SIMPLE MACHINE PROBLEMS II SOLUTIONS

SIMPLE MACHINE PROBLEMS II SOLUTIONS - SIMPLE MACHINE...

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SIMPLE MACHINE PROBLEMS II SOLUTIONS 1) a) From the diagram, the last pulley provides the support for the load. The tension in the supporting cables is 44 8 EE L E L L E += ⇒= = 8 . Hence, the mechanical advantage of the pulley system is 8. b) Assuming no friction in the pulley system, then () ( ) F FF output input input input input =⇒ = = 8 40 9 8 8 1 8 40 9 8 49 . . N N . c) The effort was increased from 49 N to 63 N which implies the extra effort must be due to overcoming friction. Hence, the frictional force must have been . 63 49 N = 14 N d) The AMA of the pulley system is 40 9 8 63 56 9 . = . The IMA of the pulley system is 8. Hence, the efficiency of the pulley system is εε =×⇒ AMA IMA 100% 56 98 100% 78% ε . 2) a) For a second class lever, the fulcrum is at one end of the lever arm. Furthermore, the input is at the other end. Hence, . Since these quantities are positive, non–zero values, then dd input output > d d d d IMA d d input output input output output output input output >⇔ > = 1 > .
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This note was uploaded on 07/27/2010 for the course MATH MATH 1025 taught by Professor Tanny during the Spring '09 term at York University.

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SIMPLE MACHINE PROBLEMS II SOLUTIONS - SIMPLE MACHINE...

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