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Force-Vector Diagram type 3 PULLEY PROBLEMS

Force-Vector Diagram type 3 PULLEY PROBLEMS. Don’t need to draw the pulley!. (A pulley in the ceiling or something; usually used to lift a huge weight.). F T. a. F T. F gR. a. F gL. Pulley Problem?.

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Force-Vector Diagram type 3 PULLEY PROBLEMS

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  1. Force-Vector Diagram type 3 PULLEY PROBLEMS Don’t need to draw the pulley! (A pulley in the ceiling or something; usually used to lift a huge weight.) FT a FT FgR a FgL

  2. Pulley Problem? • A pulley problem is just like two elevator problems put together, where one elevator is accelerating up and the other is accelerating down. • As for the forces, there are two forces in a pulley problem: • Fg (gravity) for EACH mass, and • FT (tension) along the rope, which is the same for both masses.

  3. Example 1 The givens for this problem are the masses or weights of the objects. You must determine which object masses more (or weighs more) to determine the two directions of acceleration. FT Because the left mass is lighter, the objects will accelerate as indicated. However, it could also be the other way, if the right mass was heavier. a FT FgR 165 kg or 1617 N a FgL 40 kg or 392 N

  4. Equations: • FYR = FgR - FT= ma • so,  FYR = (165)(9.8) - FT= 165a •  FYL = FT - FgL = ma • so,  FYL = FT - (40)(9.8) = 40a • Variables need to go on one side (and constants on the other) for the matrix that will be used to get the final answer...

  5. constants on other side of equation: -1617 392 FT a right eq: -1 -165 left eq: 1 -40 -1 ] [ [ ] ] [ = 631 5.98 Variables need to go on one side for the matrix that will be used to get the final answer... • - FT - 165a = - (165)(9.8) • FT - 40a = (40 )(9.8) • Put all the variables and numbers into a matrix and solve: Since FT was the first variable in the matrix, then FT 630 N and a  6.0 m/s2 (since a was the second variable in the matrix).

  6. Example 2 The givens for this example are the same masses/weights of the objects as before, only their location has switched. Recall: you must determine which object masses more (or weighs more) to determine the two directions of acceleration. FT a FT FgR a 40 kg or 392 N FgL Because the right mass is now lighter, the objects will accelerate as indicated. (Do you notice the difference?) 165 kg or 1617 N

  7. Equations: • FYR = FT - FgR= ma • so,  FYR = FT - (40)(9.8) = 40a •  FYL = FgL - FT = ma • so,  FYL = (165)(9.8) - FT = 165a • Again, variables need to go on one side (and constants on the other) for the matrix that will be used to get the final answer...

  8. constants on other side of equation: 392 -1617 FT a right eq: 1 -40 left eq: -1 -165 -1 ] [ [ ] ] [ = 631 5.98 Variables need to go on one side for the matrix that will be used to get the final answer... • FT - 40a = (40)(9.8) • - FT - 165a = -(165)(9.8) • Put all the variables and numbers into a matrix and solve: Since FT was the first variable in the matrix, then FT 630 N and a  6.0 m/s2 (since a was the second variable in the matrix).

  9. Ft Fg Real Life Example • Pulley’s are used all the time in real life, an example would be lifting the engine out of a car. • The Fg would be the weight of the engine, and the Ft would be the tension of the cable or chain holding the engine up.

  10. Start your assignment • You have the rest of the period to work on them. • REMEMBER – even though it is hard to see on this ppt, that all forces (except friction) start from the center of mass!)

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