Determining β in the L-Shaped Rod Problem

  • Thread starter Thread starter aaronfue
  • Start date Start date
  • Tags Tags
    Rod
Click For Summary
SUMMARY

The discussion centers on determining the angle β in the L-Shaped Rod Problem, specifically in relation to the angle θ. The equation derived is β = tan-1((0.24sinθ - EF)/(0.24cosθ + 0.32)), where EF is defined as 160 mm. Participants clarify that FG is not 160 mm but rather AB*sin(θ), and emphasize the need for the pulley radius to fully resolve the relationship between β and θ.

PREREQUISITES
  • Understanding of trigonometric functions and their applications in physics
  • Familiarity with Free Body Diagrams (FBD) in mechanics
  • Knowledge of L-shaped structures and their dynamics
  • Basic principles of pulley systems and their geometry
NEXT STEPS
  • Research the application of trigonometric identities in mechanical systems
  • Study the principles of Free Body Diagrams in engineering mechanics
  • Learn about the effects of pulley radius on tension and angle calculations
  • Explore advanced topics in static equilibrium involving multiple forces
USEFUL FOR

Students and professionals in mechanical engineering, physics enthusiasts, and anyone involved in solving problems related to static equilibrium and pulley systems.

aaronfue
Messages
118
Reaction score
0

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

  • statics 1.JPG
    statics 1.JPG
    23.4 KB · Views: 670
  • statics1a.JPG
    statics1a.JPG
    19.1 KB · Views: 649
Last edited:
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 13 ·
Replies
13
Views
4K
Replies
19
Views
3K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 1 ·
Replies
1
Views
9K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K