2. Pulleys Example Whiteboards - Atwood’s Machine Whiteboards - Inclined Plane M1M1 M2M2

3. Example - Find Tension, and acceleration if the coefficient of friction between the block and the plane is 0.100 32.0 kg 5.0 kg 1. Calculate obvious things like weights, friction, F || etc. Label the forces 2. Decide the direction of acceleration and + 3. Set up F = ma for each mass 4. Do Maths 0.477 m/s/s, 46.7 N

4. Whiteboards: Atwood’s Machine 11 | 2 | 3 | 4 | 52345

5. Find acceleration and tension 2.45 m/s/s, 36.8 N 5.0 kg 3.0 kg Massless frictionless pulley

6. Find acceleration and tension sometimes 5.0 kg 3.0 kg Massless frictionless pulley Step 1 - Guess the direction of acceleration - This becomes the positive direction for each mass. Uhh well um. the 5.0 kg is heavier. +a

7. Find acceleration and tension T - 29.43 N = (3.0 kg)a 5.0 kg 3.0 kg Massless frictionless pulley Step 2 - Set up the =ma for the 3.0 kg mass: T is up, and the weight is down. Down is - and up is + weight = (3.0 kg)(9.81 N/kg) = 29.43 N down (- in this case) T is up, -29.43 is down: T - 29.43 N = (3.0 kg)a +a

8. Find acceleration and tension 49.05 N - T = (5.0 kg)a 5.0 kg 3.0 kg Massless frictionless pulley Step 3 - Set up the =ma for the 5.0 kg mass: T is up, and the weight is down, but now down is + and up is - weight = (5.0 kg)(9.81 N/kg) = 49.05 N down (+ in this case) T is up (-), 49.05 N is down (+): 49.05 N - T = (5.0 kg)a +a

9. Find acceleration and tension 2.45 m/s/s 5.0 kg 3.0 kg Massless frictionless pulley Step 4 - Solve for acceleration: 49.05 N - T = (5.0 kg)a T - 29.43 N = (3.0 kg)a 49.05 N - T = (5.0 kg)a +T - 29.43 N = (3.0 kg)a T - 29.43 N + 49.05 N - T = (8.0 kg)a a = 2.4525 m/s/s +a

10. Find acceleration and tension 37 N 5.0 kg 3.0 kg Massless frictionless pulley Step 5- Solve for T: 49.05 N - T = (5.0 kg)a T - 29.43 N = (3.0 kg)a a = 2.4525 m/s/s Pick one of the formulas, and plug the acceleration in: T - 29.43 N = (3.0 kg)a T = (3.0 kg)(2.4525 m/s/s) + 29.43 N = 36.7875 N ≈ 37 N +a

11. Whiteboards: Pulleys on Inclined Planes 11 | 2 | 3 | 4 | 52345

12. Find acceleration and tension 0.289 m/s/s, 57.1 N 6.0 kg 11.0 kg 30.0 o

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

14. Find acceleration and tension T - 58.86 N = (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 6.0 kg 11.0 kg 30.0 o +a Weight = (6.0 kg)(9.81 N/kg) = 58.86 N down (-) Tension T is up (+), so we have T - 58.86 N = (6.0 kg)a

15. Find acceleration and tension 53.9550 N - T = (11.0 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: 6.0 kg 11.0 kg 30.0 o +a F || = mgsin(  ) = (11.0 kg)(9.81 N/kg)sin(30.0 o ) = 53.9550 N down (+) the plane, Tension T is up the plane (-), so we have: 53.9550 N - T = (11.0 kg)a

16. Find acceleration and tension a = -0.29 m/s/s Step 4 - Solve the math for the acceleration: 53.9550 N - T = (11.0 kg)a T - 58.86 N = (6.0 kg)a 6.0 kg 11.0 kg 30.0 o +a 53.9550 N - T = (11.0 kg)a +T - 58.86 N = (6.0 kg)a 53.9550 N - T + T - 58.86 N = (17.0 kg)a 53.9550 N - 58.86 N = (17.0 kg)a a = -0.288529412 m/s/s We guessed wrong!! it accelerates the other way!!!

17. Find acceleration and tension 57 N Step 5 - Solve for the tension: 53.9550 N - T = (11.0 kg)a T - 58.86 N = (6.0 kg)a a = -0.288529412 m/s/s 6.0 kg 11.0 kg 30.0 o +a Plug into one of the equations: T - 58.86 N = (6.0 kg)a T = 58.86 N + (6.0 kg)(-0.288529412 m/s/s) = 57.12882 N