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

3. Connected Masses What forces can you identify acting on the boxes?
Let’s call F the net force on the system

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

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 ( ) 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
Check the calculation using the 3rd mass. T2 – T1 = m2a N – 10 N = 7.5 N m2a = (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. Consider any F contributing to acceleration +
Consider any F contributing to acceleration +. Any F opposing acceleration is - . Apply Newton’s 2nd Law equation for each.
M2.
-T.
m2g
M1.
+T.
Fn.
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
30 N
7 kg
a = 4.3 m/s2

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
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 diagram for each.

25. Boxes in Contact

26. Since F is the only force acting on the two masses, it determines the acceleration of both:
                              
The force F2 acting on the smaller mass may now be determined.

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

28. 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

29. 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

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

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

32. 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

33. Inclined Pulley

34. 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

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

36. Suppose m1 = 5kg, m2 = 12-kg and the ramp angle is 20o
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

37. To practice problems go to: Hyperphysics site
To practice problems go to: Hyperphysics site. Click Mechanics, Newton’s Laws, Standard problems, then the appropriate symbol.