Download 103 PHYS - CH5 - Part2 Dr. Abdallah M. Azzeer 1

103 PHYS - CH5 - Part2
Some Basic Information
When Newton’s laws are applied, external forces are only of interest!!
Why?
Because, as described in Newton’s first law, an object will keep its
current motion unless non-zero net external force is applied.
Normal Force, n:
Tension, T:
Free-body diagram
Reaction force that reacts to gravitational force due
to the surface structure of an object. Its direction is
perpendicular to the surface.
The reactionary force by a stringy object against an
external force exerted on it.
A graphical tool which is a diagram of external forces
on an object and is extremely useful analyzing forces
and motion!! Drawn only on an object.
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103 PHYS -
Free Body Diagrams
Diagrams
Diagrams of
of vector
vector forces
forces acting
acting on
on an
an object
object
A
A great
great tool
tool to
to solve
solve aa problem
problem using
using forces
forces or
or using
using dynamics
dynamics
Select
Select aa point
point on
on an
an object
object and
and w/
w/ information
information given
given
Identify
Identify all
all the
the forces
forces acting
acting only
only on
on the
the selected
selected object
object
Define
Define aa reference
reference frame
frame with
with positive
positive and
and negative
negative axes
axes specified
specified
Draw
Draw arrows
arrows to
to represent
represent the
the force
force vectors
vectors on
on the
the selected
selected point
point
Write
Write down
down net
net force
force vector
vector equation
equation
Write
Write down
down the
the forces
forces in
in components
components to
to solve
solve the
the problems
problems
No
No matter
matter which
which one
one we
we choose
choose to
to draw
draw the
the diagram
diagram on,
on, the
the results
results
should
be
the
same,
as
long
as
they
are
from
the
same
motion
should be the same, as long as they are from the same motion
103 PHYS -
Dr. Abdallah M. Azzeer
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103 PHYS - CH5 - Part2
5.7 SOME APPLICATIONS OF NEWTON’
’S LAWS
NEWTON
NEWTON’S
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103 PHYS -
Example: Tension and
Angles
A box is suspended from the ceiling by two ropes making an angle
θ with the horizontal. What is the tension in each rope?
Draw a FBD:
T1
T2
θ
y
T1 sin θ
θ
T1 cos θ
T2 sin θ
θ
T2 cos θ
θ
m
x
mg
Since the box isn’t going anywhere, Fx,NET = 0 and Fy,NET = 0
∑ Fx = T2 cos θ - T1 cos θ = 0
T1 = T2
∑ Fy = T1 sin θ + T2 sin θ - mg = 0
T1 = T2 =
103 PHYS -
Dr. Abdallah M. Azzeer
mg
2 sin θ
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103 PHYS - CH5 - Part2
Example 5.4
A traffic light weighing 125 N hangs from a cable tied to two other
cables fastened to a support as shown in the figure. Find the tension
in the three cables.
r
r
F
∑ = m⋅a
Newton’s 2nd law
r
ur
∑F =T
1
ur ur
+T 2 +T 3
x-comp. of net force
∑F
x
=0
⇒
(
−T 1 cos 37
o
) +T
2
(
)
cos 53 o = 0 ∴ T 1 =
(5 3 ) T
co s (3 7 )
cos
o
o
2
= 0 .7 5 4 T
2
y-comp. of net force
∑F
y
=0
⇒
T 1 s in
(
(3 7 ) + T
T 2 ⎡⎣ sin 53
o
o
2
sin
(5 3 ) − m g
) + 0.754 × sin ( 37 ) ⎤⎦ = 1.25 T
o
T 2 = 100 N ;
= 0
o
2
= 125 N
T 1 = 0.754 T 2 = 75.4 N
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103 PHYS -
Example: Pushing a Box on Ice.
