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Easy statics problem -- Block being hoisted by a pulley and rope

  1. Aug 24, 2015 #1
    FullSizeRender (1).jpg Snapshot.jpg
    I drew a FBD about point A and came up with the two equations:
    ∑Fx=0 20sinθ - T-ab(cos70)=0
    ΣFy =0 -20cosθ + T-ab(sin70) -20=0

    My answer is not coming out so I either messed up on the FBD or the equations, some help would be appreciated. I'm on 3-10 by the way.
     
  2. jcsd
  3. Aug 24, 2015 #2

    BvU

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    Hard to distinguish what your equations are about; what is the 20 ?
    And with T - ab you mean Tab ?
    How come W does not feature ? nor the 80 lb ? (lbf ?)
     
  4. Aug 24, 2015 #3

    SteamKing

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    Easy statics problem ... which is why it belongs in the Intro Physics HW forum and not one of the technical engineering forums.
     
  5. Aug 24, 2015 #4
    Yes that is what I mean. I figured the force at D is 20lb because the block has 20lb of force and the cable holding it has 20lb of tension so D would as well (same cable). So I tried to resolve that 20lbs into the x and y direction.
     
  6. Aug 24, 2015 #5

    BvU

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    Sorry, only saw 3-11.
    Equations look good. Must be the solution part. What do you do to solve $$
    \quad 20\sin\theta = T_{AB}\sin(20^\circ) \\
    \ 20+ 20 \cos\theta = T_{AB}\cos(20^\circ) \quad {\rm ? }$$
     
  7. Aug 24, 2015 #6
    So then your first equation was actually $$ F_c = F_f $$ where $$ F_c = 20 lbs. $$ So $$ F_f = 20 lbs. $$
    You probably should have written that equation.
     
    Last edited: Aug 24, 2015
  8. Aug 24, 2015 #7
    I just try solving like any other system of equations. 2 unknowns and 2 equations which is leaving me with funky answers.
     
  9. Aug 24, 2015 #8
    What do you get if you solve for the extremes? Sin and cos vary between 0 and 1. You should be able to bracket your answer.
     
  10. Aug 24, 2015 #9
    You might as well revise the equations to solve for Tab. Consider replacing the fixed signs with their numbers and divide thru.
     
  11. Aug 24, 2015 #10
    Hint: you have a formula error... Sin is hypotenuse over the what? Cos is hypotenuse over the what?

    hint2: one of the angles is taken from the horizontal x axis. The other is from the vertical y axis. How does this affect if you use sin or cos...
     
    Last edited: Aug 24, 2015
  12. Aug 24, 2015 #11
    I figured it out guys, thanks for the help.
     
  13. Aug 24, 2015 #12
    And the answer is what? I got 31.32 with the fixed equations.
     
    Last edited: Aug 24, 2015
  14. Aug 24, 2015 #13
    The angle is 40degrees and Tab=37.6lb
     
  15. Aug 24, 2015 #14
    Really, I am getting degrees =34.91 and Tab 33.46? Original equation above did not divide by cos(20)

    Hmmm, not very good at using Numbers on an iPad yet. Perhaps I should try it by hand.

    Okay. got it. Tab = 37.6 and angle = 40.
     
    Last edited: Aug 24, 2015
  16. Aug 25, 2015 #15
    Edit: Add in tension equation

    Alternative solution. Rotate your x/y axis so that x is parallel to $$ T_{ab} $$

    A-C is 20 degree angle from axis.
    A-D is unknown angle

    $$ F_ {ac} = F_{ad} $$

    $$ F_{ac} cos \theta = F_{ad} cos (uk) $$
    And therefore uk angle = theta.

    And

    $$ F_{ab} = F_{ac} sin \theta + F_{ad} sin (uk) $$

    which reduces to

    $$ F_{ab} = 2 * F_{ac} sin \theta $$



    $$ F_{ab} = 2 * 20 sin (20 degrees) $$

    Voila
     
    Last edited: Aug 25, 2015
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