Transcript Static Friction

King Fahd University of
Petroleum & Minerals
Mechanical Engineering
Dynamics ME 201
BY
Dr. Meyassar N. Al-Haddad
Lecture # 11
Sir Isaac Newton
1642-1727
Proposed fundamental
laws that are the basis
of modern mechanics
3 laws of motion
law of gravitation
Newton’s laws
Law of inertia a body in motion will stay in motion and a body at rest will
stay at rest unless acted upon by a net external force.
Law of force-acceleration A particle acted upon by an unbalanced force F
experiences an acceleration a that has the same direction as the force
and a magnitude that is directly proportional to the force
F ma
Law of action-reaction for every action, there is an equal and opposite
reaction
mg = FN
Law of gravitation - all bodies are attracted to one another with a force
proportional to the product of their masses and inversely proportional
to the square of the distance between them.
m1m2
F G 2
r
13.2 The Equation of Motion
 F  ma
• Free-Body diagram
(Force Diagram)
• Kinetic diagram
(acceleration Diagram)
13.3 Equation of Motion for a System of
Particles
 F  ma
Fi  f i  mi a i
Internal forces cancel each other
F  m a
 F  ma
i
i i
G
13.4 Equations of Motion: Rectangular
Coordinates
• When the net force is
projected to separate
coordinate axes the
Newton’s second law
still holds
 F  ma
 F i   F j   F k  m( a i  a
x
y
z
F
F
F
x
 max
y
 ma y
z
 maz
x
y
j  az k)
Free Body Diagram Method
•Draw each object separately
•Draw all the forces acting on that object
•Get x and y components of all the forces to
calculate the net force
•Apply Newton’s second law to get
acceleration
•Use the acceleration in any motion analysis
and establish a Kinetic Diagram
ma
Normal & Frictional Force
Action-Reaction forces
- mg = FN
F
mg
Ff
FN
Static Friction ( ms )
The coefficient of static friction
is material dependent.
Friction
• Static friction – parallel force on the
surface when there is no relative motion
between the 2 objects
• Static friction force can vary from zero to
Maximum
Static Dynamic
Ff = msFN Ff = mkFN
Applied external force
Kinetic Friction ( mk )
• Kinetic friction – parallel force on the
surface when there is relative motion
between the 2 objects
• Kinetic friction force is always the same
F f  μ k FN
Static
• is material dependent.
Friction
• The coefficient of Kinetic friction
Ff = msFN
Dynamic
Ff = mkFN
Applied external force
Spring Force
• Spring force
Fs  ks
s
• k : spring stiffness (N/m)
• s : stretched or
compressed length
s  l  lo
lo
l
Example 13-4
•
•
•
•
m = 2 kg
y = 1m
smooth
a=?

ma
mg  Fs sin   ma
 S  12  0.752  0.75  0.5m
1 1
θ  tan
 53.13
0.75
 Nc  Fs cos θ  ks cos θ  3x 0.5x cos 53  0.9 N 
 Nc  Fs cos  0
ag
ks
sin θ  9.21m/s 2 
m
Problem
mg
y

x
N
mk N
a=?
F
y
 0;
N  mg cos   0
N  40  9.81 cos 20  0
N  369 N
F
x
 max
 mk N  mg sin   ma
 0.25(369)  40  9.81 sin 20  40ax
ax  5.66 m / s 2
Example 13-5
m A= 3 kg
m B= 5 kg
From rest
vB= ? In 2 second
  o  aB t
Block B
Same
should be ٍSame
  Fy  may
Block A
  Fy  may
2s A  s B  l
196.2  T  5aB
2 A  B  0
981  2T  3a A
2a A  aB  0
Problem
VA = ?

ac
0 = 2ac + aA
2Sc + SA = L
aA = -2ac
 F  0: N  200 * 9.81cos 30  0 N  1699 N
 F  ma : 0.5N  2T  mg sin 30  200a
y
x
c
c
F

mg
y
 maA 125 * 9.81  T  125aA
maA
T  1226.25  125aA  (2)
(2)  (1)
849.5  2(1226.25  125aA)  981  200ac
 622  200ac  250aA
 200ac  500ac
 ac  0.888(m / s 2 ) aA  1.777(m / s 2 ) T  1004 N
 VA  2a A s  4.62(m / s)
Review
•
•
•
•
Example 13.1
Example 13.2
Example 13.3
Example 13.4