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Tower Crane Forces

  1. Sep 26, 2016 #1
    1. The problem statement, all variables and given/known data
    Sin título.png
    Hello, this problem is cracking my mind for days, and I haven't solved it yet :(
    Crane dimensions: See image
    Mass of the crane: 10Tons
    Mass center: G
    Maximum Load: 4Tons
    Counterweight: P
    Weight of the counterweight: ?

    "A" says: when you have a maximum load (4Tons) at the farest left side, the minimum weight of P that lets work crane well without falling to the left. (Answer: 6.93Tons)
    "B" says: without any load, the maximum weight of the counterweight that lets work crane well without falling to the right. (Answer: 26.67Tons)

    2. Relevant equations
    The resultant force will be cero, and the resultant torque too.

    3. The attempt at a solution
    Tried to calculate the torques to find the mass of the counterweight but I didn't find the correct answer.
    I tried to use the angle of the left arm with the crane axis to make a tension T, but nothing.
    The answer will probably be easy, but I can't see it now...
     
  2. jcsd
  3. Sep 26, 2016 #2

    PhanthomJay

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    You may not be considering all the forces acting on the crane, and their location from the pivot point at the start of tipping. Please show your sum of torques calculations so that we may assist you further.
     
  4. Sep 27, 2016 #3
    I'm considering 3 forces, Load force, G force and counterweight force. Should I consider more? The distances start in the crane axis.
    Torque= 10.5(m)*4(T)*9.8(m/s²) + 0.5(m)*10(T)*9.8(m/s²) - 2.25(m)*x(T)*9.8(m/s²) =0
     
  5. Sep 27, 2016 #4

    PhanthomJay

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    you have the load, counterweight, and crane weight acting down, good. But for equilibrium in the vertical direction, you must also have one or more forces acting up. So you are missing that force or forces. What is that force and where does it act when the tower is about to tip?
     
  6. Sep 27, 2016 #5
    Well, without any up force, the crane will fall easily. What about consider an up force in the base of the crane? Now the system is in equilibrium because the module of the new force is the sum of the others, isn't it?
     
  7. Sep 27, 2016 #6

    PhanthomJay

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    Yes the up force of the ground on the crane acts on its base where it is in contact with the ground. This is called the normal force. And yes, it's magnitude must equal the sum of the down forces. Now where along the base does the normal force act when the crane starts to tip?
     
  8. Sep 27, 2016 #7
    I've just read that Pisa's tower won't fall down beacause its mass center is in the area of the base. So, if G is outside the base, crane will start to tip. It depends on the angle formed by G and the cener of the crane base. Then, the distances measuared in the torque will change, and the mass X will change too?
     
  9. Sep 27, 2016 #8

    PhanthomJay

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    The crane won't actually fall to the ground until all loads are outside the base. But it will start to tip (rotate) immediately without any counterweight.. About what point on the base?
     
  10. Sep 27, 2016 #9
    About the center of the base? When I said that crane would tip when G was outside the base, I meant this:
    15x3frt.jpg
     
  11. Sep 27, 2016 #10

    PhanthomJay

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    No that's not correct. When it starts to tip, it won't pivot about the center. It will pivot about a different point along the base. Can you visualize what that point is? And G does not have to be outside the base at all at the start of tipping, initially at that moment it stays right where it is.
     
  12. Sep 27, 2016 #11
    The possible point is maybe (-1,-5) considering G as (0,0)?
     
  13. Sep 27, 2016 #12

    PhanthomJay

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    Correct! So the normal force acts at the left lower corner of the crane at the verge of tipping. Can you solve sum of moments equal zero now to solve for the minimum counterweight load? You can sum moments about any point, but choose the point that will make the solution easier to find.
    .
     
  14. Sep 27, 2016 #13
    Wow, tried hundreds of things and nothing... Next time I will use logic instead of maths. Thank you!! I don't know if you're a teacher, but you would be a good one, didn't solve the problem but helping me to do it :D
    I guess that b) is the same but changing the inflexion point and being carefull with the forces.
     
  15. Sep 27, 2016 #14

    PhanthomJay

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    Yes, change the pivot point for part b. Good work.
     
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