1. Force, Mass, & Newton’s 1st, 2nd, 3rd Laws of Motion (4-1 through 4-5)
Status: Unit 3
Force, Mass, & Newton’s 1st, 2nd, 3rd Laws of Motion (4-1 through 4-5)
Weight – the Force of Gravity: and the Normal Force, Problem solving (4-6, 4-8, 4-7)
Solving Problems w/ Newton’s Laws, Problems w/ Friction (4-7)
Problems w/ Friction and Terminal Velocity revisited (4-7, 5-1)
11/10/2018
Physics 253

2. But First a Bit about Test One
Total Number of Exams Taken: 116
Low : 5 / Average: 30 / High: 49
Frequency
0-10:
11-20: 19
21-30: 36
31-40: 41
41-50: 17
Pick them up from the stage -- after class.
11/10/2018
Physics 253

3. Friction
To get a handle on the principles we’ve ignored friction. However this is a big oversimplification. Now we have the tools to properly deal with the force.
Friction exist between any two solid surfaces.
When one or both of the surfaces are very coarse and bumpy, like a cobble stone street impeding on the runners of a sled, it’s pretty clearly a mechanical effect.
But even for very smooth surface, friction exists due to the microscopic imperfections or roughness. Just like the cobblestones these bumps impede motion. Most likely this is an electromagnetic effect.
There are two kind so friction: kinetic and static.
11/10/2018
Physics 253

4. Kinetic Friction
When two objects are sliding past one another they are said to have kinetic fiction.
The force of friction acts opposite the direction of an object’s velocity. You’ve felt this before, perhaps when shoving a couch across a carpet.
The magnitude of the force depends on the characteristics of the two surface.
It turns out experimentally that the force of friction is
Proportional to the normal force between the two surfaces
Independent of surface area
Independent of the velocity
11/10/2018
Physics 253

5. Rubber on solid surfaces: 1 Steel on steel: 0.6
We deal with the different surfaces by imposing a constant of proportionality:
The constant or coefficient of kinetic friction depends only on the surfaces.
Caveats:
NOT a vector equation!
NOT a theory - but a model that works well
Rubber on solid surfaces: 1
Steel on steel: 0.6
Rubber on wet concrete: 0.5
Wood on wood: 0.2
Ice on ice: 0.03
Human joints: 0.01
Oiled Ball Bearings: <0.01
11/10/2018
Physics 253

6. Remember shoving the couch across the carpet
Remember shoving the couch across the carpet? You’ll also remember it’s actually harder and takes more force to get it moving.
That’s because the static friction, the force parallel to the surfaces of two motionless objects is larger than kinetic friction.
What’s more it depends on how hard you push! The couch doesn’t accelerate right a way when a small force is applied, it’s not until later when you exceed some limiting force.
Since the entire time a=0, the force of friction must be balancing and changing to match your force!
Static Friction
It turns out that the maximum force of static friction is also proportional to the normal force:
Once again the constant, in this case the coefficient of static friction, depends on the surfaces.
Since in this case the frictional force varies it’s more complete to say
11/10/2018
Physics 253

7. Rubber on solid surfaces: 1 Steel on steel: 0.6
Kinetic Coefficients: mk
Static Coefficients: ms
Rubber on solid surfaces: 1
Steel on steel: 0.6
Rubber on wet concrete: 0.5
Wood on wood: 0.2
Ice on ice: 0.03
Human joints: 0.01
Oiled Ball Bearings: <0.01
Rubber on solid surfaces: 1-4
Steel on steel: 0.7
Rubber on wet concrete: 0.7
Wood on wood: 0.4
Ice on ice: 0.1
Human Joints: 0.01
Oiled Ball Bearings: <0.01
11/10/2018
Physics 253

8. 11/10/2018
Physics 253

9. In the vertical direction there will never be acceleration
Consider the freebody diagram of the 10.0kg box with ms= 0.40 and mk= 0.30
What is the magnitude of the force of friction, Ffr, if FA = 0, 20, 38, 40 N?
In the vertical direction there will never be acceleration
For FA = 0, no force is applied, the box doesn’t move and Ffr =0.
So (FA , Ffr)=(0,0)
11/10/2018
Physics 253

