The boom weighs 2450 and is attached to a friction less pivot at its lower end. It is not uniform; the distance of its center of gravity from the pivot is 39.0 of its length.

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What is the tension in the guy wire?

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To find the tension in the guy wire, we can analyze the forces acting on the boom.

First, let's determine the length of the boom. Since the distance of the center of gravity from the pivot is 39.0 times its length, we can express this relationship as:

Distance from pivot to center of gravity = 39.0 * Length of the boom

Now, let's denote the length of the boom as "L". This means that the distance from the pivot to the center of gravity is 39.0 * L.

Next, let's draw a free-body diagram of the boom:

1. The weight of the boom (2450 N) acts vertically downward from the center of gravity.
2. The tension in the guy wire acts diagonally upward and away from the pivot.
3. There is no friction at the pivot, so no horizontal forces are present.

By looking at the diagram and analyzing the forces, we can conclude that the vertical component of the tension force should balance the weight of the boom for equilibrium.

Therefore, the vertical component of the tension force is equal to the weight of the boom:

Vertical component of tension = Weight of the boom = 2450 N

To find the tension in the guy wire, we need to calculate the total tension force using the Pythagorean theorem. The total tension force can be found by summing the vertical and horizontal components of the tension force.

Using the Pythagorean theorem:

Total tension^2 = Vertical component of tension^2 + Horizontal component of tension^2

Since there is no horizontal force acting on the boom, the horizontal component of the tension force is zero.

Thus, we have:

Total tension^2 = Vertical component of tension^2 + 0^2
Total tension^2 = Vertical component of tension^2

Simplifying:

Total tension = Vertical component of tension = Weight of the boom = 2450 N

Therefore, the tension in the guy wire is 2450 N.