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What is the force required to keep two blocks in equilibrium

  1. Sep 17, 2015 #1
    1. The problem statement, all variables and given/known data
    Blocks A and B each weight 170 lb and rest on frictionless surfaces. They are connected to one another by cable AB. Determine the force P required to hold the blocks in the equilibrium position shown and the reactions between the blocks and surfaces.

    Hint given: This problems requires two separate FBDs of blocks A and B. Isolating each block requires a closed surface that passes through the cable and the smooth surface on which it's resting. Examine block A first to determine the cable tension required to keep it in equilibrium. You can then use this information to determine the required force P to keep block B in equilibrium.

    Image attached as well as free body diagram:
    http://imgur.com/a/FV3NJ

    2. Relevant equations
    ΣFx=0
    ΣFy=0
    Basic trig

    3. The attempt at a solution
    I drew both force diagrams like the hint suggested and then started to try to find ΣFx and ∑Fy for block A and I don't think I got very far since my numbers do not seem correct (you can see in the picture). I know I need to find tension on block A then from there use that to find P, however I don't know where the angle of block B fits in. In order to keep block B in equilibrium P has to cancel out an Fx components. Just a little stuck on getting there...
     
  2. jcsd
  3. Sep 17, 2015 #2

    haruspex

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    I don't understand your FBD for A. You seem to show the weight acting at 45 degrees, and don't show a normal force.
    Please correct that and post whatever equations you get.
     
  4. Sep 17, 2015 #3
    ImageUploadedByPhysics Forums1442535634.008757.jpg
     
  5. Sep 17, 2015 #4

    haruspex

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    Your equations are wrong. You still seem to have the weight of the block acting at 45 degrees to the vertical. That seems to violate the definition of vertical.
     
  6. Sep 17, 2015 #5
    I went to my engineering learning center today and asked one of the coaches where the 45 degree force would go and he told me there... I agree with mrblanco's placement of the normal vector, I just didn't put it in because our statics textbook seems to never mention putting them in even in the examples.
     
  7. Sep 17, 2015 #6

    haruspex

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    The normal force is fine. It's the force due to gravity that's wrong.
    You must not leave out the normal force in general. In many problems it is simply equal and opposite to a gravitational force, so you can ignore both (other than how the normal force feeds into the frictional force). But that does not apply here. All three forces, tension, gravity and normal force, must be considered and represented correctly in the equations.
     
  8. Sep 17, 2015 #7
    IMG_20150917_232538.jpg
    Alright here is my updated FBD.

    Would the angle of the normal vector be 45 degrees then? Not sure where that fits in.
     
  9. Sep 17, 2015 #8

    haruspex

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    Yes, the normal force is called that because it is normal (i.e. at right angles) to the surface. The surface is at 45 degrees to the horizontal so the normal is at 45 degrees to the vertical.
    Your diagram is right now, but your equations are truncated in the image. Anyway, posting handwritten working as images is discouraged for several reasons. Please take the trouble to type your equations in. Preferably use LaTeX, or use the subscript and superscript features (X2, X2) in the formatting line above the post area.
     
  10. Sep 17, 2015 #9
    Sorry, it was just a little easier to show the FBD in a picture.

    The equations I got were:
    ΣFx=-Ncos(45°)+TABcos(32°)=0
    ΣFy=-170+Nsin(45°)+TABsin(32°)=0
     
  11. Sep 18, 2015 #10

    haruspex

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    Yes. So now you can find TAB and plug that into the equations for the other block. (Please post those equations.)
     
  12. Sep 18, 2015 #11
    Do I just leave N as undefined? Or is there some way of calculating it?
     
  13. Sep 18, 2015 #12

    haruspex

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    You have two equations and two unknowns: N and TAB. Using the equations to eliminate the unknown you don't care about (N) to find the one you do care about (TAB) is just standard algebraic procedure ('simultaneous equations').
     
  14. Sep 18, 2015 #13
    TAB=(170/cos(32°)+sin(32°)) which is ≈ 123.37

    As for the equations for Block B:
    P=xi+0j (we are finding x)
    TAB=-TABcos(32°)i-TABsin(32°)j
    Weight of Block=-170j

    Adding them together:
    ΣFx=x-TABcos(32°)=0
    ΣFy=-TABsin(32°)-170=0
     
    Last edited: Sep 18, 2015
  15. Sep 18, 2015 #14

    haruspex

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    Your Fy equation for B is clearly wrong: two negatives adding up to make zero. What have you left out (again)?
    But you don't need this equation, do you?
     
  16. Sep 18, 2015 #15
    The normal vector... whoopsie

    The problem in addition to asking for TAB and P asks for RA and RB so don't I need all the components?
     
  17. Sep 18, 2015 #16

    haruspex

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    Yes, you will.
     
  18. Sep 18, 2015 #17
    Alright so

    ΣFx=x-TABcos(32°)=0
    ΣFy=-TABsin(32°)-170+N=0? Since it's completely in the y direction
     
  19. Sep 18, 2015 #18

    haruspex

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    Yes.
     
  20. Sep 18, 2015 #19
    I put the answers in for TAB and P and got them correct, yay!
    Now I just have to plug everything back in and find the magnitudes.

    Thank you so much, I understand so much better now.

    Do you think you could look at my other post I have? It's much more simpler than this one.
     
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