Equilibrium sum of forces question

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Equilibrium question with distance and weight

Homework Statement


Would someone be able to tell me if my attempt at this is correct. It seems too easy. If it is not could you help?


Homework Equations


Sum of the forces in the X direction
Sum of the forces in the Y direction
Mid point = DF

The Attempt at a Solution


attached
 
Attachments
  • Textbook diagram of a collar A on a horizontal rod with force P pulling left, a cord running over a pulley at B to a 50 lb load C hanging 20 in. below, and distance x marked from the wall to the cord.
    Equilibrium001.jpg
    32.8 KB · Views: 470
  • Handwritten free-body diagram of a bar AB pinned at A with an applied 40 lb horizontal load P, a 50 lb weight C hanging from B, and reactions Ax, Ay, Bx, By; below, equilibrium equations and a boxed solution with Ax = 40 lb, By = 50 lb, and moment about point A = 1.6 in.
    Equilibrium002.jpg
    22 KB · Views: 496
Last edited:
I am just seeing if i am on the right path. A simple yes or no would be enough.
Thank you
 
No, it's not quite right. Hint: What's the tension in the rope? Then you can sum the forces on the sliding collar.
 
Thanks Doc Al
I will look at it again when i get home from work.
 
I was going about this the wrong way, making it too complicated.
1st what I did was get the angle where the rope attached to collar A and the rod, γ = cos-1 (40/50), γ=36.87 ; the to find x I just used tan36.87= 20/x  x = 20/tan36.87 so the result is x = 26.67in.
I find that I over think a problem too much but I guess that is because I find it hard to find a starting point.

Thanks