Tower Crane Forces

  • Thread starter Rafa54G2
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  • #1
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Homework Statement


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)

Homework Equations


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

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...
 

Answers and Replies

  • #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.
 
  • #3
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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
 
  • #4
PhanthomJay
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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
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?
 
  • #5
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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?
 
  • #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?
 
  • #7
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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?
 
  • #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?
 
  • #9
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About the center of the base? When I said that crane would tip when G was outside the base, I meant this:
15x3frt.jpg
 
  • #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.
 
  • #11
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The possible point is maybe (-1,-5) considering G as (0,0)?
 
  • #12
PhanthomJay
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The possible point is maybe (-1,-5) considering G as (0,0)?
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.
.
 
  • #13
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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.
 
  • #14
PhanthomJay
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Yes, change the pivot point for part b. Good work.
 

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