1. Mar 10, 2006 #1

   finlejb

   

   A 1.95 kg ball is attached to the bottom end of a length of 11.0 lb (48.9 N) fishing line. The top end of the fishing line is held stationary. The ball is released from rest while the line is taut and horizontal At what angle (measured from the vertical) will the fish line break?
   
   I've got the equation [tex] T = mg(3-2cos\theta)[/tex] but when I solve that for [tex] \theta[/tex] with T = 48.9, it says it's wrong. any thoughts?

    

   finlejb, Mar 10, 2006
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3. Mar 10, 2006 #2

   Doc Al

   

   Staff: Mentor

   Show how you arrived at that equation.

    

   Doc Al, Mar 10, 2006
4. Mar 10, 2006 #3

   Andrew Mason

   

   Science Advisor

   Homework Helper

   What is the acceleration of the ball as it moves in this path? What does it depend on? What provides this acceleration? Hint: it is not just gravity that provides the acceleration here.
   
   AM

    

   Andrew Mason, Mar 10, 2006
5. Oct 5, 2008 #4

   hsiao

   

   Andrew, I don't see any other source of acceleration. Explain.

    

   hsiao, Oct 5, 2008
6. Oct 6, 2008 #5

   Andrew Mason

   

   Science Advisor

   Homework Helper

   The ball is falling (gravity) and it is moving in a circular arc.
   
   AM

    

   Andrew Mason, Oct 6, 2008
7. Oct 6, 2008 #6

   Doc Al

   

   Staff: Mentor

   The ball is attached to a fishing line.

    

   Doc Al, Oct 6, 2008

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