1. Lecture 7
Goals:
Solve 1D and 2D problems with forces in equilibrium and non-equilibrium (i.e., acceleration) using Newton’ 1st and 2nd laws.
Differentiate between Newton’s 1st, 2nd and 3rd Laws
Use Newton’s 3rd Law in problem solving
Assignment: HW4, (Chapters 6 & 7, due 10/1, Wednesday)
Finish Chapter 7
1st Exam Thursday, Oct. 2nd from 7:15-8:45 PM Chapters 1-7 1

2. No Net Force, No acceleration…a demo exercise
In this demonstration we have a ball tied to a string undergoing horizontal UCM (i.e. the ball has only radial acceleration)
1 Assuming you are looking from above, draw the orbit with the tangential velocity and the radial acceleration vectors sketched out.
2 Suddenly the string brakes.
3 Now sketch the trajectory with the velocity and acceleration vectors drawn again.

3. Friction revisited: Static friction
Static equilibrium: A block with a horizontal force F applied,
As F increases so does fs
S Fx = 0 = -F + fs  fs = F
S Fy = 0 = - N + mg  N = mg F m1
FBD fs N mg

4. Static friction, at maximum (just before slipping)
Still equilibrium: A block, mass m, with a horizontal force F applied,
Direction: A force vector  to the normal force vector N and the vector is opposite to the velocity.
Magnitude: fS is proportional to the magnitude of N
fs = ms N
ms called the “coefficient of static friction” N F m fs mg

5. Kinetic or Sliding friction (fk < fs)
Dynamic equilibrium, moving but acceleration is still zero
As F increases fk remains nearly constant
(but now there acceleration is acceleration) m1
FBD fk N mg
S Fx = 0 = -F + fk  fk = F
S Fy = 0 = - N + mg  N = mg v F
fk = mk N

6. Case study ... big F Dynamics: x-axis i : max = F  KN
y-axis j : may = 0 = N – mg or N = mg
so F Kmg = m ax fk v i j N F
max fk
K mg mg

7. Case study ... little F Dynamics: x-axis i : max = F  KN
y-axis j : may = 0 = N – mg or N = mg
so F Kmg = m ax fk v j N F i fk
max
K mg mg

8. Sliding Friction: Quantitatively
Direction: A force vector  to the normal force vector N and the vector is opposite to the velocity.
Magnitude: fk is proportional to the magnitude of N
fk = k N ( = Kmg in the previous example)
The constant k is called the “coefficient of kinetic friction”
As the normal force varies so does the frictional force

9. Additional comments on Friction:
The force of friction does not depend on the area of the surfaces in contact (a relatively good approximation if there is little surface deformation)
Logic dictates that S > K for any system

10. Coefficients of Friction
Material on Material
s = static friction
k = kinetic friction
steel / steel
0.6
0.4
add grease to steel
0.1
0.05
metal / ice
0.022
0.02
brake lining / iron
0.3
tire / dry pavement
0.9
0.8
tire / wet pavement
0.7

11. An experiment S Fy = 0 = T – m1g S Fx = 0 = -T + fs = -T + mS N
Two blocks are connected on the table as shown. The
table has unknown static and kinetic friction coefficients.
Design an experiment to find mS & mK. m2
m2g N T
Static equilibrium: Set m2 and add mass to m1 to reach the breaking point.
Requires two FBDs fS T m1
m1g
Mass 1
S Fy = 0 = T – m1g
T = m1g = mS m2g  mS = m1/m2
Mass 2
S Fx = 0 = -T + fs = -T + mS N
S Fy = 0 = N – m2g

12. An experiment S Fy = 0 = T – m1g S Fx = 0 = -T + ff = -T + mk N
Two blocks are connected on the table as shown. The
table has unknown static and kinetic friction coefficients.
Design an experiment to find mS & mK. m2
m2g N T
Dynamic equilibrium: Set m2 and adjust m1 to find place when
a = 0 and v ≠ 0
Requires two FBDs fk T m1
m1g
Mass 1
S Fy = 0 = T – m1g
T = m1g = mk m2g  mk = m1/m2
Mass 2
S Fx = 0 = -T + ff = -T + mk N
S Fy = 0 = N – m2g

