Equilibrium and impulsive forces (crane and demolition ball problem)

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Homework Help Overview

The discussion revolves around a problem involving a crane and a demolition ball, specifically focusing on equilibrium and impulsive forces. Participants are exploring concepts related to potential and kinetic energy, tension in the rope, and the calculation of height and velocity in the context of the ball's motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to calculate acceleration using Newton's second law and are discussing the relationship between potential and kinetic energy. Some are questioning the assumptions made about energy conservation and the role of tension in the system. Others are exploring geometric relationships to determine height and angles.

Discussion Status

The discussion is ongoing, with various participants offering different perspectives and calculations. Some have provided specific numerical attempts, while others express uncertainty about their results. There is a recognition of the need to reconcile calculated values with expected outcomes, particularly regarding the final velocity before impact.

Contextual Notes

Participants note the challenge of fitting their calculations to the specified final velocity of 6.7 m/s and a reasonable height, indicating potential constraints in the problem setup or missing information.

rotiff
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Homework Statement


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The Attempt at a Solution


I tried to find the acceleration using F= ma

Resultant force = Tension - mg
so a = (10295.2-8730)/890 = 1.758 m/s^2 but I don't think it works that way...

please help.
 
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The height which from the ball falls, and its mass, gives a potential energy to the ball. This results in an identical kinetic energy at 0-level. The angle and length of rope should give the height. Then impulse can be calculated.
 
I don't think so, because at the vertical position the total energy would equal kinetic energy + some potential energy therefore mgh = 0.5 mv^2 + mgh

And both velocity & height are missing from the right side of the equation.
 
I think the height can be calculated by angle and length, but possibly that is not what is asked for. My reply might be uninformative. Tension is not a topic of mine. Nor is geometry, really.
 
Can you please attempt it? I have solved other parts of this question and I will see if your calculated height fits in with the rest.
 
At vertical position, the ball can fall no further in this system: equals no potential energy
 
I tried it and the final velocity came up as 17 m/s, I know this is wrong because in the next part it says SHOW that the final velocity just before impact = 6.7 m/s
 
This is ridiculous... I'm sure the answer should be fairly easy.
 
The 32deg. form the top of a triangle. The two other angles are 180-32=148 deg. combined, or 74 deg. each. Find length of base of triangle. Probably some cosinus or sinus calculating. I don't know. Then there's another small triangle below, whith 90 and 90-74=16 deg. angles. Base from other triangle is one side, short side of this small triangle is your heigth, and basis for potential/kinetic energy
 
  • #10
Thank you but I still think there is some other solution that jiggles with Newton's laws. If someone figures it please post.
 
  • #11
has anyone got any idea, please? At this point I tried everything and I just can't get my answer to fit both a final velocity of 6.7 m/s and a reasonable height
 
  • #12
I figured this and thought I'd post the solution:

Height at elevated position = 15 cos 32 = 12.72m

Therefore h = 15-12.72 = 2.28m
 

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