Determining β in the L-Shaped Rod Problem

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The discussion revolves around determining the angle β in the L-shaped rod problem, which is influenced by the angle θ and the geometry of the system. The equation derived for β is β = tan-1((0.24sinθ - EF)/(0.24cosθ + 0.32)), but the value of EF needs clarification. Participants note that FG is not 160 mm, but rather EF is defined as 160 mm, while FG is calculated using AB*sin(θ). The relationship between β and the pulley radius is also highlighted, suggesting that without the radius, β remains indeterminate. The conversation emphasizes the importance of accurately defining all parameters to solve for β effectively.
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Homework Statement



The position of the L-shaped rod shown is controlled by a cable attached at point B. Determine the angle, β, in terms of θ.

The Attempt at a Solution



From the FBD attached, I was able to get:

β = tan-1(\frac{0.24sinθ - EF}{0.24cosθ+0.32})

How can I determine what EF is? Is it possible to get an actual number? FG = 0.160
 

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Is the beta angle supposed to be the angle the cable BE makes to the vertical?

Then I think there is something not quite right about your diagram ...
F is vertically above the pivot of the beam and level with the top of the pulley?
It is 160mm horizontally from F to the center of the pulley.

It looks like you'll need the diameter of the pulley.
 
aaronfue said:
FG = 0.160.
aaronfue: FG is not 160 mm. FG is simply AB*sin(theta), whereas EF = 160 mm.

Here are my current thoughts. Theta can be any value; it is given. Given theta, then I think beta can be any value, depending on the pulley radius. You are given the pulley axle horizontal location, which is EF = 160 mm.

But are you given the pulley radius? If not, I am currently thinking beta will also be a function of pulley radius, r.
 
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