1. Physics 215 – Fall 2014Lecture 05-21 Welcome back to Physics 215 Today’s agenda: More on free-body diagrams and force components Applying Newton’s laws

2. Physics 215 – Fall 2014Lecture 05-22 Current homework assignment HW4: –Knight textbook -- Ch.5: 40, 42, 44, 56 –Ch.6: 26, 28 –Due Friday, Sept. 26 th in recitation

3. Physics 215 – Fall 2014Lecture 05-23 First Law:In the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with constant velocity. Second Law:F net = ∑ F on object = m a Third Law:F AB = - F BA (“action = reaction”) [regardless of type of force and of motion of objects in question] Newton’s Laws

4. Physics 215 – Fall 2014Lecture 05-24 1.the net force on the person must be zero 2.the two forces form a Newton’s third-law pair 3.neither of the above explanations 4.both of the above explanations Consider a person sitting on a chair. We can conclude that the downward weight force on the person (by the Earth) and the upward normal force on the person (by the chair) are equal and in opposite direction, because

5. Physics 215 – Fall 2014Lecture 05-25 Free-body diagrams: Person sitting on chair ChairPersonEarth (incomplete)

6. Physics 215 – Fall 2014Lecture 05-26 1.Cart A doesn’t move and Cart B moves backwards 2.Cart B doesn’t move and Cart A moves backwards 3.Both carts move in opposite directions 4.Neither cart moves There are two people facing each other, each on a separate cart. If person A pushes on person B, while person B does nothing, what will be the resulting motion of the carts?

7. Physics 215 – Fall 2014Lecture 05-27 A force P is applied by a hand to two blocks which are in contact on a frictionless, horizontal table as shown in the figure. The blocks accelerate to the right. Block A has a smaller mass than block B. (a) Draw free body diagrams for each block. (b) Which block experiences the larger net force? (c) Suppose that initially the mass of block A were half that of block B. If in a subsequent experiment the mass of block A were doubled, by what factor would the acceleration change assuming the pushing force remained constant? A B

8. Physics 215 – Fall 2014Lecture 05-28 1.zero 2.straight down 3.down and to the left (along the incline) 4.not zero, but neither of 2 or 3 A block is held in place on a friction- less incline by a massless string, as shown. The acceleration of the block is

9. Physics 215 – Fall 2014Lecture 05-29 1.a normal force given by vector A. 2.a normal force given by vector B. 3.the force given by vector A, but it’s not a normal force. 4.the force given by vector B, but it’s not a normal force. A block is held in place on a frictionless incline by a massless string, as shown. The force on the block by the incline is

10. Physics 215 – Fall 2014Lecture 05-210 Normal force Always perpendicular to surface of contact Generic name given to contact force between 2 objects

11. Physics 215 – Fall 2014Lecture 05-211 Other forces Besides normal force what other forces are present for block on inclined plane?

12. Physics 215 – Fall 2014Lecture 05-212 Free-body diagram: Block on frictionless incline Show all forces exerted on the block. Do not show forces exerted by the block on anything else.

13. Physics 215 – Fall 2014Lecture 05-213 Geometry…   N W T vertical component of N BP is N BP cos  horizontal is N BP sin(  ) x y

14. Physics 215 – Fall 2014Lecture 05-214 Force components: Block on frictionless incline F = 0 implies all components of F are zero! Horizontal and vertical

15. Physics 215 – Fall 2014Lecture 05-215 When the block is held in place as shown, is the magnitude of the normal force exerted on the block by the incline 1.greater than the magnitude of the weight force (on the block by the Earth), 2.equal to the magnitude of the weight, or 3.less than the magnitude of the weight. 4.Can’t tell.

16. Physics 215 – Fall 2014Lecture 05-216 Solving system Vertical equilibrium: N BP cos(  ) + W BE = 0 Horizontal equilibrium: N BP sin(  ) + T BR = 0 2 equations  solve for N BP and T BE in terms of W BE and 

17. Physics 215 – Fall 2014Lecture 05-217 What have we learned? If at rest – net force = 0, since a = 0 ! Write down FBD – identify all forces present Take force components in 2 (in 2D) directions to find unknowns Don’t plug in numbers until end

18. Physics 215 – Fall 2014Lecture 05-218 The string that holds the block breaks, so there is no more tension force exerted on the block. Will the magnitudes of either of the other two forces on the block (i.e., the weight and the normal force) change? 1.Both weight and normal force will change. 2.Only the weight force will change. 3.Only the normal force will change. 4.Neither one of the two forces will change.

