1. Physics 121, Sections 9, 10, 11, and 12 Lecture 8
Today’s Topics:
Homework 3: Due Friday Sept. 23 @ 6:00PM
Ch.3: # 64, 75, and 81.
Ch.4: # 4, 8, 21, 25, 36, 40, and 51.
Chapter 4: Motion in 2-D
More examples of FBD’s
Examples with friction 1

2. Consider the following two cases
An Example
Consider the following two cases

3. An Example The Free Body Diagrams mg FB,T= N mg Ball Falls
For Static Situation
N = mg

4. The action/reaction pair forces
An Example
The action/reaction pair forces
FB,E = -mg
FB,T= N
FT,B= -N
FB,E = -mg
FE,B = mg
FE,B = mg

5. Lecture 8, ACT 1 Gravity and Normal Forces
A woman in an elevator is accelerating upwards
The normal force exerted by the elevator on the woman is,
A) greater than
B) the same as
C) less than
the force due to gravity acting on the woman

6. Lecture 8, ACT 1b Gravity and Normal Forces
A woman in an elevator is accelerating upwards
The normal force exerted by the elevator on the woman is,
A) greater than
B) the same as
C) less than
the force the woman exerts on the elevator.

7. The Free Body Diagram Newton’s 2nd Law says that for an object F = ma.
Key phrase here is for an object.
So before we can apply F = ma to any given object we isolate the forces acting on this object:

8. The Free Body Diagram... Consider the following case
What are the forces acting on the plank ?
P = plank
F = floor
W = wall
E = earth
FW,P
FP,W
FP,F
FP,E
FF,P
FE,P

9. The Free Body Diagram... Consider the following case
What are the forces acting on the plank ?
FW,P
FP,W
FP,F
FP,E
FF,P
FE,P
Isolate the plank from
the rest of the world.

10. The Free Body Diagram...
The forces acting on the plank should reveal themselves...
FP,W
FP,F
FP,E

11. Aside... In this example the plank is not moving...
And it is not accelerating!
So FNET = ma becomes FNET = 0
This is the basic idea behind statics.
FP,W
FP,F
FP,E
FP,W + FP,F + FP,E = 0

12. Example Example of a dynamics problem:
A box of mass m = 2kg slides on a horizontal frictionless floor. A force Fx = 10N pushes on it in the i direction. What is the acceleration of the box?
F = Fx i
a = ? m j i

13. Example... Draw a picture showing all of the forces j FB,F F i FB,E
FF,B
FE,B

14. Example... Draw a picture showing all of the forces.
Isolate the forces acting on the block. j
FB,F F i
FB,E = mg
FF,B
FE,B

15. Example... Draw a picture showing all of the forces.
Isolate the forces acting on the block.
Draw a free body diagram. j
FB,F i F mg

16. Example... Draw a picture showing all of the forces.
Isolate the forces acting on the block.
Draw a free body diagram.
Solve Newton’s equations for each component.
FX = maX
FB,F - mg = maY j
FB,F i F mg

17. Example... FX = maX So aX = FX / m = (10 N)/(2 kg) = 5 m/s2.
FB,F - mg = maY
But aY = 0
So FB,F = mg.
The vertical component of the force of the floor on the object (FB,F ) is often called the Normal Force (N).
Since aY = 0 , N = mg in this case. N j FX i mg

18. Example Recap
N = mg FX j
aX = FX / m i mg

19. Tools: Ropes & Strings Can be used to pull from a distance.
Tension (T) at a certain position in a rope is the magnitude of the force acting across a cross-section of the rope at that position.
The force you would feel if you cut the rope and grabbed the ends.
An action-reaction pair. T
cut T T

20. Tools: Ropes & Strings...
The direction of the force provided by a rope is along the direction of the rope: T
Since ay = 0 (box not moving), m
T = mg mg

21. Lecture 8, Act 2: Force and acceleration
A fish is being yanked upward out of the water using a fishing line that breaks when the tension reaches 180 N. The string snaps when the acceleration of the fish is observed to be is 12.2 m/s2. What is the mass of the fish?
snap !
(a) kg
(b) kg
(c) kg
a = 12.2 m/s2
m = ?