A skater is pushing a heavy box (mass m = 100 kg) across a
sheet of ice (horizontal & frictionless). He applies a force of 50 N
in the x direction. If the box starts at rest, what is its speed v after
being pushed a distance d = 10 m ?
v
v=0
F
F
m
a
m
a
x
d
103 PHYS -
Dr. Abdallah M. Azzeer
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103 PHYS - CH5 - Part2
• Start with F = ma.
–
a = F / m.
– Recall that v 2 = v02 + 2ax
v=
– So v = 2Fd / m
2Fd
m
Plug in F = 50 N, d = 10 m, m = 100 kg:
v = 3.2 m/s.
v
F
m
a
x
d
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103 PHYS -
Example
A fish is being yanked upward out of the water using a fishing line that breaks
when the tension reaches 180 N. The string snaps when the acceleration of the
fish is observed to be is 12.2 m/s2. What is the mass of the fish?
snap !
a = 12.2
m/s2
m=?
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Dr. Abdallah M. Azzeer
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103 PHYS - CH5 - Part2
• Draw a Free Body Diagram
T
Use Newton’s 2nd law
in the upward direction:
a = 12.2 m/s2
m=?
Σ F = ma
T - mg = ma
mg
T = ma + mg = m(g+a)
m=
T
g+a
m=
180 N
( 9.8 + 12.2 ) m/s 2
103 PHYS -
= 8.2 kg
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example
In
In the
the Figure
Figure shown,
shown, aa passenger
passenger of
of mass
mass m ==
72.2
72.2 kg
kg stands
stands on
on aa platform
platform scale
scale in
in an
an
elevator
elevator cab.
cab. We
We are
are concerned
concerned with
with the
the scale
scale
readings
when
the
cab
is
stationary,
and
when
readings when the cab is stationary, and when
itit is
is moving
moving up
up or
or down.
down. (a)
(a) Find
Find aa general
general
solution
solution for
for the
the scale
scale reading,
reading, whatever
whatever the
the
vertical
vertical motion
motion of
of the
the cab.
cab.
103 PHYS -
Dr. Abdallah M. Azzeer
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103 PHYS - CH5 - Part2
N − mg = ma
N = m ( g +a)
SOLUTION:
Positive direction for a is up
(b) What does the scale read if the cab is stationary or moving
upward at a constant 0.50 m/s?
N = ( 72.2 kg ) ( 9.8 m / s 2 + 0 ) = 708 N
(c) What does the scale read if the cab accelerates upward at 3.20 m/s2
and downward at 3.20 m/s2?
N = ( 72.2 kg ) ( 9.8 m / s 2 + 3.20 m / s 2 ) = 939 N
N = ( 72.2 kg ) ( 9.8 m / s 2 − 3.20 m / s 2 ) = 477 N
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103 PHYS -
(d) During the upward acceleration in part (c), what is the magnitude
Fnet of the net force on the passenger, and what is the magnitude ap,cab of
the
acceleration as measured in the frame of the cab? Does
r passenger's
v
Fnet = ma p ,cab ?
Fnet = N − Fg = 939 N − 708 N = 231 N
The acceleration ap,cab of the passenger relative to the frame of the
cab is zero. Thus, in the noninertial frame of the accelerating cab,
Fnet is not equal to map,cab, and Newton's second law does not hold
103 PHYS -
Dr. Abdallah M. Azzeer
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103 PHYS - CH5 - Part2
Example
A block of mass m slides down a frictionless ramp that makes angle
θ with respect to the horizontal. What is its acceleration a ?
Define convenient axes parallel and
perpendicular to plane:
Acceleration a is in x direction only.
m
a
y
θ
Geometry Review
N
y
y
θ
θ
θ
mg cos θ
θ
x
x
ma
a
θ
mg
g
x
mg sin θ
θ
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103 PHYS -
Consider x and y components separately:
x: mg sin θ = ma.
a = g sin θ
y: N - mg cos θ = 0.
103 PHYS -
Dr. Abdallah M. Azzeer
N = mg cos θ
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103 PHYS - CH5 - Part2
Atwood’s Machine:
Masses m1 and m2 are attached to an ideal massless string and
hung as shown around an ideal massless pulley.
