Transcript Lecture 5.2

Welcome back to Physics 215
Today’s agenda:
• More on free-body diagrams
and force components
• Applying Newton’s laws
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Current homework assignment
• HW4:
– Knight textbook -- Ch.5: 40, 42, 44, 56
– Ch.6: 26, 28
– Due Friday, Sept. 26th in recitation
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Newton’s Laws
First Law:
In the absence of external forces,
an object at rest remains at rest
and an object in motion continues
in motion with constant velocity.
Second Law: Fnet = ∑Fon object = m a
Third Law:
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FAB = - FBA (“action =
reaction”)
[regardless of type of force and of
motion of objects in question]
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Consider a person sitting on a chair. We can
conclude that the downward weight force on
the person (by the Earth) and the upward
normal force on the person (by the chair) are
equal and in opposite direction, because
1. the net force on the person must be zero
2. the two forces form a Newton’s third-law
pair
3. neither of the above explanations
4. both of the above explanations
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Free-body diagrams: Person sitting on chair
Chair
Person
Earth
(incomplete)
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There are two people facing each other, each
on a separate cart. If person A pushes on
person B, while person B does nothing, what
will be the resulting motion of the carts?
1. Cart A doesn’t move and Cart B moves
backwards
2. Cart B doesn’t move and Cart A moves
backwards
3. Both carts move in opposite directions
4. Neither cart moves
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A force P is applied by a hand to two blocks which are in contact on a frictionless,
horizontal table as shown in the figure. The blocks accelerate to the right. Block A
has a smaller mass than block B. (a) Draw free body diagrams for each block. (b)
Which block experiences the larger net force? (c) Suppose that initially the mass of
block A were half that of block B. If in a subsequent experiment the mass of block A
were doubled, by what factor would the acceleration change assuming the pushing
force remained constant?
B
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A block is held in place on a frictionless incline by a massless string,
as shown. The acceleration of the
block is
1.
2.
3.
4.
zero
straight down
down and to the left (along the incline)
not zero, but neither of 2 or 3
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A block is held in place on a frictionless incline
by a massless string, as shown. The
force on the block by the incline is
1.
2.
3.
4.
a normal force given by vector A.
a normal force given by vector B.
the force given by vector A, but it’s not a
normal force.
the force given by vector B, but it’s not a
normal force.
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Normal force
• Always perpendicular to surface of
contact
• Generic name given to contact force
between 2 objects
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Other forces
Besides normal force what other
forces are present for block on
inclined plane?
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Free-body diagram: Block on frictionless incline
• Show all forces exerted on the block.
• Do not show forces exerted by the block on anything else.
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Geometry…
y
vertical component
of NBP is NBPcos(q)
horizontal is NBPsin(q)
N
q
x
T
q
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W
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Force components: Block on frictionless incline
F = 0 implies all components
of F are zero! Horizontal and
vertical
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When the block is held in place
as shown, is the magnitude of
the normal force exerted on the
block by the incline
1. greater than the magnitude of the weight
force (on the block by the Earth),
2. equal to the magnitude of the weight, or
3. less than the magnitude of the weight.
4. Can’t tell.
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Solving system
• Vertical equilibrium:
NBPcos(q) + WBE = 0
• Horizontal equilibrium:
NBPsin(q) + TBR = 0
2 equations  solve for NBP and TBE in
terms of WBE and q
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What have we learned?
• If at rest – net force = 0, since a = 0 !
• Write down FBD – identify all forces
present
• Take force components in 2 (in 2D)
directions to find unknowns
• Don’t plug in numbers until end
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The string that holds the block breaks,
so there is no more tension force
exerted on the block.
Will the magnitudes of either of the other two forces on
the block (i.e., the weight and the normal force) change?
1.
2.
3.
4.
Both weight and normal force will change.
Only the weight force will change.
Only the normal force will change.
Neither one of the two forces will change.
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Discussion
• Say normal does not change. What will
be the new net force? Same as old, but
“minus” the tension, i.e. straight to the
left (and not zero).