22. Tools: Pegs & Pulleys Used to change the direction of forces.
An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: F1
ideal peg
or pulley
| F1 | = | F2 | F2

23. Tools: Pegs & Pulleys Used to change the direction of forces.
An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude:
FW,S = mg mg T m
T = mg

24. Lecture 8, Act 3: Force and acceleration
A block weighing 4 lbs is hung from a rope attached to a scale. The scale is then attached to a wall and reads 4 lbs. What will the scale read when it is instead attached to another block weighing 4 lbs? ? m m m
(1)
(2)
(a) 0 lbs (b) 4 lbs (c) 8 lbs.

25. < Example with pulley F
A mass M is held in place by a force F. Find the tension in each segment of the rope and the magnitude of F.
Assume the pulleys massless and frictionless.
Assume the rope massless.
< M T5 T4 T3 T2 T1 F
We use the 5 step method.
Draw a picture: what are we looking for ?
What physics idea are applicable ? Draw a diagram and list known and unknown variables.
Newton’s 2nd law : F=ma
Free-body diagram for each object

26. < Pulleys: continued F FBD for all objects T4 F=T1 T2 T3 M T5 T4 T3Mg

27. Pulleys: finally
Step 3: Plan the solution (what are the relevant equations)
F=ma , static (no acceleration: mass is held in place) M T5 Mg T4
F=T1 T2 T3
T5=Mg
T1+T2+T3=T4
F=T1 T2 T3 T5
T2+T3=T5

28. Pulleys: really finally!
Step 4: execute the plan (solve in terms of variables)
We have (from FBD):
T5=Mg
F=T1
T2+T3=T5
T1+T2+T3=T4
Pulleys are massless and frictionless
< M T5 T4 T3 T2 T1 F
T2=T3
T1=T3
T2+T3=T5 gives T5=2T2=Mg
T2=Mg/2
F=T1=Mg/2
T1=T2=T3=Mg/2 and T4=3Mg/2
T5=Mg
and
Step 5: evaluate the answer (here, dimensions are OK and no numerical values)

29. Exercise: Inclined plane
A block of mass m slides down a frictionless ramp that makes angle  with respect to horizontal. What is its acceleration a ? m a 

30. Inclined plane...
Define convenient axes parallel and perpendicular to plane:
Acceleration a is in x direction only. i j m a 

31. Inclined plane... Consider x and y components separately:
i: mg sin  = ma a = g sin 
j: N - mg cos  = N = mg cos  ma i j
mg sin 
mg cos  N mg 

32. Angles of an Inclined plane
ma = mg sin  mg N 
 + f = 90 f 

33. New Topic: Friction What does it do? It opposes motion!
How do we characterize this in terms we have learned?
Friction results in a force in a direction opposite to the direction of motion! ma
FAPPLIED
fFRICTION mg N i j

34. Friction... Force of friction acts to oppose motion:
Parallel to surface.
Perpendicular to Normal force. j N F i ma fF mg

35. Model for Sliding Friction
The direction of the frictional force vector is perpendicular to the normal force vector N.
The magnitude of the frictional force vector |fF| is proportional to the magnitude of the normal force |N |.
|fF| = K | N | ( = K|mg | in the previous example)
The “heavier” something is, the greater the friction will be...makes sense!
The constant K is called the “coefficient of kinetic friction”.

36. Model... Dynamics: i : F KN = m a j : N = mg so F Kmg = m a j N F

37. Lecture 8, ACT 4 Friction and Motion
A box of mass m1 = 1 kg is being pulled by a horizontal string having tension T = 40 N. It slides with friction (mk= .5) on top of a second box having mass m2 = 2 kg, which in turn slides on an ice rink (frictionless).
What is the acceleration of the second box ?
(a) a = 0 m/s2 (b) a = 2.5 m/s2 (c) a = 10 m/s2
slides with friction (mk=0.5 ) T m1
a = ? m2
slides without friction

38. Inclined Plane with Friction:
Draw free-body diagram: ma
KN j N  mg  i
next

39. Inclined plane... Consider i and j components of FTOT = ma :
i mg sin KN = ma  i j mg N
KN ma
mg sin 
mg cos 
j N = mg cos 
mg sin Kmg cos  = ma
a / g = sin Kcos 
next

40. Example Pulleys and Friction
Three blocks are connected on the table as shown. The table has a coefficient of kinetic friction of 0.350, the masses are m1 = 4.00 kg, m2 = 1.00kg and m3 = 2.00kg. m1 T1 m2 m3
What is the magnitude and direction of acceleration on the three blocks ?
What is the tension on the two cords ?

