1. Tension in a rope has two properties: * it is evenly distributed throughout the connected objects * its always directed away from any object in the system (you can’t push a rope) 50 N 11 lbs 50 N

2. Frictionless, low mass pulleys change only the direction of a tension, not its magnitude. Tension still is evenly distributed throughout the rope as it goes over the pulley.

3. Reaction forces

4. The acceleration’s direction is NOT always the same as the direction of it’s velocity. The acceleration is in the direction of the net force on a body.  F = ma = W= mg

5. Even in projectile motion, the acceleration follows the force, NOT the direction of motion, The accelereation is constantly 9.8 m/s 2 downward as well, because the only force acting is the weight (downward)

6. Equilibrium  F = ma  F x = ma x  F y = ma y If the 3 forces were connected head to tail, they would form ………..

7. The Fool-Proof Newtonian Force Method for Solving Problems a)Circle the 1 body you are going to analyze b)Reduce it to a point and draw a FREE BODY DIAGRAM (show each force as an arrow and label it) c)Do  F=ma in one direction at a time (don’t mix x and y in the same equation) d)Repeat as often as necessary to get an equation with the one variable for which you are looking. Try other directions and other bodies. e)Solve using algebra. Harder problems may involve solving simultaneous equations

8. Rope Tension: Hanging Basket

9. A Block & Tackle

10. Equilibrium  F =0 Forms Closed Triangle

11. Find how hard the elephant must pull to keep the ring in equilibrium. 90  135  At this angle equilibrium is impossible to achieve!!!! See p. 91. The Equilibrant must be 500N at 36.9 degrees west of South

12. T1 T2 T3 If each block is 1 kg, find the tension in each rope.

13. 800 kg 1 kg 1000 kg 1kg F No friction If the whole system accelerates at 1 m/s 2, find the force on each object. What would happen if d and b were both eggs that broke under 2N of force? d c b a

14. ICE no friction M m Assume a perfect pulley: massless and no friction. Why must M accelerate no matter how small m is? Why must m fall at less than 9/8 m/s 2 ? On what factors will the acceleration of the system depend? Derive a formula for the tension in the rope that depends only on m, M and g. On what factors will the tension in the rope depend?

15. ICE no friction M m +T +N -W  F x = Ma x -T +W  F y = ma y W - T = ma y mg - T = ma y T = Ma x These Ts must be opposite in sign! a = (mg- T)/m T = M (mg- T)/m FOIL and solve for T

16. T = M (mg- T)/m T = (Mmg –MT)/m T = Mg –MT/m T + MT/m = Mg T(1 +M/m) = Mg T = Mg/(1+M/m) T = g Mm FOIL distribute m into each term above Factor out T Divide both sides by ( ) Clean up by multiplying by m/m (M+m)

17. Wood  = 0.1 M m Assume a perfect pulley Now let’s add friction. Derive a formula for the acceleration that depends only on , m, M and g. -f +T -N +W  F x = Ma x T - f = Ma x T - uN = Ma x T - uMg = Ma x -T +W  F y = ma y W - T = ma y mg - T = ma y

18. T - uMg = Ma x mg - T = ma y mg - ma y = T mg -ma - uMg = Ma mg - uMg = Ma + ma factoring g(m – uM) = (M + m) a factoring a = g(m – uM) / (M + m) Check dimensional consistency Check motion if M = 0, a = g Wood  = 0.1 M m

19. M m Assume a perfect pulley Now let’s add friction. Derive a formula for the acceleration that depends only on , m, M and g.

20. Wood  = 0.1 M m Assume a perfect pulley Using the opposite sign convention +f -T +N -W  F x = Ma x f – T = Ma x uN – T = Ma x uMg - T = Ma x +T -W  F y = ma y T - W = ma y T - mg = ma y

21. uMg – T = Ma x T – mg = ma y T = ma + mg uMg – (ma + mg) = Ma uMg - ma - mg = Ma factoring g( uM - m) = (M + m) a factoring a = g(uM - m) / (M + m) Check dimensional consistency Check motion if m = 0, a = g Wood  = 0.1 M m uMg - mg = Ma + ma

22. Analysis of results a = g(m – uM) / (M + m) This will give + acceleration (down and to the right) if m > uM, and a = 0 if m = uM. The function is undefined if m < uM since negative a makes no physical sense; it will never fall “up” and to the left. The reality is a will stay at zero if a≤ 0.

23. Two dimensional forces must be summed up in the x direction alone, the y direction alone, then finally connected head to tail and added with the pythagorean theorem

24. 7 kg 2N 3N 8N Find the direction and magnitude of the acceleration of the box.

25. Equilibrium  F = ma  F x = ma x  F y = ma y If the 3 forces were connected head to tail, they would form ………..

26. Equilibrium  F =0 Forms Closed Triangle

27. Real World  Vector Diagram

28. Vector Diagram  Resultant Equilibrium  F =0 Forms Closed Triangle

29. Inclined Plane