Determining β in the L-Shaped Rod Problem

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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([itex]\frac{0.24sinθ - EF}{0.24cosθ+0.32}[/itex])

How can I determine what EF is? Is it possible to get an actual number? FG = 0.160
 
Attachments
  • Hand-drawn free-body diagram of an L-shaped rod with pivot A, reaction forces Ax and Ay, cable tension T at angle β from vertical, load P at end D, angle θ, and labels AB sinθ, AB cosθ, and (AB+BC) sinθ + CD cosθ.
    statics 1.JPG
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  • Engineering statics diagram of an L-shaped rod pinned at A and held by a cable from B passing over a pulley at E with tension T; dimensions 160 mm, 320 mm, 240 mm, 300 mm and 80 mm are labelled, with load P applied at end D and angle θ marked at A.
    statics1a.JPG
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Last edited:
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.