How to Calculate the Tension in a Cable Supporting a Lever Rod?

AI Thread Summary
To calculate the tension in the cable supporting a lever rod, consider the lever rod's length (L = 7.34 m), weight (W = 69 N), and its position in equilibrium at an angle of 55 degrees from the vertical. The cable is attached 2.15 m from the end of the rod, creating a distance (d = L - b) from the pivot. The problem involves summing moments about the pivot point, leading to the equation T(L - b) sin(beta) - Fg(L/2) sin(alpha) = 0. After applying the correct torque equations, the tension in the cable is calculated to be approximately 42.53 N. Understanding the relationship between the angles and distances is crucial for solving similar problems.
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http://img6.imageshack.us/img6/8772/leverrodtension3db.th.png
How do I find the tension of the string? (The image is from a homework problem)
I want to know how to approach it. (hence, I left the numbers out)
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Here's the info they give in hopes of more response:

Consider a lever rod of length L = 7.34 m,
weight W = 69 N and uniform density. As
shown on the picture below, the lever rod is
pivoted on one end and is supported by a
cable attached at a point b = 2.15 m from the other end.
The lever rod is in equilibrium at angle beta =
55 degrees from the vertical wall. The cable makes
angle alpha = 70 degrees with the rod.

What is the tension of the supporting cable? Answer in units of N.
</edit>
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All the variables in the picture are given. Now in terms of the variables, how do I solve it??
I think this is a torque and moment of inertia problem. But what equations will be used?
I came up with the following by comparing it was a somewhat similar problem. However, the "b" variable is throwing me off as to how that is utilized.
Help much appreciated. Thanks.

Is the following even close to helping finding T?

TL sin(beta) - Fg(L/2)sin(alpha) = O
 
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I would begin by summing the moments about the hinge
 
b is simply there to give you the distance of the string from the pivot (d=L-b) it is at this point that a moment equal but oppposite to the moment caused by the weight of the bar.
 
I've added the actual problem. I'm in need of desperate help since my exam is tomorrow and I want to know how to do this problem -- although it probably won't be on it (I hope).

Help is much appreciated.:approve:
 
Maybe T(L-b) sin(beta) - Fg(L/2)sin(alpha) = O
 
yeah that looks right
 
Aneleh said:
yeah that looks right

Thanks for the confirmation. I worked it out and the answer came out correct. (42.5330474 or 42.53 N)
 
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