ap physics 1 exam review - mrs. libretto's physics...
TRANSCRIPT
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AP PHYSICS 1 EXAM REVIEW By Karyn Libretto (Northport High School)
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Format of Exam
¨ 3 hours longs ¤ Part 1 (90 minutes)
n 50 multiple- choice questions n Calculator & entire Reference Table allowed
¤ 10 minute break ¤ Part 2 (90 minutes)
n 5 free-response questions n 1 experimental design question n 1 qualitative / quantitative translation n 3 short-answer question
n Calculator & entire Reference Table allowed n The two sections are weighted equally
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Content of the Exam
¤ Newtonian Mechanics n Kinematics n Newton’s Laws of Motion n Torque n Rotational Motion & Angular Momentum n Gravitation & Circular Motion n Work, Energy & Power n Linear Momentum n Oscillations (SHM), Mechanical Waves & Sound n Introduction to Electric Circuits
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¨ We have A LOT to cover so please help me set the pace! ¤ If you understand and want me to move faster say
so! ¤ If you don’t understand tell me to slow down or
repeat. n I don’t mind!
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Topic #1: Methods & Tools
!
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Metric & Prefix Conversions
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A student wishing to determine experimentally the acceleration g due to gravity has an apparatus that holds a small sphere above a recording plate, as shown at right. When the sphere is released, a timer automatically begins recording the time of fall. The timer automatically stops when the sphere strikes the recording plate. The student measures the time of fall for different values of the distance D shown at right and records the data in the table. These data points are also plotted on the graph.
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Graphing: Best Fit Line
¨ Line or curve with equal amount of points located above and below.
a) On the grid above, sketch the smooth curve that best represents the student’s data.
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Graphing: Best Fit Line
¨ Line or curve with equal amount of points located above and below.
a) On the grid above, sketch the smooth curve that best represents the student’s data.
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Graphing: Linearization
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Graphing: Linearization
¨ The student can use these data for distance D and time t to produce a second graph from which the acceleration g due to gravity can be determined.
b) If only the variables D and t are used, what quantities should the student graph in order to produce a linear relationship between the two quantities?
Since distance & time are related by
d = ½ gt2, you should graph
distance vs (time)2
(You may also graph √d vs t)
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Graphing: Linearization
Time2 (s2) 0.196 .1024 .2116 .3481 .3969
Dist
ance
(m)
2.5
2.0
1.5
1.0
0.5
0 0.1 0.2 0.3 0.4 0.5 0.6 Time2 (s2)
c) On the grid below, plot the data points for the quantities you have identified in part (b), and sketch the best straight-line fit to the points. Label your axes and show the scale that you have chosen for the graph.
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Graphing: Physical Significance of Slope & Area under Curve
Velo
city
(m/s
)
Time (s)
slope =
ΔyΔx
= ΔvΔt
= ?
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Graphing: Physical Significance of Slope & Area under Curve
Velo
city
(m/s
)
Time (s)
slope =
ΔyΔx
= ΔvΔt
= a = acceleration
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Graphing: Physical Significance of Slope & Area under Curve
Velo
city
(m/s
)
Time (s)
AreaΔ = 1/2 b i h = ?
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Graphing: Physical Significance of Slope & Area under Curve
Velo
city
(m/s
)
Time (s)
AreaΔ = 1/2 b i h
AreaΔ = 1/2 t i v = displacement
(d = v i t)
v = vi + vf
2 =
vf2
= 1/2 vf
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Graphing: Physical Significance of Slope & Area under Curve
Dist
ance
(m)
Time (s)
Pote
ntia
l Ene
rgy
(J)
Height (m)
Pres
sure
(Pa)
Volume (m3)
Forc
e (N
)
Time (s)
Forc
e (N
)
Spring Stretch (m) En
ergy
per
Pho
ton
(J)
Frequency (Hz)
?
? ?
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Graphing: Physical Significance of Slope & Area under Curve
Dist
ance
(m)
Time (s)
Pote
ntia
l Ene
rgy
(J)
Height (m)
Pres
sure
(Pa)
Volume (m3)
Forc
e (N
)
Time (s)
Forc
e (N
)
Spring Stretch (m) En
ergy
per
Pho
ton
(J)
Frequency (Hz)
? ?