Find the accelerations, of the masses.
What is the tension in the string T ?
Assume m1> m2 : the direction of
Motion is shown
T1
T2
m1
m2
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103 PHYS -
• Draw free body diagrams for each object
• Define the acceleration direction
• Applying Newton’s Second Law:
Free Body Diagrams
For m1
m1 g -T1 = m1a
For m2
T2 - m2g = m2a
But T1 = T2 = T
a
(a)
(b)
m1g
• Two equations & two unknowns
– we can solve for both unknowns (T and a).
103 PHYS -
Dr. Abdallah M. Azzeer
T2
a
+
+
since pulley is ideal
m1 g -T = m1 a
T - m 2 g = m2 a
T1
m2g
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103 PHYS - CH5 - Part2
• add (b) + (a):
(m1 - m2 ) g= (m1+ m2 ) a
a=
( m1 − m2 )
g
( m1 + m2 )
• Substitute the value of a in eq. (b) , we obtained;
T = 2m1m2 / (m1 + m2 ) g
Is the result reasonable? Check limiting cases!
Special cases:
i.) m1 = m2 = m
D
a = 0 and T = mg.
OK!
ii.) m2 or m1 = 0
D
|a| = g and T= 0.
OK!
• Atwood’s machine can be used to determine g (by measuring the
acceleration a for given masses).
103 PHYS -
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103 PHYS -
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Dr. Abdallah M. Azzeer
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103 PHYS - CH5 - Part2
Attached bodies on two inclined planes
smooth peg
•Find the accelerations, of the masses.
Fist chose the direction of motion. Then
draw free body diagram for each mass
separately
y
x
θ1
θ2
All surfaces frictionless
N
y
m
N
2
T
x
m
θ2
1
T
a
m2
m1
a
m2 g
θ1
From the free body diagrams for each body, and the
chosen coordinate system for each block, we can
apply Newton’s Second Law:
m1 g
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103 PHYS -
y
(2)
a
adding (1) and (2) gives:
m1gsin θ1 - m2gsin θ2 = (m1+m2 )a
a=
m 1 sin θ1 - m2 sin θ2
g
m1 + m 2
T
x
y
x
T2 a
1
N
(1)
m
Taking “x” components:
For m1
m1g sin θ1 - T = m1 a
For m2
T - m2g sin θ2 = m2 a
θ1
m1 g
N
m
2
θ2
m2 g
103 PHYS -
Dr. Abdallah M. Azzeer
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103 PHYS - CH5 - Part2
smooth peg
m sin θ1 - m2 sin θ2
a= 1
g
m1 + m 2
m2
m1
θ1
Special Case 1:
If θ1 = 0 and θ2 = 0, a = 0.
θ2
All surfaces frictionless
m2
m1
Boring
Special Case 2:
If θ1 = 90 and θ2 = 90,
T
Atwood’s Machine
a=
( m1 − m 2 )
g
( m1 + m 2 )
T
m1
m2
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103 PHYS -
Special Case
3:
If θ1 = 0 and θ2 = 90,
m2
a =
g
( m1 + m2 )
m1
m2
Lab configuration
Two-body dynamics
Three blocks of mass 3m, 2m, and m are connected by strings and
pulled with constant acceleration a. What is the relationship between
the tension in each of the strings?
a
3m
(a) T1 > T2 > T3
103 PHYS -
Dr. Abdallah M. Azzeer
T3
2m
T2
(b) T3 > T2 > T1
m
T1
(c) T1 = T2 = T3
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103 PHYS - CH5 - Part2
Solution
• Draw free body diagrams!!
T3
3m
T3 = 3ma
T2 - T3 = 2ma
T3
T2 = 2ma +T3 > T3
T1 - T2 = ma
T2
T1 = ma + T2 > T2
2m
m
T2
T1
T1 > T2 > T3
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Dr. Abdallah M. Azzeer
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