1. Lecture 8: Forces & The Laws of Motion

2. Questions of Yesterday 1) You must apply a force F 1 to begin pushing a crate from rest across the floor, you must apply a force F 2 to keep the crate moving at a constant velocity once its in motion. Which statement is true? a) F 1 = F 2 b) F 1 > F 2 c) F 1 < F 2 2) When do action and reaction pairs of forces not cancel one another out? a) when one of the objects is accelerating b) when both objects are accelerating c) never

3. Newton’s Second Law The acceleration a of an object is directly proportional to the net force acting on it and inversely proportional to its mass a =  F/m  F = ma SI Unit of force = Newton 1 N = 1 kg*m/s 2 F and a are vectors! Vector equation!  F x = ma x  F y = ma y  F z = ma z

4. Free Body Diagrams FgFg N N’N’ Diagram of a single object with all the forces acting ON it DO NOT INCLUDE: -Forces that the object exerts on other objects or the Earth -Other objects that the object is in contact with Fg’Fg’

5. Free Body Diagrams Tennis ball undergoing projectile motion FgFg

6. Free Body Diagrams Tennis ball while a tennis player is serving it. The player swings his racket, hitting the 60 g tennis ball with a force 60 N at an angle 30 o with the horizontal. FgFg What is the magnitude and direction of the net force on the ball? What is the resultant acceleration? FAFA 30 o

7. Free Body Diagrams A block sliding down a frictionless inclined plane. FgFg In what direction?   N Is the block accelerating?

8. Free Body Diagrams A block sliding down a frictionless inclined plane. FgFg N   + y + x - x - y Choose coordinate system to be in direction of incline plane! F gy F gx F gx = F g sin  F gy = F g cos 

9. Free Body Diagrams What is the acceleration of the block? FgFg N 30 o  + y + x - x - y F gy F gx F gx = F g sin  F gy = F g cos  3 kg  F x = ma x  F y = ma y  F y = ma y = 0 N - F gy = 0  F x = ma x ma x = F gx

10. FBDs and using Newton’s Laws How much do you weigh in a moving elevator? A 60-kg person stands on a scale in an elevator Draw a free body diagram for the person What does the scale read when the elevator is: a) at rest? b) Ascending with a speed of 2.0 m/s? c) Descending with a speed of 2.0 m/s? d) Ascending with an acceleration of 2.0 m/s 2 ? e) Ascending with an acceleration of 2.0 m/s 2 ?

11. Practice Problem A train has a mass of 5.0*10 6 kg and is moving at 90 km/h. The engineer applies the brakes, resulting in a net backward force of 2.0*10 6 N on the train. The brakes are held on for 30.0 s. a) What is the final speed of the train? b) How far does the train travel in this period?

12. Tension Free Body Diagrams and Newton’s 2nd law only use forces acting directly on an object Draw a free body diagram for the block What about the rope? FAFA a

13. Tension Tension at one end of the rope is equal to the other Tension is CONSTANT all along the rope Tension = force exerted along a rope FAFA N FgFg T T’T’ 2nd law: ma = T’ + F A = 0 Applied force is acts “directly” on block through the rope 3rd law: T = -T’

14. Tension: Practice Problem Find the acceleration of each object and the tension in the rope 5 kg 10 kg

15. Objects in Equilibrium When is ∑F = 0 ? Conditions for equilibrium: ∑F x = 0 ∑F y = 0 When is a = 0 ? An object at rest or moving with a constant velocity is in EQUILIBRIUM

16. Objects in Equilibrium Conditions for equilibrium: ∑F x = 0 ∑F y = 0 Find the tension in the two wires that support the light fixture 40 o

17. Questions of the Day You must apply a force F to push your physics book across your desk at a constant velocity. 1a) The net force acting on the book is… a) F b) between 0 and F c) greater than F d) 0 1b) Are other forces acting on the book in the horizontal direction? a) YES b) NO c) not enough information to know 3) A large crate is at rest in the bed of a truck. As the truck accelerates the crate remains at rest relative to the truck. In what direction is the net force on the crate? a) the same direction as the truck’s acceleration b) opposite the direction of the truck’s acceleration c) the net force is zero