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Pulleys

Pulleys. Example Whiteboards - Atwood’s Machine Whiteboards - Inclined Plane. M 1. M 2. Step 1 - guess direction of + acceleration (This becomes the + direction for each mass). +a. For the 5.0 kg mass, the tension T is up (-). T (-). The weight of 5*9.8 = 49 N is down (+):. +a.

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Pulleys

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  1. Pulleys Example Whiteboards - Atwood’s Machine Whiteboards - Inclined Plane M1 M2

  2. Step 1 - guess direction of + acceleration (This becomes the + direction for each mass) +a For the 5.0 kg mass, the tension T is up (-) T (-) The weight of 5*9.8 = 49 N is down (+): +a +49 N Example - Find Tension, and acceleration if k = .11: 32.0 kg Step 2 - Calculate or express <F> = ma for every mass object 5.0 kg And our formula becomes: 49 N - T = (5.0 kg)a

  3. Step 1 - guess direction of + acceleration (This becomes the + direction for each mass) -34.496 N +a And, assuming it is moving to the right, a frictional force of .11*32*9.8 = 34.496 N to the left (-) +a Example - Find Tension, and acceleration if k = .11: 49 N - T = (5.0 kg)a 32.0 kg T Step 2 - Calculate or express <F> = ma for every mass object The 32.0 kg mass has the tension T to the right (+) 5.0 kg So our equation becomes: T - 34.496 N = (32.0 kg)a

  4. Add them Together! + And now it’s math time!!!! Two equations, two unknowns! 49 N - T = (5.0 kg)a T - 34.496 N = (32.0 kg)a 49 N - T + T - 34.496 N = (5.0 kg)a + (32.0 kg)a 49 N - 34.496 N = (37 kg)a a = (49 N - 34.496 N)/37 kg = .392 m/s/s And plug into an equation to find T: T - 34.496 N = (32.0 kg)a T = (32.0 kg)(.392 m/s/s) + 34.496 N = 47.04 N

  5. Whiteboards: Atwood’s Machine 1 | 2 | 3 | 4 | 5 TOC

  6. +a +a Uhh well um. the 5.0 kg is heavier. Massless frictionless pulley Find acceleration and tension Step 1 - Guess the direction of acceleration - This becomes the positive direction for each mass. 3.0 kg 5.0 kg W Green, Bananas never do

  7. Massless frictionless pulley Find acceleration and tension Step 2 - Set up the <F>=ma for the 3.0 kg mass: T is up , and the weight is down. Down is - and up is + +a 3.0 kg 5.0 kg +a weight = (3.0 kg)(9.8 N/kg) = 29.4 N down (- in this case) T is up, -29.4 is down: T - 29.4 N = (3.0 kg)a W T - 29.4 N = (3.0 kg)a

  8. Massless frictionless pulley Find acceleration and tension Step 3 - Set up the <F>=ma for the 5.0 kg mass: T is up , and the weight is down, but now down is + and up is - +a 3.0 kg 5.0 kg +a weight = (5.0 kg)(9.8 N/kg) = 49 N down (+ in this case) T is up (-), 49 N is down (+): 49 N - T = (5.0 kg)a W 49 N - T = (5.0 kg)a

  9. Massless frictionless pulley Find acceleration and tension Step 4 - Solve for acceleration: 49 N - T = (5.0 kg)a T - 29.4 N = (3.0 kg)a +a 3.0 kg 5.0 kg +a 49 N - T = (5.0 kg)a +T - 29.4 N = (3.0 kg)a T - 29.4 N + 49 N - T = (8.0 kg)a a = 2.45 m/s/s W 2.45 m/s/s

  10. Massless frictionless pulley Find acceleration and tension Step 5- Solve for T: 49 N - T = (5.0 kg)a T - 29.4 N = (3.0 kg)a a = 2.45 m/s/s +a 3.0 kg 5.0 kg +a Pick one of the formulas, and plug the acceleration in: T - 29.4 N = (3.0 kg)aT = (3.0 kg)(2.45 m/s/s) + 29.4 N = 36.75 N = 37 N W 37 N

  11. Whiteboards: Pulleys on Inclined Planes 1 | 2 | 3 | 4 | 5 TOC

  12. +a s = 0 k = 0 Find acceleration and tension 11.0 kg 30.0o 6.0 kg +a Step 1 - Guess the direction of acceleration. Let’s guess this way. (it’s wrong) W Hmmm. Coconuts?

  13. +a s = 0 k = 0 Find acceleration and tension 11.0 kg 30.0o 6.0 kg +a Step 2 - Set up the equation for the 6.0 kg mass. T is positive (up), and the weight of the mass is down Weight = (6.0 kg)(9.8 N/kg) = 58.8 N down (-) Tension T is up (+), so we have T - 58.8 N = (6.0 kg)a W T - 58.8 N = (6.0 kg)a

  14. +a s = 0 k = 0 Find acceleration and tension 11.0 kg 30.0o 6.0 kg +a Step 3 - Set up the equation for the 11.0 kg mass. Remember, down the plane is now positive. You have the tension T up (-) the plane, and the parallel component of gravity down (+) the plane: F|| = mgsin() = (11.0 kg)(9.8 N/kg)sin(30.0o) = 53.9 N down (+) the plane, Tension T is up the plane (-), so we have: 53.9 N - T = (11.0 kg)a W 53.9 N - T = (11.0 kg)a

  15. +a s = 0 k = 0 Find acceleration and tension 11.0 kg 30.0o 6.0 kg +a Step 4 - Solve the math for the acceleration: 53.9 N - T = (11.0 kg)a T - 58.8 N = (6.0 kg)a 53.9 N - T = (11.0 kg)a +T - 58.8 N = (6.0 kg)a 53.9 N - T + T - 58.8 N = (17.0 kg)a 53.9 N - 58.8 N = (17.0 kg)a a = -0.2882 m/s/s We guessed wrong!! it accelerates the other way!!! W a = -0.2882 m/s/s

  16. +a s = 0 k = 0 Find acceleration and tension 11.0 kg 30.0o 6.0 kg +a Step 5 - Solve for the tension: 53.9 N - T = (11.0 kg)a T - 58.8 N = (6.0 kg)a a = -0.2882 m/s/s Plug into one of the equations: T - 58.8 N = (6.0 kg)a T = 58.8 N + (6.0 kg)(-0.2882 m/s/s) = 57.1 N W 57.1 N

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