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Pulleys and Pegs

Pulleys and Pegs. Pulleys and Pegs. pulley. LINK. weight. weight. http://www.mathsnet.net/asa2/2004/m14pulley.html. Pulleys and Pegs. 5kg. A peg changes the direction of the force. 10kg. A simple pulley system.

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Pulleys and Pegs

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  1. Pulleys and Pegs

  2. Pulleys and Pegs pulley LINK weight weight http://www.mathsnet.net/asa2/2004/m14pulley.html

  3. Pulleys and Pegs 5kg A peg changes the direction of the force 10kg

  4. A simple pulley system To solve pulley/peg problems the key fact is that the tension in the string is equal throughout The bodies will experience equal [opposite] acceleration Problems are solve by considering the forces on the masses separately 4 kg 12 kg

  5. “Calculate the acceleration and the tension in the string” Same acceleration 4 kg 4 kg 12 kg 12 kg Example First: Draw Force diagrams T T a ms-2 Same string, same tension a ms-2 117.6 N 39.2 N N2L : F = ma weight = 12g = 12 x 9.8 = 117.6 N N2L : F = ma weight = 4g = 4 x 9.8 = 39.2 N

  6. “Calculate the acceleration and the tension in the string” 4 kg 12 kg Example Left Weight First: Draw Force diagrams Resultant force (in direction of acceleration) a ms-2 T T a ms-2 = 117.6 - T N2L : F = ma 117.6 - T = 12a Right Weight Resultant force (in direction of accel.) 117.6 N 39.2 N = T - 39.2 N2L : F = ma T - 39.2 = 4a

  7. “Calculate the acceleration and the tension in the string” 4 kg 12 kg Example Left Weight 117.6 - T = 12a First: Draw Force diagrams Right Weight a ms-2 T T a ms-2 T - 39.2 = 4a 117.6 - T = 12a + T - 39.2 = 4a 117.6 - 39.2 = 16a 117.6 N 39.2 N 78.4 = 16a T - 39.2 = 4a T=39.2 + 4a = 39.2 + 4 x4.9 = 58.8 N a = 78.4 / 16 = 4.9 ms-2

  8. Example 2 - find tension and acceleration a ms-2 T 5kg 5g = 49 N T a ms-2 10g = 98 N First: Draw Force diagrams 5kg 10kg A peg changes the direction of the force 10kg

  9. Example 2 - find tension and acceleration a ms-2 T 5kg 5g = 49 N T a ms-2 10g = 98 N Force diagrams Top Weight Resultant force (in direction of acceleration) = T 98 -T = 10a N2L : F = ma T = 5a T = 5a 98 -5a = 10a 98 = 15a Bottom Weight 10kg a = 98 / 15 = 6.5 ms-2 Resultant force (in direction of accel.) = 98 -T T = 5a N2L : F = ma 98 -T = 10a T = 5x 6.5 = 32.5 N

  10. The cable does not stretch There is no friction here Modeling Assumptions The cable/string is weightless m1 The surface is smooth (questions with friction will always have the word “rough” in them and talk about the coefficient of friction) Air resistance is negligible m2

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