19. Physics 215 – Fall 2014Lecture 05-219 Discussion Say normal does not change. What will be the new net force? Same as old, but “minus” the tension, i.e. straight to the left (and not zero). So what would the block do? Accelerate. OK -- Straight to the left? NO, can’t be correct answer. Acceleration is down the incline, so normal gets smaller.

20. Physics 215 – Fall 2014Lecture 05-220 Geometry (2)   N W component of W BE at 90 0 to plane is W BE cos  component along plane is W BE sin(  ) 

21. Physics 215 – Fall 2014Lecture 05-221 Force components: Block on frictionless incline The string that holds the block breaks. What does happen? Take components now along and perpendicular to incline

22. Physics 215 – Fall 2014Lecture 05-222 Solving for second case Since acceleration is down incline, component of net force at 90 degrees to slope is zero. N BP + W BE cos(  ) = 0 Using second law applied to components along incline: a = W BE sin(  )/m block

23. Physics 215 – Fall 2014Lecture 05-223 Conclusion Normal forces can change when small changes are made to the situation. Can choose any 2 directions to find force components, but it pays to pick ones that simplify equations

24. Physics 215 – Fall 2014Lecture 05-224 Suppose we chose to resolve forces vertically and horizontally? Vertically: (1)Ncos(  ) + W = -ma sin(  ) Horizontally: (2)Nsin(  ) = ma cos(  ) Multiply (1) by cos(  ), (2) by sin(  ) and add:  N + Wcos(  ) = 0 as before!!

25. Physics 215 – Fall 2014Lecture 05-225 If the block lies 0.5 m up such a plane inclined at an angle of 30  how long will it take for the block to reach the ground? (take g = 10 m/s 2 )

26. Physics 215 – Fall 2014Lecture 05-226 “real” inclined plane For a real incline, we expect for small angles that the block still does not move when string breaks – why? Friction – force up slope that opposes motion – more later…

27. Physics 215 – Fall 2014Lecture 05-227 Summary To solve problems in mechanics, identify all forces and draw free body diagrams for all objects If more than one object, use Newton’s Third law to reduce number of independent forces Use Newton’s Second law for all components of net force on each object Choose component directions to simplify equations

28. Physics 215 – Fall 2014Lecture 05-228 Weight, mass, and acceleration What does a weighing scale ``weigh’’? Does it depend on your frame of reference? Consider elevators….

29. Physics 215 – Fall 2014Lecture 05-229 A person is standing on a bathroom scale while riding an express elevator in a tall office building. When the elevator is at rest, the scale reads about 160 lbs. While the elevator is moving, the reading is frequently changing, with values ranging anywhere from about 120 lbs to about 200 lbs. At a moment when the scale shows the maximum reading (i.e., 200 lbs) the elevator 1.must be going up 2.must be going down 3.could be going up or going down 4.I’m not sure.

30. Physics 215 – Fall 2014Lecture 05-230 Motion of elevator (if a < 0 ) Moving upward and slowing down, OR Moving downward and speeding up. Motion of elevator (if a > 0 ) Moving upward and speeding up, OR Moving downward and slowing down.

31. Physics 215 – Fall 2014Lecture 05-231 A person is standing on a bathroom scale while riding an express elevator in a tall office building. When the elevator is at rest, the scale reads about 160 lbs. While the elevator is accelerating, a different reading is observed, with values ranging anywhere from about 120 lbs to about 200 lbs. At a moment when the scale shows the maximum reading (i.e., 200 lbs) the acceleration of the elevator is approximately 1.1 m/s 2 2.2.5 m/s 2 3.5 m/s 2 4.12.5 m/s 2

32. Physics 215 – Fall 2014Lecture 05-232 Conclusions Scale reads magnitude of normal force |N PS | Reading on scale does not depend on velocity (principle of relativity again!) Depends on acceleration only * a > 0  normal force bigger * a < 0  normal force smaller

33. Physics 215 – Fall 2014Lecture 05-233 Reminder of free-fall experiment Objects fall even when there is no atmosphere (i.e., weight force is not due to air pressure). When there is no “air drag” things fall “equally fast.” i.e. same acceleration From Newton’s 2 nd law, a = W/m is independent of m -- means W = mg The weight (i.e., the force that makes objects in free fall accelerate downward) is proportional to mass.

34. Physics 215 – Fall 2014Lecture 05-234 Inertial and gravitational mass Newton’s second law: F = m I a For an object in “free fall” W = m G g If a independent of m I, must have m I = m G  Principle of equivalence

35. Physics 215 – Fall 2014Lecture 05-235 Reading assignment Forces, Newton’s Laws of Motion, Weight & Friction Ch. 5-7 in textbook