1. Journal 9/22/16 Objective Tonight’s Homework
Today we’re going to deal with friction. What does this term mean to you?
Objective Tonight’s Homework
To define friction and how it is calculated
pp 99: 10 pp 105: 7, 8, 9

2. Notes on Friction
There are a number of interesting forces in the world. The first we’re going to discuss is friction.
Friction is a special force that always acts in the opposite direction of the movement of the object.
Example:
friction net force
If all the forces on an object push it one way, friction will act in the opposite direction.

3. Notes on Friction
So what causes friction? It involves looking at objects on the molecular level.
Up close, even the smoothest of objects are actually quite rough.
As two objects try to move past each other, the grooves and bumps hit, lock, and scrape, slowing down movement. Since this always slows an object down, it’s always in the opposite direction from that which the object is moving.

4. Notes on Friction We define this friction in two ways: Static Friction
This type happens when 2 objects are sitting still. As they’re sitting, the grooves and bumps have already interlocked. If you try to get an object to start moving, you have to use a certain amount of force to break the objects out from their interlocked state. This usually results in a sudden jerk as movement starts. Once the objects are moving, static friction has been overcome and is no longer an issue.
This kind of friction explains how you can have a box on a hill without the box just sliding down.

5. Notes on Friction Sliding Friction
This kind of friction occurs as two objects move past each other. The strength of the force usually measures less than that of static friction, but can still be quite strong.
Sliding friction is a set amount for each object. If the force you pull with is less than the force of friction, the object will slow down.
If the force you pull with is equal to the force of friction, the object will move with constant velocity.
If the force you pull with is greater than the force of friction, the object will accelerate.

6. Equations Force of Friction E: V: S: Ff = µFN
Ff : The force of friction in Newtons
µ : A unitless number called the “Coefficient of Friction”. This number is like a percent that tells how rough the surface of the object is.
FN: The amount of force directly pushing the 2 objects together. This is weight plus any downward force.
Used to find the friction opposing the motion of 2 objects. (Note that if there is an additional force upward or at an angle, you need to subtract the vertical component from the normal force.)

7. Notes on Friction Example (easy):
A block of wood sits on a table. If the coefficient of friction is 0.34 and the block weighs 20 N, what is the force of friction?
20N

8. Notes on Friction Examples (easy):
A block of wood sits on a table. If the coefficient of friction is 0.34 and the block weighs 20 N, what is the force of friction?
Our equation says Ff = µFN.
Ff = ? µ = FN = 20 N
Ff = (0.34)(20 N) Ff = 6.8 N
This means we will have to apply a force of at least 6.8 N to the side before the block moves.
20N

9. Notes on Friction Example (medium):
A man pushes on a block of wood with a force of N at a constant velocity of 15 m/s. If the block weighs 51 N, what is the coefficient of friction between the block and the table?
51 N
Ff N

10. Notes on Friction Example (medium):
A man pushes on a block of wood with a force of N at a constant velocity of 15 m/s. If the block weighs 51 N, what is the coefficient of friction between the block and the table?
Since the block is moving at a constant velocity, the force of friction equals the pulling force.
Ff = 22.4 N µ = ? FN = 51 N
22.4 N = µ (51 N) µ = 0.439
51 N
Ff N

11. Notes on Friction Example (hard):
A man is pulling a box across the ground at constant velocity with a rope. If the angle of the rope is 30 degrees and the pulling force is 100 N, what is the weight of the box if µ is 0.5?
100 N Ff
30°

12. Notes on Friction Example (hard):
A man is pulling a box across the ground at constant velocity with a rope. If the angle of the rope is 30 degrees and the pulling force is 100 N, what is the weight of the box if µ is 0.5?
This is more complex. Since we have constant velocity, we know that the pulling force equals the friction force. If we can get the friction force, we can get the normal force (or weight of the box). To get the pulling force, we have to break the angled pulling vector into parts.
100 N Ff
30°

13. Notes on Friction Example (hard):
A man is pulling a box across the ground at constant velocity with a rope. If the angle of the rope is 30 degrees and the pulling force is 100 N, what is the weight of the box if µ is 0.5?
x = 100 cos(30) = 86.6 N
y = 100 sin(30) = 50 N
We can solve this triangle using sines and cosines. Doing so, we get the components above. “X” is our pulling force, so that’s what’s equal to Ff. “Y” is lifting the box up, reducing friction so FN is actually weight – “Y”.
100 N Ff
30°
100 N
30°

14. Notes on Friction Example (hard):
A man is pulling a box across the ground at constant velocity with a rope. If the angle of the rope is 30 degrees and the pulling force is 100 N, what is the weight of the box if µ is 0.5?
So Ff = “x” Ff = 86.6 N
Ff = µFN N = (0.5)(FN)
86.6 N = (0.5)(“weight” - y)
173.2 N = “weight” – y
“weight” = N + 50 N
“weight” = N (sig figs) weight = 200 N
100 N Ff
30°

15. Exit Question Can you ever have a negative coefficient of friction?
a) Yes
b) No
c) It depends
d) Not enough information