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35. (a) With the pivot at the hinge, Eq. 12-9 yields cos 0 a TL F y θ −= . This leads to T = ( F a /cos )( y/L ) so that we can interpret F a /cos as the slope on the tension graph (which we estimate to be 600 in SI units). Regarding the F h graph, we use Eq. 12 − 7 to get F h = T cos − F a = ( − F a )( y/L ) − F a after substituting our previous expression. The result implies that the slope on the
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
- Spring '08