10. The applied force is 20N so the box will not move.
Now consider FA = 20N, the force of static friction will oppose an applied force up to a max of:
The applied force is 20N so the box will not move.
From Newton’s 2nd Law:
So (FA , Ffr)=(20N,20N)
The same argument holds for 38N so (FA , Ffr)=(38N,38N)
Last consider FA = 40N. This will start the box in motion since it exceeds the maximum force from static friction.
Now we need to switch to kinetic friction.
In this case: (FA , Ffr)=(40N,29N)
11/10/2018
Physics 253

11. Let’s collect our points: (FA , Ffr)=(0,0) (FA , Ffr)=(20N,20N)
We return to Newton’s law to calculate the acceleration of the moving box:
Let’s collect our points: (FA , Ffr)=(0,0)
(FA , Ffr)=(20N,20N)
(FA , Ffr)=(38N,38N)
(FA , Ffr)=(40N,29N)
11/10/2018
Physics 253

12. How do we measure ms?
Consider a truck carrying a crate up a 10-degree hill.
The coefficient of static friction between the truck bed and the crate is ms = 0.35.
Find the maximum acceleration that the truck can reach before the crate begins to slip backward.
11/10/2018
Physics 253

13. For it to be accelerated some force must act upon it.
The crate will not slip if it’s acceleration is equal to that of the truck.
For it to be accelerated some force must act upon it.
One of these is certainly the static friction force. This force must be directed upward to keep the crate from sliding backwards.
As the acceleration of the truck increases so must the frictional force.
Once Ffr(max) = msFN is reached the crate will start sliding.
This will be the point at which the maximum acceleration of the crate and truck is obtained.
11/10/2018
Physics 253

14. Concentrating on the x comp.:
Note that the free-body diagram has been drawn for the moment at which slippage occurs.
Concentrating on the x comp.:
But, we can’t go any further without FM. Concentrating on the y-direction.
Substituting this 2nd result into the 1st:
11/10/2018
Physics 253

15. Note how the mass cancels (this is a hallmark of friction).
Substituting for g, ms, and the trig functions we see:
If we just consider any angle q then,
No acceleration will be tolerated when the maximum acceleration =0.
To find the coefficient of
static friction just put an
object on a plane and tilt it!
11/10/2018
Physics 253

16. How do we measure ms?
Consider the skier at rest accelerating down a hill. If mk=0.10 what is her acceleration and speed after 4.0 s?
11/10/2018
Physics 253

17. First the Free-body Diagram
Note:
Three forces
Axes require only
one decomposition
11/10/2018
Physics 253

18. Next Decomposition FN=+FNj Ffr=-Ffri= -mkFNi FG =+mgsinqi - mgcosqj
11/10/2018
Physics 253

19. Now Apply Newton’s 2nd Law
Note that the acceleration
doesn’t depend on mass!
It’s the same for anyone.
11/10/2018
Physics 253

20. What about determining ms?
Well, just consider what happens if the skier is moving at constant velocity:
That is, the slope at which the skier moves at constant velocity will give you the coefficient of kinetic friction.
This will be true for any system.
Note the similarity between ms and mk!
11/10/2018
Physics 253

21. A Problem with It All!
Consider a box of mass m1=10.0kg resting on a plane inclined at 37o. The 1st box is connected via a pulley by a perfect cord to a hanging box of mass m2.
If ms = 0.40 what range of masses for m2 will keep the system at rest?
If ms = 0.40 and m1=m2=10.0kg, what is the acceleration of the system?
11/10/2018
Physics 253

22. M1<<M2
M1>>M2
11/10/2018
Physics 253

23. Now Apply Newton’s 2nd Law
11/10/2018
Physics 253

24. 11/10/2018
Physics 253

25. 11/10/2018
Physics 253

26. 11/10/2018
Physics 253

27. Next we’ll start exploring circular motion and gravity.
Summary
We’ve explored Newton’s Laws in a number of situations and with conventional forces.
There is a clear path from force to acceleration and to the equations of motions given by SF=ma
Next we’ll start exploring circular motion and gravity.
11/10/2018
Physics 253