Solving a Weighty Problem: Calculating Tension in Steel I-Beam Cables

[BlackMamba]

Hello again,

I'm stuck on a new problem. Again, I thought I had the correct equation but apparently not as my answer is not correct.

Here's the problem: A steel I-beam has a weight of 9.10 N and is being lifted at a constant velocity. What is the tension in each cable attached to its ends?

Here is what the picture looks like: http://www.webassign.net/CJ/4-50.gif

This is the equation I came up with to solve for this problem but apparently it isn't the correct one.

F = (2mg)cos70.0 degrees

But like I said, when solving for the above equation my answer is not correct.

Any help or direction given would be greatly appreciated.

 

[Pyrrhus]

Well you know the weight of the Beam, i suppose this will be using uniform gravitational field, so it will be in its center of gravity, and you got two tension each with the same vertical and horizontal components, calculate the Tensions vertical components and equal them to the weight of the beam.

Also Remember Newton's 1st Law:

\sum_{i=1}^{n} \vec{F}_{i} = 0 \rightarrow \vec{V} = constant

 

[M98Ranger]

Hello there,

I can understand your frustration with this problem. It can be tricky to find the correct equation for a specific situation. Let's break down the problem and see if we can find the correct equation together.

First, we have a steel I-beam with a weight of 9.10 N. This means that the weight of the beam is acting downwards, and we need to find the tension in the cables that are attached to its ends. We can assume that the cables are pulling upwards to counteract the weight of the beam.

Next, we need to consider the angle of the cables. In the picture, we can see that the cables are at a 70.0 degree angle from the horizontal. This means that the force of the cables will be split into vertical and horizontal components.

To find the tension in each cable, we can use the equation T = mg + ma, where T is the tension, m is the mass of the beam, g is the acceleration due to gravity, and a is the acceleration of the beam (which is 0 since it is being lifted at a constant velocity). However, we need to take into account the angle of the cables.

To do this, we can use trigonometry. The vertical component of the tension will be Tsin70.0 degrees, and the horizontal component will be Tcos70.0 degrees. Since the beam is being lifted at a constant velocity, the acceleration is 0, so we can disregard the ma term in the equation.

Putting all of this together, we can write the equation as T = mg + Tsin70.0 degrees. Now we just need to solve for T. Rearranging the equation, we get T = mg/(1-sin70.0 degrees). Plugging in the values, we get T = 9.10 N/(1-0.9397) = 143.2 N. This is the tension in each cable attached to the ends of the beam.

I hope this helps you understand the problem better and find the correct equation. Keep in mind that practice makes perfect, so don't get discouraged if it takes a few tries to get the right answer. Good luck!