?
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Graphing: Physical Significance of Slope & Area under Curve
Dist
ance
(m)
Time (s)
Pote
ntia
l Ene
rgy
(J)
Height (m)
Pres
sure
(Pa)
Volume (m3)
Forc
e (N
)
Time (s)
Forc
e (N
)
Spring Stretch (m) En
ergy
per
Pho
ton
(J)
Frequency (Hz)
? ?
?
-
Graphing: Physical Significance of Slope & Area under Curve
Dist
ance
(m)
Time (s)
Pote
ntia
l Ene
rgy
(J)
Height (m)
Pres
sure
(Pa)
Volume (m3)
Forc
e (N
)
Time (s)
Forc
e (N
)
Spring Stretch (m) En
ergy
per
Pho
ton
(J)
Frequency (Hz)
? ?
Work on Gas
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Graphing: Physical Significance of Slope & Area under Curve
Dist
ance
(m)
Time (s)
Pote
ntia
l Ene
rgy
(J)
Height (m)
Pres
sure
(Pa)
Volume (m3)
Forc
e (N
)
Time (s)
Forc
e (N
)
Spring Stretch (m) En
ergy
per
Pho
ton
(J)
Frequency (Hz)
?
Work on Gas
Impulse
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Graphing: Physical Significance of Slope & Area under Curve
Dist
ance
(m)
Time (s)
Pote
ntia
l Ene
rgy
(J)
Height (m)
Pres
sure
(Pa)
Volume (m3)
Forc
e (N
)
Time (s)
Forc
e (N
)
Spring Stretch (m) En
ergy
per
Pho
ton
(J)
Frequency (Hz)
?
Work on Gas
Impulse
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Graphing: Physical Significance of Slope & Area under Curve
Dist
ance
(m)
Time (s)
Pote
ntia
l Ene
rgy
(J)
Height (m)
Pres
sure
(Pa)
Volume (m3)
Forc
e (N
)
Time (s)
Forc
e (N
)
Spring Stretch (m) En
ergy
per
Pho
ton
(J)
Frequency (Hz)
Work on Gas
Impulse Work or PEs
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Graphing: Physical Significance of Slope & Area under Curve
Dist
ance
(m)
Time (s)
Pote
ntia
l Ene
rgy
(J)
Height (m)
Pres
sure
(Pa)
Volume (m3)
Forc
e (N
)
Time (s)
Forc
e (N
)
Spring Stretch (m) En
ergy
per
Pho
ton
(J)
Frequency (Hz)
Work on Gas
Impulse Work or PEs
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Graphing: Physical Significance of Slope & Area under Curve
Dist
ance
(m)
Time (s)
Pote
ntia
l Ene
rgy
(J)
Height (m)
Pres
sure
(Pa)
Volume (m3)
Forc
e (N
)
Time (s)
Forc
e (N
)
Spring Stretch (m) En
ergy
per
Pho
ton
(J)
Frequency (Hz)
Work on Gas
Impulse Work or PEs fo
Φ
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Graphing: Linearization
Dist
ance
(m)
2.5
2.0
1.5
1.0
0.5
0 0.1 0.2 0.3 0.4 0.5 0.6 Time2 (s2)
. . . and back to our graph
What is the physical significance of our slope?
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Graphing: Physical Significance of Slope d) Using the slope of your graph in part (c), calculate the acceleration g due to gravity in this experiment.
¨ Linear general equation: y = mx + b ¨ Specific equation: d = (4.94 m/s2) t2 + 0m ¨ Mathematical model: d = ½ gt2 (d = ½ at2)
¨ So. . . g = 2 (slope) = 2 (4.94 m/s2) = 9.88 m/s2 ( 9 – 11 m/s2 was acceptable on exam)
slope =
ΔyΔx
= 1.0 m - 0.2 m
.212 s2 - .05 s2 = 4.94 m/s2
slope =
ΔyΔx
= dt2
= 1/2 g
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Error
¨ Systematic Error - an error associated with a particular instrument or experimental technique that causes the measured value to be off by the same amount each time.
