Download Physic 231 Lecture 9

Physic 231 Lecture 9
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Main points of last lecture:
Atwood’s machine.
Normal force
Newton’s 3d Law:
– “When a body exerts a force
on another, the second body
exerts an equal oppositely
directed force on the first
body.”
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Main points of today’s lecture:
Examples of 3d Law
Frictional forces:
– kinetic friction:
f k = µk N
– static friction
f s < µs N
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More examples.
Example
v
60 N
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N12
v
N 21
10 kg
5 kg
A block with mass 5 kg and a second block with mass 10 kg are
supported by a frictionless surface. A force of 60 N is applied to the 10
kg mass. What is the force of the 5 kg block on the 10 kg block?
x components : (drop subscript x)
body 2 : N 21 = m2a; N 21 = − N12
body 1 : F + N12 = m1a; F = − N12 + m1a
F = N 21 + m1a = m2a + m1a = (m2 + m1 )a
a = F / (m2 + m1 )
N12 = − N 21 = − m2 F / (m2 + m1 ) = −20 N
N12 = 20 N to the left
If objects move together, the relevant mass is their
total mass.
Example
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Two skaters, an 82 kg man and a 48 kg woman, are standing on ice.
Neglect any friction between the skate blades and the ice. The woman
pushes the man with a force of 45 N due east. Determine the
accelerations (magnitude and direction) of the man and the woman.
x components (West is positive)
− 45N
aman =
= 0.55m / s 2 east
82kg
45N
awoman =
= 0.94m / s 2 west
48kg
East
quiz
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Two skaters, an 100 kg man and a 50 kg woman, are standing on ice.
Neglect any friction between the skate blades and the ice. By pushing
the man, the woman is accelerated at 2 m/s2 in the direction of due
west. What is the corresponding acceleration of the man?
– a) 4 m/s2 due east
– b) 1 m/s2 due east
– c) 2 m/s2 due east
– d) 1.5 m/s2 due east
East
Friction
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Friction impedes the motion of one object along the surfaces of another.
It occurs because the surfaces of the two objects temporarily stick
together via “microwelds”. The frictional force can be larger if the two
surfaces are at rest with respect to each other.
v
Experimentally we have two cases:
N
v
v
v
fk
– kinetic friction:
f k = µk N
– static friction
f s < µs N
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v
fs
v
N
v
F
The coefficient of static friction exceeds that for kinetic friction:
µ s > µk
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Frictional forces always oppose the motion of one surface with respect
to the other.
Example
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Consider the figure below, with M1=105 kg and M2=44.1 kg. What is
the minimum static coefficient of friction necessary to keep the block
from slipping.
T = M2g
T ≤ µs N = µs M1g
M 2 g ≤ µ s M1g
M2
≤ µs
M1
Example
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Consider the figure below, with M1=105 kg. If the two mass move
with constant speed and the coefficient of kinetic friction is 0.2, what
is M2?
T = M 2g
T = µ k N = µ k M1g
M 2 g = µ k M1g
M 2 = µ k M1 = 21 kg
Example
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Chucky puts a block on an incline plane as shown below. He increases
the angle to 40o, at which point the block begins to slide. What is the
static coefficient of friction?
v
fs
v
N
1.
2.
y
x
0
40040v W
400
static equilibrium
static friction
θ=40o
W
y
Wx
3.
4.
Draw the forces.
Choose an appropriate
coordinate system.
Calculate the components.
Use Newton’s 2nd law to get µs
W , N , f s are magnitudes of forces
x : ∑iFi , x = 0 = f s − W sin (θ )
⇒ µ s N > f s = W sin (θ )
y : ∑iFi , y = 0 = N − W cos(θ )
⇒ N = W cos(θ )
⇒ µ sW cos(θ ) > W sin (θ )
⇒ µ s > tan (θ )
at max, µ s = tan (40o ) = .84
Example
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The block shown below starts sliding down the ramp. Assuming the
coefficient of kinetic friction µk = 0.3, how long does it take for the
block to travel 2m to the bottom of the ramp ?
v
fk
v
N
∆
1.
2.
3.
4.
my
2
s=
x
400
400
v
W
Draw the forces.
Choose an appropriate
coordinate system.
Calculate the components.
Use Newton’s 2nd to get t
W , f k , N are magnitude of forces
x : ∑iFi , x = ma x = f k − W sin (θ )
⇒ µk N − mg sin (θ ) = ma x
kinetic friction
µk = 0.3
vo=0
θ=40o
y : ∑iFi , y = 0 = N − W cos(θ )
⇒ N = mg cos(θ )
⇒ µk g cos(40o ) − g sin (40o ) = a x
⇒ a x = −4.05m / s 2
∆x =
1 2
2 ∆x
axt ⇒ t =
=
2
ax
− 4m
= 1s
2
− 4.05m / s
Example
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Two packing crates of masses 10.0 kg and 5.00 kg are connected by a
light string that passes over a frictionless pulley as in the figure below.
The 5.00-kg crate lies on a smooth incline of angle 40.0°. Find the
acceleration of the 5.00-kg crate and the tension in the string.
T − M1g = M1a 1
( )
T − M 2 g sin 40o = M 2 a 2
a 1 = −a 2
M1
(
( ))
− M a + M g = M a + M g sin (40 )
M g − M g sin (40 )
= 4 .4 m / s
a =
T = M1 (a1 + g ) = M 2 a 2 + g sin 40o
M2
o
1 2
1
2 2
2
o
2
1
(
2
M1 + M 2
)
T = M1 g − 4.4m / s 2 = 54 N
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