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Force Systems

Force Systems. Combination Systems – connected masses Horizontal Pulley Atwood’s Machine. For any force system you must sum forces. F net = S F = F 1 + F 2 … ma = F 1 + F 2 …. Connected Masses. F net = m tot a. F net must accelerate the entire system’s mass together.

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Force Systems

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  1. Force Systems • Combination Systems – connected masses • Horizontal Pulley • Atwood’s Machine

  2. For any force system you must sum forces. Fnet = SF = F1 + F2 … ma = F1 + F2 …

  3. Connected Masses

  4. Fnet = mtota Fnet must accelerate the entire system’s mass together.

  5. Given the masses and Fnet: sketch free body diagrams for each mass ignore vertical forces. Assign 1 direction as positive (right). Write the Fnet equation for each, find acceleration.then isolate each masses to find T1 & T2. m1a = T1. m2a = T2 - T1. m3a = F – T2.

  6. Add the equations. Factor & make cancellations. m1a = T1. m2a = T2 - T1. m3a = F – T2. m1a + m2a + m3a = T1 + T2 – T1 + F – T2. a (m1 + m2 + m3) = F.

  7. Equation to solve problems in connected masses. a = Fnet. m1 + m2 + m3

  8. Ex 1: Connected Masses: Given a Fnet of 20N and masses of 4, 3, and 1 kg, find the acceleration of the system and the tension in each cord. 20 N (4 + 3 + 1) kg a = 2.5 m/s2.

  9. Use the free body diagram & known acceleration to find the tension in each cord. 4 kg m1a = T1 = (4 kg)(2.5 m/s2) = 10 N. m3a = F -T2 Fnet - m3a = T2. 20N - (1 kg)(2.5 m/s2) = 17.5 N 1 kg

  10. Check the calculation using the 3rd mass.T2 – T1 = m2a 17.5 N – 10 N = 7.5 Nm2a = (3 kg)(2.5 m/s2) = 7.5 N.It is correct!!

  11. m1a = T1. m2a = T2 - T1. • T1 = 10 N • T2 = 17.5 N

  12. Horizontal Pulley.

  13. The masses accl together, the tension is uniform, accl direction is positive. Sketch the free body diagram each mass.

  14. -T. M2. m2g Consider any F contributing to acceleration +. Any F opposing acceleration is - .Apply Newton’s 2nd Law equation for each. Fn. +T. M1. m1g m1a = T m2a = m2g - T

  15. Add the equations: m1a + m2a = T + m2g – T T cancels. m1a + m2a = m2g Factor a & solve

  16. a = m2g m1 + m2 Solve for a, and use the acceleration to solve for the tension pulling one of the masses. m1a = T

  17. Ex 2: Horizontal Pulley: Given a mass of 4 kg on a horizontal frictionless surface attached to a mass of 3 kg hanging vertically, calculate the acceleration, and the tension in the cord.Compare the tension to the weight of the hanging mass, are they the same?

  18. a = m2g m1 + m2 a = 4.3 m/s2 30 N 7 kg

  19. m1a = T • (4 kg)(4.3 m/s2) • T = 17 N • mg = 30 N, tension less than weight.

  20. Atwood’s Machine Use wksht.

  21. Sketch the free body

  22. The equation • m1a = T – m1g • m2a = m2g – T • a (m1 +m2) = m2g - m1g • a = m2g - m1g • m tot

  23. Given Atwood’s machine, m1 = 2 kg, m2 = 4 kg. Find the acceleration and tension. • a = 3.3 m/s2. • T = 26 N

  24. Sketch the free body diagramfor each.

  25. Boxes in Contact

  26. Using the previously determined accl, the force F2 acting on the smaller mass is F2 = m2a

  27. By Newton’s 3rd Law, F2 acts backward on m1.The force on m1 is: The net force, F1, on m1 is: m1 F2 F

  28. Given a force of 10N applied to 2 masses, m1 =5 kg and m2 =3kg, find the accl and find F2 (the contact force) between the boxes. a = 1.25m/s2 F2 = 3.75 N

  29. Given a force of 100 N on 100 1 kg boxes, what is the force between the 60th and 61st box. 100-N 1-kg

  30. Find a for system.F2 must push the remaining 40 boxes or 40 kg. 40 N.

  31. Ignoring friction, derive an equation to solve for a and T for this system: Begin by sketching the free body diagram Write the equations for each box Add them. Solve for accl

  32. Inclined Pulley

  33. Given a 30o angle, and 2 masses each 5-kg,find the acceleration of the system, and the tension in the cord. a= 2.45m/s2. T =36.75 N

  34. 1. Derive an equation for the same inclined pulley system including friction between m1 and the ramp.

  35. Suppose m1 = 5kg, m2 = 12-kg and the ramp angle is 20o. Find the m that would make the system acceleration 4 m/s2.What will be the tension in the cord? • The m = 0.7 • T = 69 N

  36. To practice problems go to:Hyperphysics site.Click Mechanics, Newton’s Laws, Standard problems, then the appropriate symbol. • http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html#mechcon

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