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Forces and wrecking ball Question

  1. Dec 17, 2005 #1
    hey Im new here and i need help with this problem:

    a 1240 kg wrecking ball is pulled back with a horizontal force of 5480 N before being swung against the side of a building. a) What angle does the wrecking ball make with the vertical when it is pulled back? b) What is the tension in the ball's supporting cable when it is at this angle?
     
  2. jcsd
  3. Dec 17, 2005 #2

    Tide

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    If you're new here then you may not have noticed the requirement that you show some of your own work before seeking help. So, tell us what you've done so far! :)
     
  4. Dec 17, 2005 #3
    sorry i wasn't aware

    well i think i have to do ((1240N)(9.8m/s^2))/cosx)=5480 and then try to solve for the angle but that cant be what im suppose to do since that since the number i would be doing cosine inverse for is greater than 1 and for part b im just not sure
     
    Last edited: Dec 17, 2005
  5. Dec 17, 2005 #4

    Tide

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    There are three forces acting on the ball. Ultimately, you want to find the tension in the cable. Since all three forces add to zero, you will need to find the resultant vector of the weight (vertical) and the applied force (horizontal) displacing the ball. What angle does the resultant force make with respect to the vertical? It will help to draw a picture!

    For part (b), the tension exactly balances the other two forces. You should be able to determine the magnitude of the vector resulting from adding the force of the weight (vertical) with applied force (horizontal).
     
  6. Dec 17, 2005 #5
    im still not sure

    so for a would i do (1240 kg)(9.8m/s^2)=12152
    then (5480^2)+(12152^2)=resultant^2 which would have the res= 13330.457
    then cosx=12152/13330.457... x= 24.27 degrees

    and then for part b would it just be 13330.457

    Is this right or am I WAY off
     
  7. Dec 17, 2005 #6

    Tide

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    Looks good!
     
  8. Dec 17, 2005 #7
    But wait up. To pull the ball back in the first place, you have to apply a force greater than the weight of the wrecking ball, no?

    So, 1240*-9.8 = -12152N

    And, since you are only applying 5480 N, -12152 + 5480 is -6672 (which would be the force holding the wrecking ball in place (something like friction, since the ball is hanging down). Therefore, the angle would be 90 (vertical).

    Also, realize that this is not going to be a right triangle (you're using Pythagoreans Theorm). It's going to be a curve, or arc.

    Correct me if I'm wrong!

    Thanks
     
    Last edited: Dec 17, 2005
  9. Dec 17, 2005 #8

    Tide

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    No. You do not have to apply a force equal to the weight of the ball to displace it from its equilibrium point. Most of the weight is being supported by the cable.

    Also, the problem specifically stated that the applied force is horizontal. The weight is directed vertically. Therefore, Pythagoras works just fine.
     
  10. Dec 18, 2005 #9
    Thanks

    Thanks for your help
     
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