PHY183-Lecture13 - February 6, 2006 Physics for...

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Unformatted text preview: February 6, 2006 Physics for Scientists&Engineers 1 1 Physics for Scientists & Physics for Scientists & Engineers 1 Engineers 1 Spring Semester 2006 Lecture 13 February 6, 2006 Physics for Scientists&Engineers 1 2 Example: Still Rings (1) Example: Still Rings (1) ! Gymnast of mass 55 kg hangs from still rings. ! Question 1: What is the tension in each rope? Example February 6, 2006 Physics for Scientists&Engineers 1 3 Example: Still Rings (2) Example: Still Rings (2) ! Answer: no forces in x-direction; in y-direction: Example F y , i = T 1 + T 2 i ! " mg = ! Both ropes support gymnast equally T 1 = T 2 ! T T + T ! mg = " T = 1 2 mg = 1 2 (55 kg) # (9.81 m/s 2 ) = 270 N February 6, 2006 Physics for Scientists&Engineers 1 4 Example: Still Rings (3) Example: Still Rings (3) ! Question 2: What changes if the rings do not hang straight down, but at an angle relative to the ceiling? Example ! Equilibrium conditions: ! Combine both equations: ! As angle gets smaller, T gets bigger! F x , i = T 1 cos ! " T 2 cos ! i # = $ T 1 = T 2 = T F y , i = T 1 sin ! + T 2 sin ! i " # mg = 2 T sin " mg = T = mg 2sin ! ! x y ! ! ! < 90 February 6, 2006 Physics for Scientists&Engineers 1 5 Two Pulleys (1) Two Pulleys (1) ! Mass m hung from a system of two pulleys February 6, 2006 Physics for Scientists&Engineers 1 6 Two Pulleys (2) Two Pulleys (2) ! Note: tension in string is same everywhere! ! Free-body diagram of block: ! Free-body diagram of pulley B above block: ! Combine these two equations: F g ! T 3 = " T 3 = F g = mg 2 T 1 ! T 3 = " T 1 = 1 2 T 3 T 1 = 1 2 T 3 = 1 2 mg February 6, 2006 Physics for Scientists&Engineers 1 7 Two Pulleys (3) Two Pulleys (3) ! We found: ! => Important result: If you use this pulley system, you only need to apply a force equal to 1/2 of the weight of...
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PHY183-Lecture13 - February 6, 2006 Physics for...

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