13. An experiment (with a ≠ 0)
Two blocks are connected on the table as shown. The
table has unknown static and kinetic friction coefficients.
Design an experiment to find mS & mK. m2
m2g N T
Non-equilibrium: Set m2 and adjust m1 to find regime where a ≠ 0
Requires two FBDs fk T m1
m1g
Mass 1
S Fy = m1a = T – m1g
T = m1g + m1a = mk m2g – m2a  mk = (m1(g+a)+m2a)/m2g
Mass 2
S Fx = m2a = -T + fk = -T + mk N
S Fy = 0 = N – m2g

14. Sample Problem
You have been hired to measure the coefficients of friction for the newly discovered substance jelloium. Today you will measure the coefficient of kinetic friction for jelloium sliding on steel. To do so, you pull a 200 g chunk of jelloium across a horizontal steel table with a constant string tension of 1.00 N. A motion detector records the motion and displays the graph shown. What is the value of μk for jelloium on steel?

15. S Fx =ma = F - ff = F - mk N = F - mk mg S Fy = 0 = N – mg
Sample Problem
S Fx =ma = F - ff = F - mk N = F - mk mg
S Fy = 0 = N – mg
mk = (F - ma) / mg & x = ½ a t2  0.80 m = ½ a 4 s2
a = 0.40 m/s2
mk = ( · 0.40 ) / (0.20 ·10.) = 0.46

16. Another experiment
A block is connected to a horizontal massless string. The table has coefficients of kinetic & static friction (mK & mS). There is a unknown mass m and you apply a variable force T (by pulling on the rope) as shown in the plot.
T (N) 50 m1 T 40 30 20
(A) On the next slide are tables and plots of velocity vs. time
(B) Can you deduce the various coefficients of friction and the mass ? 10 10 20 30 40
t (sec)

17. The Experimental Data time (sec) vel. #1 m/s vel. #2 m/s vel. #3 m/s
t < 30 s puts constraints on ms
(Static equilibrium)
t > 30 s reflects mk
(Non-equilibrium)
Const. accel. at (T= 40 N) and (T=50 N) second times
speed (m/s)
T (N) 10 40 30 50 20 5 10 15 20 25
time (sec)
vel. #1 m/s
vel. #2 m/s
vel. #3 m/s 10
0.0 20 30 40
5.1
4.9
5.0 50
20.2
19.8
20.0 10 20 30 40
t (sec)

18. Another experiment N f T mg
A block is connected by a horizontal massless string. The table has coefficients of kinetic & static friction (mK & mS). There is a unknown mass m and you apply a variable force as shown in the plot. N
FBD m1
Static case (30 N & less)
S Fx = 0 = -T + f = -T + mN
S Fy = 0 = - N + mg
T = m m g , 2 unknowns f T mg

19. Another experiment N S Fx = max = -T + f = -T + mK N S Fy = 0 = N – mg
A block is connected by a horizontal massless string. The table has coefficients of kinetic & static friction (mK & mS). There is a unknown mass m and you apply a variable force as shown in the plot.
FBD f N
(B) Non-equilibrium
S Fx = max = -T + f = -T + mK N
S Fy = 0 = N – mg
max = -T + mK m g
Using information at right you can identify 2 equations and 2 unknowns m1 T mg
Notice that at 30 s < t < 40 s
T= 40 N ax= Dv/Dt = -5/10 m/s2
and at 40 s < t < 50 s
T= 50 N ax= Dv/Dt =-15/10 m/s2

20. Inclined plane with “Normal” and Frictional Forces
At first the velocity is v up along the slide
Can we draw a velocity time plot?
What the acceleration versus time?
“Normal” means
perpendicular
Normal
Force
Friction Force
Sliding Down fk
Sliding Up v q q
mg sin q
Weight of block is mg
Note: If frictional Force = Normal Force  (coefficient of friction)
Ffriction =  Fnormal = m mg sin q then zero acceleration

21. The inclined plane coming and going (not static): the component of mg along the surface > kinetic friction
Fx = max = mg sin q ± uk N > 0
Fy = may = -mg cos q + N
Putting it all together gives two different accelerations,
ax = g sin q ± uk g cos q. A tidy result but ultimately it is the process of applying Newton’s Laws that is key.

22. Lecture 6 Assignment: HW4, (Chapters 6 & 7, due 10/1, Wednesday)
Finish Chapter 7
1st Exam Thursday, Oct. 2nd from 7:15-8:45 PM Chapters 1-7 1