• So what would the block do?
Accelerate. OK -- Straight to the left?
NO, can’t be correct answer.
• Acceleration is down the incline, so
normal gets smaller.
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Geometry (2)
N
q
q
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component of WBE
at 900 to plane
is WBEcos(q)
component along
plane is WBEsin(q)
Wq
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Force components: Block on frictionless incline
• The string that holds the block breaks.
• What does happen?
• Take components now along and
perpendicular to incline
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Solving for second case
• Since acceleration is down incline,
component of net force at 90 degrees to
slope is zero.
NBP + WBEcos(q) = 0
• Using second law applied to components
along incline:
a = WBEsin(q)/mblock
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Conclusion
• Normal forces can change when small
changes are made to the situation.
• Can choose any 2 directions to find
force components, but it pays to pick
ones that simplify equations
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Suppose we chose to resolve forces
vertically and horizontally?
•
Vertically:
(1) Ncos(q) + W = -masin(q)
•
Horizontally:
(2) Nsin(q) = ma cos(q)
•
Multiply (1) by cos(q), (2) by sin(q) and
add:
 N + Wcos(q) = 0 as before!!
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If the block lies 0.5 m up such a plane inclined
at an angle of 30 how long will it take for the
block to reach the ground? (take g = 10 m/s2)
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“real” inclined plane
• For a real incline, we expect for small
angles that the block still does not move
when string breaks – why?
• Friction – force up slope that opposes
motion – more later…
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Summary
• To solve problems in mechanics, identify all
forces and draw free body diagrams for all
objects
• If more than one object, use Newton’s Third
law to reduce number of independent forces
• Use Newton’s Second law for all components
of net force on each object
• Choose component directions to simplify
equations
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Weight, mass, and acceleration
•
What does a weighing scale
``weigh’’?
•
Does it depend on your frame of
reference?
•
Consider elevators….
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A person is standing on a bathroom scale while riding
an express elevator in a tall office building. When the
elevator is at rest, the scale reads about 160 lbs.
While the elevator is moving, the reading is frequently
changing, with values ranging anywhere from about
120 lbs to about 200 lbs.
At a moment when the scale shows the maximum
reading (i.e., 200 lbs) the elevator
1.
2.
3.
4.
must be going up
must be going down
could be going up or going down
I’m not sure.
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Motion of elevator
Motion of elevator
(if a < 0 )
(if a > 0 )
• Moving upward and
slowing down,
• Moving upward and
speeding up,
OR
OR
• Moving downward
and speeding up.
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• Moving downward
and slowing down.
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A person is standing on a bathroom scale while riding
an express elevator in a tall office building. When the
elevator is at rest, the scale reads about 160 lbs.
While the elevator is accelerating, a different reading is
observed, with values ranging anywhere from about
120 lbs to about 200 lbs.
At a moment when the scale shows the maximum
reading (i.e., 200 lbs) the acceleration of the elevator is
approximately
1.
2.
3.
4.
1 m/s2
2.5 m/s2
5 m/s2
12.5 m/s2
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Conclusions
• Scale reads magnitude of normal force |NPS|
• Reading on scale does not depend on
velocity (principle of relativity again!)
• Depends on acceleration only
* a > 0  normal force bigger
* a < 0  normal force smaller
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Reminder of free-fall experiment
• Objects fall even when there is no atmosphere
(i.e., weight force is not due to air pressure).
• When there is no “air drag” things fall “equally
fast.” i.e. same acceleration
• From Newton’s 2nd law, a = W/m is
independent of m -- means W = mg
• The weight (i.e., the force that makes objects in
free fall accelerate downward) is proportional to
mass.
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Inertial and gravitational mass
• Newton’s second law:
F = mI a
• For an object in “free fall”
W = mG g
• If a independent of mI, must have
mI = mG
 Principle of equivalence
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Reading assignment
• Forces, Newton’s Laws of Motion,
Weight & Friction
• Ch. 5-7 in textbook
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