Download Sliding Friction

Journal 9/27/16
Today we’re going to deal with friction. What does this
term mean to you?
Objective
To define friction and how it
is calculated
Tonight’s Homework
pp 99: 10 pp 105: 7, 8, 9
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.
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.
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.
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.
Equations
Force of Friction E: Ff = µFN
V: 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.
S: 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.)
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
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?
20N
Our equation says Ff = µFN.
Ff = ?
Ff = (0.34)(20 N)
µ = 0.34
FN = 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.
Notes on Friction
Example (medium):
51 N
A man pushes on a block
of wood with a force of
Ff
22.4 N
22.4 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?
Notes on Friction
Example (medium):
51 N
A man pushes on a block
of wood with a force of
Ff
22.4 N
22.4 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
22.4 N = µ (51 N)
µ=?
FN = 51 N
µ = 0.439
Notes on Friction
Example (hard):
A man is pulling a box
100 N
across the ground at
Ff
30°
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?
Notes on Friction
Example (hard):
A man is pulling a box
100 N
across the ground at
Ff
30°
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.
Notes on Friction
Example (hard):
A man is pulling a box
100 N
across the ground at
Ff
30°
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
x = 100 cos(30) = 86.6 N
30°
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”.
Notes on Friction
Example (hard):
A man is pulling a box
100 N
across the ground at
Ff
30°
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
86.6 N = (0.5)(FN)
86.6 N = (0.5)(“weight” - y)
173.2 N = “weight” – y
“weight” = 173.2 N + 50 N
“weight” = 223.2 N (sig figs) weight = 200 N
Exit Question
Can you ever have a negative coefficient of friction?
a) Yes
b) No
c) It depends
d) Not enough information