1 / 33

Force Systems accelerate together

Force Systems accelerate together. Combination Systems – connected masses Horizontal Pulley Atwood’s Machine. Connected Masses. F net = m tot a for the system a = F net . m 1 + m 2 + m 3. A constant net force, F, accelerates the entire system’s mass.

asteely
Télécharger la présentation

Force Systems accelerate together

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

  2. Connected Masses

  3. Fnet = mtota for the systema = Fnet. m1 + m2 + m3 . A constant net force, F, accelerates the entire system’s mass.

  4. Each box has separate forces acting on it. Sketch free body diagrams for each mass ignore vertical forces. Assign 1 direction as positive (right). Write the Fnet equation for each, find acceleration.Isolate each masses to find T1 & T2. m1a = T1. m2a = T2 - T1. m3a = F – T2.

  5. 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. a = Fnet. m1 + m2 + m3

  6. Find system acceleration: a = Fnet m1 + m2 + m3. 20 N (4 + 3 + 1) kga = 2.5 m/s2.

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

  8. 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!!

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

  10. Horizontal Pulley.

  11. The masses accl together, the tension is uniform, accl direction is positive.

  12. -T. M2. m2g Sketch free body diagrams for each mass separately. Write Newton’s 2nd Law equation for each. Fn. +T. M1. m1g m1a = T m2a = m2g - T

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

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

  15. 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?

  16. a = 4.2 m/s2T = 16.8 Nmg = 30 N, it is less than the tension.

  17. Ex: Given horizontal pulley system where m1 = 2 kg, m2 = 5 kg, and m, the coefficient of friction between m1 and the counter is 0.35, sketch the free body diagrams for m1 and m2, calculate the acceleration of the system, and find T.

  18. 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

  19. Atwood’s Machine Use wksht.

  20. Given Atwood’s machine, m1 = 2 kg, m2 = 4 kg. Find the acceleration and tension.

  21. Sketch the free body diagramfor each.

  22. Boxes in Contact

  23. Using the previously determined accl, the force F2 acting on the smaller mass is F2 = m2a By Newton’s 3rd Law, F2 acts backward on m1. The net force, F1, on m1 is: m1 F2 F

  24. 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. F2 = m2a a = 1.25m/s2 F2 = 3.75 N

  25. 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

  26. Find a for system. a = 1m/s2.F2 must push the remaining 40 boxes or 40 kg.F = 40 kg (1m/s2.) 40 N.

  27. 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

  28. Inclined Pulley

  29. 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

  30. Derive an equation for the same inclined pulley system including friction.

More Related