1. Newton’s Laws
Applications!!!

2. Friction Friction is the force that opposes a sliding motion.
Friction is due to microscopic irregularities in even the smoothest of surfaces.
Friction is highly useful. It enables us to walk and drive a car, among other things.
Friction is also dissipative. That means it causes mechanical energy to be converted to heat. We’ll learn more about that later.

3. Microscopic View N Fpush f W (friction) Big view:
Surfaces look perfectly smooth.
Fpush f
(friction)
Small view:
Microscopic irregularities resist movement.
Friction may or may not exist between two surfaces. The direction of friction, if it exists, is opposite to the direction object will slide when subjected to a horizontal force. It is always parallel to the surface.

4. Friction depends on the normal force.
The friction that exists between two surfaces is directly proportional to the normal force.
Increasing the normal force increases friction; decreasing the normal force decreases friction.
This has several implications, such as…
Friction on a sloping surface is less than friction on a flat surface (since the normal force is less on a slope).
Increasing weight of an object increases the friction between the object and the surface it is resting on.
Weighting down a car over the drive wheels increases the friction between the drive wheels and the road (which increases the car’s ability to accelerate).

5. Static Friction
This type of friction occurs between two surfaces that are not slipping relative to each other.
fs  sN
fs : static frictional force (N)
s: coefficient of static friction
N: normal force (N)

6. fs < msN is an inequality!
The fact that the static friction equation is an inequality has important implications.
Static friction between two surfaces is zero unless there is a force trying to make the surfaces slide on one another.
Static friction can increase as the force trying to push an object increases until it reaches its maximum allowed value as defined by ms.
Once the maximum value of static friction has been exceeded by an applied force, the surfaces begin to slide and the friction is no longer static friction.

7. Static friction and applied horizontal force
Physics N W
Force Diagram
surface
fs = 0
There is no static friction since there is no applied horizontal force trying to slide the book on the surface.

8. Static friction and applied horizontal force
Physics N W
Force Diagram F fs
surface
0 < fs < msN and fs = F
Static friction is equal to the applied horizontal force, and there is no movement of the book since SF = 0.

9. Static friction and applied horizontal force
Physics N W
Force Diagram F fs
surface
fs = msN and fs = F
Static friction is at its maximum value! It is still equal to F, but if F increases any more, the book will slide.

10. Static friction and applied horizontal force
Physics N W
Force Diagram fk F
surface
fs = msN and fs < F
Static friction cannot increase any more! The book accelerates to the right. Friction becomes kinetic friction, which is usually a smaller force.

11. Static friction on a ramp
surface fs N
Without friction, the book will slide down the ramp. If it stays in place, there is sufficient static friction holding it there.
Physics
W = mg q
Wx = mgsinq and N = mgcosq
At maximum angle before the book slides, we can prove that ms = tanq.

12. Static friction on a rampx
surface fs
Assume q is maximum angle for which book stays in place. N
SF = 0
Wx = fs
mgsinq = msmgcosq
ms = sinq/cosq = tanq
Physics q Wx
W = mg q
fs = mgsinq and N = mgcosq
At maximum angle before the book slides, we can prove that ms = tanq.

13. Kinetic Friction
This type of friction occurs between surfaces that are slipping past each other.
fk = kN
fk : kinetic frictional force (N)
k: coefficient of kinetic friction
N: normal force (N)
Kinetic friction (sliding friction) is generally less than static friction (motionless friction) for most surfaces.

14. Sample Problem
A 10-kg box rests on a ramp that is laying flat. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30.
What is the maximum horizontal force that can be applied to the box before it begins to slide?
What force is necessary to keep the box sliding at constant velocity?

15. Sample Problem
A 10-kg wooden box rests on a ramp that is lying flat. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is What is the friction force between the box and ramp if
no force horizontal force is applied to the box?
a 20 N horizontal force is applied to the box?
a 60 N horizontal force is applied to the box?

16. Problem
A 10-kg wooden box rests on a wooden ramp. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is What is the friction force between the box and ramp if
the ramp is at a 25o angle?
the ramp is at a 45o angle?
what is the acceleration of the box when the ramp is at 45o?

17. Terminal Velocity
In reality objects in free fall DO HAVE resistance from the air
As this resistance increases during a fall, the acceleration of the object decreases
The force of resistance will eventually equal the gravitational force (balanced forces)

18. Terminal Velocity
When FR = Fg, the acceleration will be zero and the object will fall at a constant velocity
This constant, maximum speed is called terminal velocity
Terminal velocity depends on the shape and area of the object as well as the density of air

19. Terminal Velocity Vterminal2 = 2mg / pCdA Where: m = mass
g = acceleration due to gravity
p = density of air
Cd = drag coefficient
A = cross-sectional area

20. Tension
Tension is a pulling force that arises when a rope, string, or other long thin material resists being pulled apart without stretching significantly.
Tension always pulls away from a body attached to a rope or string and toward the center of the rope or string.

