Transcript Physics 141 Mechanics Yongli Gao Lecture 4 Motion in 3-D

Physics 141
Mechanics
Lecture 7
Newton’s Laws and Friction Yongli Gao
• Often we have to consider geometrical constrains to
solve a problem.
Example: Friction and Massive String
Two blocks are held by a string and through a
pulley, both of negligible mass, at opposite sides of a
slope of angle q.
a) Find the acceleration if there’s no friction.
b) The same with friction.
c) What will happen if the
string has non-negligible
M
g
mass?
q
m
Solution:
T’
y
a) We have to choose the
x
coordinates intelligently.
T
mg
As shown in the free body
q
diagrams, because of the
x’
string, we have the
Mg
geometrical constraint
(for M) ax  ax' (for m) and T=T’. The rest must
be from Newton’s laws.
For M  Fix  T  Mgsin q  Ma x
i
for m  Fix '  mg  T'  ma x'  mg  T  max
i
Eliminate T from the above equations, we have
m  Msinq
(M  m)ax  (m  Msin q )g  ax 
g
Mm
Discussion:
If m>Msinq, ax>0, M accelerates up.
If m<Msinq, ax<0, M accelerates down.
b) With friction and assume there is already relative
motion M upward, we have for M
T  Mgsin q  fk  Max ,, N  Mgcosq  0,, fk   k N
 T  Mgsin q  k Mg cosq  Max
For m still mg  T  max
Eliminate T we get
m  M(sin q   k cosq )
ax 
g
m M
If the initial motion is M downward, the friction will
be at the opposite direction,
m  M(sin q  k cosq )
ax 
g
m M
If initially M and m are at rest, suppose the
acceleration is upward, then fs is downward,
m  M(sin q   s cosq )
ax 
g
mM
Here we must have
m
m  M(sin q   s cosq )  0   s 
 tanq
M cosq
If the acceleration is downward, then fs is upward,
m
m  M(sin q   s cosq )  0   s  tan q 
M cosq
We have assumed cosq>0, or q<90°.
If m/M>sinq, it’s only possible to move downward
from rest. If m/M<sinq, it’s only possible to move
upward from rest. Both would have to pass a
threshold determined by scosq.
c) If the string has nonT
y
T’
negligible mass ms, the
x
tension will not be the
same in the string, and
q
mg
the force terms due to the
x’
Mg
portion of the string on the slope and that vertical
will have to be added. Let the string length be l,
then the density of the string is r=ms/l.
T  (m  rx' )g  (m  rx' )ax
T  (M  r (l  x' ))gsin q  (M  r (l  x' ))ax
(m  r x' )  (M  r (l  x' ))sin q
 ax 
g
M  m  rl
For simplicity we have ignored friction. We see that
now the acceleration is no longer constant.
Example: Damped Motion
What will be the motion of a particle of mass m if
the only force on it is the damping force f=-av,
where a is a constant?
Solution:
Applying Newton’s 2nd law, dv
f  ma  m  av
dt
Since the change of velocity is along itself, the
direction is not altered and the only change is the
magnitude. We may choose the velocity direction as
the axis and treat it as a 1-D problem dv   a v
dt
m
Trial solution
dv
 t
v  v0e   v0 e  t  v
dt
which leads to
a at / m
v   v0e
m
Fundamental Forces
• We have used F as a generic symbol for force.
Other symbols, such as W for weight, T for tension,
f for friction, N for normal force, etc. These forces
are but reflection of fundamental forces.
• There are four fundamental forces: gravitational,
electromagnetic, weak, and strong.
Intensity
range
Gravitational
weak
long
Electromagnetic medium
long
Weak
weak
short (10-18 m)
Strong
strong
short (10-15 m)
Aspects of Fundamental Forces
• All the forces we encounter are but presentations of
the four fundamental forces. Clearly weight W
belongs to gravitational interaction. The other
forces, including tension T, normal force N, friction
f, etc., are electromagnetic interaction in nature
because they involve deformation of solids or atoms
which are bounded by electromagnetic interaction.
The strong interaction bounds the charged
elementary particles together in nucleus in spite of
the strong Coulomb repulsion, and weak interaction
is reflected in the decay of some elementary
particles. The electromagnetic and weak interaction
have been unified by Glashow, Salam, and Weinberg
in 1979.