Finding the tension in a cable of a bridge

[arared]

1. Compute the cable tension at the midpoint, the quarterpoint and support for a 1600 foot span with a sag at the midpoint of 100 feet, when the cable is loaded with 10 kips/foot spread uniformly along a horizontal line.

Midpoint Cable Tension=?
a)3200 pounds
b) 16000 pounds
c) 32,000 kips
d) 63,000 kips

Homework Equations

The Attempt at a Solution

I have no idea what to do. the only things that come to mind are the vertical and horizontal forces. Also setting the torque to 0. No real idea I repeat.

 

[PhanthomJay]

1. Compute the cable tension at the midpoint, the quarterpoint and support for a 1600 foot span with a sag at the midpoint of 100 feet, when the cable is loaded with 10 kips/foot spread uniformly along a horizontal line.

Midpoint Cable Tension=?
a)3200 pounds
b) 16000 pounds
c) 32,000 kips
d) 63,000 kips

Homework Equations

The Attempt at a Solution

I have no idea what to do. the only things that come to mind are the vertical and horizontal forces. Also setting the torque to 0. No real idea I repeat.

Hello, arared, and wecome to PF!
Cables suspended between 2 points, under their own weight, take the shape of the catenary, which often can be approximated by a parabolic curve. Cables that are uniformly loaded by external distributed vertical forces, which appears to be your case, do take on a parabolic shape. The horizontal tension is easily calculated by a simple formula, and the other tensions can be calculated from that value. Are you given any data other than what is given, or are you being asked to derive the formula? I could give you the formula for the midpoint tension, but that would give you the answer, so I have to keep it a secret for now, until you clarify the assignment.

 

[piercas]

The calculation for the tension in a cable of a bridge involves using the principles of statics and equilibrium. The key equation to use is the equation for the sum of forces in the horizontal direction, which states that the sum of all the horizontal forces acting on an object must be equal to zero for the object to be in equilibrium.

To find the tension at the midpoint, quarter point, and support, we need to first determine the distribution of the load along the cable. In this case, the load is spread uniformly along a horizontal line, which means that the load is evenly distributed over the entire length of the cable.

Next, we can use the equation for the sum of forces in the horizontal direction to set up a system of equations. We know that the cable will experience a tension force in both the horizontal and vertical directions, and we also know that the horizontal component of the tension force will be equal to the load acting on the cable. Using this information, we can set up the following equations:

At the midpoint:
Tension in the horizontal direction = Load acting on the cable = 10 kips/foot x (1600 ft/2) = 8,000 kips
Tension in the vertical direction = 0 (since the cable is not being pulled up or down)

At the quarter point:
Tension in the horizontal direction = Load acting on the cable = 10 kips/foot x (1600 ft/4) = 4,000 kips
Tension in the vertical direction = 0 (since the cable is not being pulled up or down)

At the support:
Tension in the horizontal direction = Load acting on the cable = 10 kips/foot x 1600 ft = 16,000 kips
Tension in the vertical direction = 0 (since the cable is not being pulled up or down)

Therefore, the correct answer is option b) 16,000 pounds for the midpoint, quarter point, and support. It is important to note that the units are in pounds, not kips, as stated in the options. This is a common mistake when dealing with units of force.

In summary, to find the tension in a cable of a bridge, we need to consider the distribution of the load, use the equation for the sum of forces in the horizontal direction, and set up a system of equations to solve for the tension at different points along the cable.