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Crane Problem: Find Horizontal and Vertical Components

  1. Mar 21, 2008 #1
    hey guys, i wasnt sure if you wanted me to consolidate both of my threads into one thread, so i just opted to make two to avoid a lot of clutter :P

    please feel free to merge the threads if you find it appropriate or contact me if there is an issue regarding the thread(s).

    1. The problem statement, all variables and given/known data
    A crane of mass 2,330 kg supports a load of 10,200 kg as in the figure below. The crane is pivoted with a frictionless pin at and rests against a smooth support at .

    [​IMG]

    (a) Find the horizontal component of the reaction force at .
    Please round your answer to three significant figures.


    (b) Find the vertical component of the reaction force at .
    Please round your answer to three significant figures.


    (c) Find the horizontal component of the reaction force at .
    Please round your answer to three significant figures.

    (d) Find the vertical component of the reaction force at .


    2. Relevant equations
    i believe we need to use
    torque=Fd
    somehow incorporate vectors, and free body diagrams


    3. The attempt at a solution
    i was thinking this might be more of a conceptual question and i thought that the force at the point a would just be the weight of thhe crane, however on my online homework i was incorrect.

    also all 4 questions are the same so if someone could explain at least one part to me and maybe the conceptual aspect behind it, that would really be awesome.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 22, 2008 #2
    first FBD, a pin joint can only resist forces in the horizontal and vertical directions(point A), the smooth support provides just a horizontal componet,

    then you take the sum of the moments at point B, to find Ay force at A vertical component, then sum of moments at A to get Bx, and for Ax sum of the force in x-direction
     
  4. Mar 22, 2008 #3
    Show us your FBD.
     
  5. Mar 22, 2008 #4
    well at point A i wold assume it to be:

    EFx: 0
    EFy: -m1g-m2g (m1 being the mass of the crane and m2 being the mass of the load)

    then for pt B

    EFx: 0
    EFy: 0

    i know my free body diagrams are incorrect because on one hand i cant think of how to put together all the external forces. like for instance, i know there has to be forces at point B, but i dont know where to put them at all. i know that after i correct my free body diagrams, i will have the correct answer.

    cheers
    tron
     
  6. Mar 30, 2008 #5
    I have the answer

    Take Nax, Nay, Nbx, and Nby for the forces exerted in the x and y direction by points a and b (since these are normal forces)
    Do a free body diagram of the forces in the Y direction

    Sum of forces in Y direction: 0=Nay-Mcg-m1g

    you know Mcg and m1g so just solve for Nay
    (should be around 120000)

    Then do the sum of the torques using a as the pivot point
    sum of the torques: 0=(1meter)Nbx-(2meters)Mcg-(6meters)m1g

    once again you know Mcg and m1g so just solve for Nbx
    (should be around 600000)

    Nby is zero cause it exerts no vertical resistive force

    Nax actually equals Nbx. The force exerted by B to the right has to equal the force exerted by A to the left to establish equilibrium since the weight of the crane and the mass on it are in the vertical direction and don't affect this.
     
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