximately 3.77 × 10^3 N.