41. T12
T23 m1 T1 m2 m3
T12
T23
N=-m2g
m1g
T12
m3g
T23
T12
T23 a m2 m1 m3 a
mk m2g a
m2g
T12 - m1g = - m1a
T23 - m3g = m3a
-T12 + T23 + mk m2g = - m2a
SOLUTION: T12 = = 30.0 N , T23 = 24.2 N , a = 2.31 m/s2 left for m2

42. Static Friction...
So far we have considered friction acting when something moves.
We also know that it acts in un-moving “static” systems:
In these cases, the force provided by friction will depend on the forces applied on the system. j N F i fS mg

43. Static Friction... Just like in the sliding case except a = 0.
i : F fS = 0
j : N = mg
While the block is static: fS F (unlike kinetic friction) j N F i fS mg

44. Static Friction...
The maximum possible force that the friction between two objects can provide is fMAX = SN, where s is the “coefficient of static friction”.
So fS  S N.
As one increases F, fS gets bigger until fS = SN and the object “breaks loose” and starts to move. j N F i fS mg

45. Static Friction...
S is discovered by increasing F until the block starts to slide:
i : FMAX SN = 0
j : N = mg
S FMAX / mg j N
FMAX i
Smg mg

46. Additional comments on Friction:
Since f = N , the force of friction does not depend on the area of the surfaces in contact.
By definition, it must be true that S > K for any system (think about it...).
Graph of Frictional force
vs Applied force:
fF = SN
fF = KN fF
Active Figure
fF = FA FA

47. Lecture 8, ACT 5: Forces and Motion
A box of mass m =10.21 kg is at rest on a floor. The static coefficient of friction between the floor and the box is ms = 0.4
A rope is attached to the box and pulled at an angle of q = 30o above horizontal with tension T = 40 N.
Does the box move ?
(a) yes (b) no (c) too close to call T q m
static friction (ms = 0.4 )
next

48. Static Friction: We can also consider S on an inclined plane.
In this case, the force provided by friction will depend on the angle  of the plane. 
next

49. (Newton’s 2nd Law along x-axis)
Static Friction
The force provided by friction, fF , depends on . fF
ma = 0 (block is not moving)
mg sin ff i j N 
(Newton’s 2nd Law along x-axis) mg 
next

50. Static Friction
We can find s by increasing the ramp angle until the block slides:
mg sin ff
In this case:
ff SN  Smg cos M
SN i j
mg sin MSmg cos M N
M mg 
Stan M
next

51. Problem: Box on Truck
A box with mass m sits in the back of a truck. The static coefficient of friction between box and truck is S
What is the maximum acceleration a that the truck can have without the box slipping? S m a
next

52. Problem: Box on Truck Draw Free Body Diagram for box:
Consider case where fF is max... (i.e. if the acceleration were any larger, the box would slip). N j i
fF = SN mg
next

53. Problem: Box on Truck Use FTOT = ma for both i and j components
i SN = maMAX
j N = mg
aMAX = S g N j
aMAX i
fF = SN mg
next

54. Lecture 8, ACT 6 Forces and Motion
An inclined plane is accelerating with constant acceleration a. A box resting on the plane is held in place by static friction. What is the direction of the static frictional force? S a
(a) Ff Ff
(b) Ff
(c)
next

55. Problem: Putting on the brakes
Anti-lock brakes work by making sure the wheels roll without slipping. This maximizes the frictional force slowing the car since S > K .
The driver of a car moving with speed vo slams on the brakes. The static coefficient of friction between wheels and the road is S . What is the stopping distance D? ab vo
v = 0 D
next

56. Problem: Putting on the brakes
Use FTOT = ma for both i and j components
i SN = ma
j N = mg
a = S g N j a i
fF = SN mg
next

57. Problem: Putting on the brakes
As in the last example, find ab = Sg.
Using the kinematic equation: v2 - v02 = 2a( x -x0 )
In our problem: v02 =  2ab( D ) vo ab
v = 0 D
next

58. Problem: Putting on the brakes
In our problem: v02 =  2ab( D )
Solving for D:
Putting in ab = Sg ab vo
v = 0 D
end

59. Recap of today’s lecture
Homework 3: Due Friday Sept. 23 @ 6:00PM
Ch.3: # 64, 75, and 81.
Ch.4: # 4, 8, 21, 25, 36, and 40.
Chapter 4: Motion in 2-D
More examples of FBD’s
Examples with friction 27