21. A physical picture of tension
Imagine tension to be the internal force preventing a rope or string from being pulled apart. Tension as such arises from the center of the rope or string. It creates an equal and opposite force on objects attached to opposite ends of the rope or string.

22. Tension examples
Note that the pulleys shown are magic! They affect the tension in any way, and serve only to bend the line of action of the force.

23. Sample problem
A 1,500 kg crate hangs motionless from a crane cable. What is the tension in the cable? Ignore the mass of the cable.
Suppose the crane accelerates the crate upward at 1.2 m/s2. What is the tension in the cable now?

24. Springs (Hooke’s Law)
The magnitude of the force exerted by a spring is proportional to the amount it is stretched.
F = kx
F: force exerted by the spring (N)
k: force constant of the spring (N/m or N/cm)
x: displacement from equilibrium (unstretched and uncompressed) position (m or cm)
The direction of the force is back toward the equilibrium (or unstretched) position.

25. Sample problem
A 1.50 kg object hangs motionless from a spring with a force constant of k = 250 N/m. How far is the spring stretched from its equilibrium length?

26. Sample problem
A 1.80 kg object is connected to a spring of force constant 120 N/m. How far is the spring stretched if it is used to drag the object across a floor at constant velocity? Assume the coefficient of kinetic friction is 0.60.

27. Sample problem
A 5.0 kg object (m1) is connected to a 10.0 kg object (m2) by a string. If a pulling force F of 20 N is applied to the 5.0 kg object as shown,
A) what is the acceleration of the system?
B) what is the tension in the string connecting the objects?
(Assume a frictionless surface.)

28. Gravity
A very common accelerating force is gravity. Here is gravity in action. The acceleration is g.

29. Slowing gravity down
The pulley lets us use gravity as our accelerating force… but a lot slower than free fall. Acceleration here is a lot lower than g.

30. Magic pulleys on a flat table
Magic pulleys bend the line of action of the force without affecting tension. T
m2g N
m1g
SF = ma
m2g + T – T = (m1 + m2)a
a = m2g/(m1+m2) -x x m1
Frictionless table m2

31. Sample problem
Mass 1 (10 kg) rests on a frictionless table connected by a string to Mass 2 (5 kg). Find
the acceleration of each block.
the tension in the connecting string. m1 m2

32. Sample problem
Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). Find the minimum coefficient of static friction for which the blocks remain stationary. m1 m2

33. Sample problem - solution
SF = 0
m2g - T + T – fs = 0
fs = m2g
msN = m2g
msm1g = m2g
ms = m2/m1 = 0.50 T
m2g N
m1g fs m1 m2

34. Note: we know from previous problem that the static friction is not enough to hold the blocks in place!
Sample problem
Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). If ms = 0.30 and mk = 0.20, what is
the acceleration of each block?
the tension in the connecting string? m1 m2

35. Sample problem – solution (a)
SF = ma
m2g - T + T – fk = ma
m2g - mkm1g = (m1 + m2)a
a = (m2- mkm1) g/(m1 + m2)
a = 2.0 m/s2 T
m2g N
m1g fk m1 m2

36. Sample problem – solution (b)
Using that the acceleration is 2.0 m/s2 from part a)
Sample problem – solution (b)
Using block 2
SF = ma
m2g - T = m2a
T = m2(g – a)
T = 40 N
Using block 1
SF = ma
T - fk = m1a
T = m1(a + mkg)
T = 40 N T
m2g N
m1g fk m1 m2

37. Magic pulleys on a ramp q
It’s a little more complicated when a magic pulley is installed on a ramp.
m1g N T
m2g
m1gsinq
m1gcosq
SF = ma
m2g -T + T – m1gsinq = (m1+m2)a
m2g – m1gsinq = (m1+m2)a
a = (m2 – m1sinq)g/(m1+m2) m1 m2 q

38. Sample problem
Two blocks are connected by a string as shown in the figure. What is the acceleration, assuming there is no friction?
10 kg
5 kg
45o

39. Sample problem - solution
m1g N T
m2g
m1gsinq
m1gcosq
SF = ma
m2g -T + T – m1gsinq = (m1+m2)a
m2g – m1gsinq = (m1+m2)a
a = (m2 – m1sinq)g/(m1+m2)
a = [(5 – 10sin45o)(9.8)]/15
a = m/s2
10 kg
5 kg
45o

40. Sample problem - solution
m1g N T
m2g
m1gsinq
m1gcosq
SF = ma
m2g -T + T – m1gsinq = (m1+m2)a
m2g – m1gsinq = (m1+m2)a
a = (m2 – m1sinq)g/(m1+m2)
a = [(5 – 10sin45o)(9.8)]/15
a = m/s2
10 kg
5 kg
45o
How would this change if there is friction on the ramp?