Physics Help Forum Question No.3
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 Jun 15th 2017, 01:32 AM #1 Junior Member   Join Date: Apr 2017 Location: Turkey Posts: 20 Question No.3 Hello again my friends Could any one help me to find the suitable procedure to solve the attached problem ? i will thankful for him. Best regards Razi Attached Thumbnails
 Jun 15th 2017, 06:35 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,280 Use energy principles, as the problem suggested. The work done by the force of the wind equals the maximum potential energy gain of the ball. The work done by the wind is the integral oftegh force times the distance the ball travels in the direction of that force: $\displaystyle W = \int _0 ^ \Theta \vec F \cdot d \vec S$ where S is the path length of the ball. Using $\displaystyle S = L \theta$, and then $\displaystyle \vec F \cdot d \vec S = |F| \cos \theta d \theta$, the work equation becomes: $\displaystyle W = \int _0 ^ \Theta |F| \cos \theta d \theta$ where $\displaystyle \Theta$ is the maximum angle it can attain. Set this equal to the gain in PE: $\displaystyle \Delta PE = mgH$ The rest is simply converting $\displaystyle \sin \Theta$ into units of L ang H, and manipulating the equations. Hope this helps. topsquark likes this.
 Jun 15th 2017, 09:21 AM #3 Junior Member   Join Date: Apr 2017 Location: Turkey Posts: 20 Dear ChipB Thanks for replay , i have did your suggestion and i got this result : H = F * L * sin(theta) / (m*g) Now , how could eliminate sin(theta) ? and convert it to function of L ? Thanks in advance and best regards Razi
 Jun 15th 2017, 12:39 PM #4 Senior Member     Join Date: Aug 2008 Posts: 113 $H = L-L\cos{\theta} \implies \cos{\theta} = \dfrac{L-H}{L}$ recall $\sin{\theta} = \sqrt{1-\cos^2{\theta}}$ topsquark likes this.
 Jun 15th 2017, 01:11 PM #5 Junior Member   Join Date: Apr 2017 Location: Turkey Posts: 20 I did it , thank you sooooooooooooooooooooooooo much Best regards Razi

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