# Newtonâ€™s Laws of Motion - and Friction

Document Sample

```					Newton’s Laws of Motion
- and Friction

No. 27 : a friction burn
Friction - Reminder
An object is kept in equilibrium (forces will balance)
when
F  R
F = Friction;   R - normal   Reaction;  is the coefficient of friction
(material dependant)

In   limiting equilibrium (the body is about to move)
then
F = R
Also [new bit],
Once the object is moving the friction stays constant
at
F = R
A friction and N2L problem (1)
A particle on a rough surface weighing 5kg is being pulled by
force of 80N at 20o the the horizontal.
The coefficient of friction is 0.3
(between the particle and the surface).
Find : the normal reaction, the friction force and the acceleration
First: Draw a Force diagram                  = ma
N2L : F
R         80 N              weight = 5g
a ms   -2
= 5 x 9.8
20o                            = 49 N
F
Resolve Vertically

49 N                R + 80 sin 20 = 49
R + 27.4 = 49
R = 49 - 27.4 = 21.6 N
A friction and N2L problem (2)
A particle on a rough surface weighing 5kg is being pulled by
force of 80N at 20o the the horizontal.
The coefficient of friction is 0.3
(between the particle and the surface).
Find : the normal reaction, the friction force and the acceleration
First: Draw a Force diagram         Once the object is moving
21.6 N                   the friction stays constant at
80 N
a ms   -2
F = R
F               20o
F = 0.3 x 21.6
49 N
= 6.5 N
A friction and N2L problem (3)
A particle on a rough surface weighing 5kg is being pulled by
force of 80N at 20o the the horizontal.
The coefficient of friction is 0.3
(between the particle and the surface).
Find : the normal reaction, the friction force and the acceleration
First: Draw a Force diagram               = ma
N2L : F
21.6 N       80 N         Resultant = 5 x a = 5a N
a ms   -2

6.5 N                20o           Resolve Horizontally
Resultant = 80 cos 20 - 6.5 N
= 75.2 - 6.5 = 68.7 N
49 N
Resultant = 5a
68.7 = 5a
a = 68.7 / 5 = 13.7 ms-2
Friction and Slopes
“An object on a rough plane”
Normal
Reaction
The friction force will act
in the opposite direction to
Shallow                       the one it wants to move
slope
Weight

An object is kept in equilibrium (forces will balance)
when
F  R
Friction and Slopes
“An object on a rough plane”

As the object starts to slip,
Normal                    the friction will act against
Reaction                  it to try to slow it down

Steep
slope        Weight

Once the object is moving the friction stays constant
at
F = R
Friction and Slopes - Example (1)
E.g a 5 kg sled sliding down a hill The coefficient of friction is 0.3
Find : the normal reaction, the friction force and the acceleration
a ms-2                           = ma
N2L : F
weight = 5g
R                 F
= 5 x 9.8
= 49 N

Resolve Perpendicular to slope
30o             No resultant force -
49 N          acceleration is parallel to slope
60o
R = 49 sin 30

= 24.5 N
Friction and Slopes - Example (2)
E.g a 3 kg sled sliding down a hill The coefficient of friction is 0.3
Find : the normal reaction, the friction force and the acceleration
a ms-2                  Once the object is moving,
the friction stays constant
24.5              F
at
F = R

F = R

49 N                       = 0.3 x 24.5
60o
= 7.35 N
Friction and Slopes - Example (3)
E.g a 3 kg sled sliding down a hill The coefficient of friction is 0.3
Find : the normal reaction, the friction force and the acceleration
a ms-2                    Resolve Parallel to slope

24.5              7.35N        A resultant force must exist
to make the object accelerate

Resultant force = 49 cos 30 - 7.35
= 42.45 - 7.35
30o                          = 35.1 N
49 N             = ma
N2L : F
60o                        Resultant = 5 x a = 5a N

So,   35.1 = 5a
a = 35.1 / 5 = 7.0 ms-2
Down the slope
A rough slope and N2L problem (1)
An object of mass 2.8kg is pulled up a rough slope inclined at
40o by a light string parallel to the slope.
The coefficient of friction is 0.3
Find : the normal reaction and the tension in the string if …
(a) it moves up the slope with constant speed
(b) it accelerates at 1.5 ms-1
First: Draw a Force diagram
a ms-2
R
T
Friction acts against
40o                  the motion
F
= ma
N2L : F
40o
weight = 2.8g
= 2.8 x 9.8                              27.44 N
= 27.44 N
A rough slope and N2L problem (2)
An object of mass 2.8kg is pulled up a rough slope inclined at
40o by a light string parallel to the slope.
The coefficient of friction is 0.3
Find : the normal reaction and the tension in the string if …
(a) it moves up the slope with constant speed
(b) it accelerates at 1.5 ms-1
Force diagram

Resolve Perpendicular to slope              R
No resultant force -                                       T
Any acceleration is parallel to slope

F
R = 27.44 sin 50                          50o
40o
= 21.0 N                           27.44 N
A rough slope and N2L problem (3)
An object of mass 2.8kg is pulled up a rough slope inclined at
40o by a light string parallel to the slope.
The coefficient of friction is 0.3
Find : the normal reaction and the tension in the string if …
(a) it moves up the slope with constant speed
(b) it accelerates at 1.5 ms-1
Force diagram
Once the object is moving,
the friction stays constant
at                              21.0 N
F = R                                         T
F = 0.3 x 21.0 = 6.3 N
Resolve Parallel to slope
No resultant force exists               F
50o
- no acceleration                       40o
T = 27.44 cos 50 + F                        27.44 N
= 17.6 + 6.3 = 23.9 N
A rough slope and N2L problem (4)
An object of mass 2.8kg is pulled up a rough slope inclined at
40o by a light string parallel to the slope.
The coefficient of friction is 0.3
Find : the normal reaction and the tension in the string if …
(a) it moves up the slope with constant speed
(b) it accelerates at 1.5 ms-1
Force diagram
Resolve Parallel to slope
A resultant force must exist                    21.0 N
to make the object accelerate                                  T
Resultant force up slope
1.5 ms -2

= T - 27.44 cos 50 – F
= T – 17.6 – 6.3 = T – 23.9N            6.3 N 50o
N2L : F = ma                               40o
Resultant = 2.8 x 1.5 = 4.2 N                   27.44 N
4.2 = T – 23.9         T = 4.2 + 23.9 = 28.1 N
Activity / Homework

•   Mechanics 1 book

• Exercise F [page 95]
•Q   1 , 3, 5
• Exercise H [page 100]
•Q   1, 2
• and lots more if you want to do it well

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 9 posted: 12/23/2009 language: English pages: 15
How are you planning on using Docstoc?