¤ affects the accuracy of results ¤ can be eliminated by fixing source of error ¤ shows up as non-zero y-intercept on a graph
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Error
¨ Random Uncertainty - an uncertainty produced by unknown and unpredictable variations in the experimental situation.
¤ affects the precision of results ¤ can be reduced by taking repeated trials but not eliminated ¤ shows up as error bars on a graph
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Error
e) State one way in which the student could improve the accuracy of the results if the experiment were to be performed again. Explain why this would improve the accuracy.
¨ The student could increase the number of trials for each distance value and then averages the results. ¤ This reduces personal and random errors
¨ The student may perform the experiment in a vacuum to reduce air resistance. ¤ This reduces systematical errors.
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Kinematics
¨ Average Speed (v)– the rate of change of distance
¨ Average Velocity ( )– the rate of change of displacement
v = d
t v =
v + v02
v = x - x0t - t0
= Δx
Δt
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Kinematics
¨ When is the distance traveled by an object equal to its displacement?
¨ When is the average speed of an object equal to
the magnitude of its average velocity?
¨ When can an object be traveling at a constant speed but not at a constant velocity?
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Kinematics
¨ When is the distance traveled by an object equal to its displacement?
When it travels in a straight line ¨ When is the average speed of an object equal to
the magnitude of its average velocity?
¨ When can an object be traveling at a constant speed but not at a constant velocity?
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Kinematics
¨ When is the distance traveled by an object equal to its displacement?
When it travels in a straight line ¨ When is the average speed of an object equal to
the magnitude of its average velocity? When it travels in a straight line
¨ When can an object be traveling at a constant speed but not at a constant velocity?
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Topic 1: Newtonian Mechanics
¨ When is the distance traveled by an object equal to its displacement?
When it travels in a straight line ¨ When is the average speed of an object equal to
the magnitude of its average velocity? When it travels in a straight line
¨ When can an object be traveling at a constant speed but not at a constant velocity?
When it is turning
Kinematics
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Topic 1: Newtonian Mechanics
¨ Two friends bicycle 3.0 kilometers north and then turn to bike 4.0 kilometers west in 30. Minutes. ¤ What is their average speed?
¤ What is their average velocity?
Kinematics
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Topic 1: Newtonian Mechanics
¨ Two friends bicycle 3.0 kilometers north and then turn to bike 4.0 kilometers west in 30. Minutes. ¤ What is their average speed?
¤ What is their average velocity?
Kinematics
v = d
t =
7.0 km0.50 hr
= 14 km/hr
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Topic 1: Newtonian Mechanics
¨ Two friends bicycle 3.0 kilometers north and then turn to bike 4.0 kilometers west in 30. Minutes. ¤ What is their average speed?
¤ What is their average velocity?
Kinematics
v = d
t =
7.0 km0.50 hr
= 14 km/hr
v = Δx
t =
x2 + y2
t =
(4.0 km)2 + (3.0 km)2
0.50 hr =
5.0 km0.50 hr
= 10. km/hr at 53° west of north
θ = tan−1 O
A = tan−1
4.0 km3.0 km
= 53°
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Topic 1: Newtonian Mechanics
Kinematics ¨ Acceleration (a) – the rate of change of velocity
a = v - v0t - t0
= Δv
Δt
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Topic 1: Newtonian Mechanics
¨ What are the three different ways an object can accelerate?
¨ If an object is accelerating, is its. . . ¤ speed changing? ¤ velocity changing?
¨ Can an object have a velocity but no acceleration?
¨ Can an object have acceleration but no velocity?
Kinematics
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Topic 1: Newtonian Mechanics
¨ What are the three different ways an object can accelerate? Speed up, slow down, & turn
¨ If an object is accelerating, is its. . . ¤ speed changing? ¤ velocity changing?
¨ Can an object have a velocity but no acceleration?
¨ Can an object have acceleration but no velocity?
Kinematics
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Topic 1: Newtonian Mechanics
¨ What are the three different ways an object can accelerate? Speed up, slow down, & turn
¨ If an object is accelerating, is its. . . ¤ speed changing? maybe ¤ velocity changing?
¨ Can an object have a velocity but no acceleration?
¨ Can an object have acceleration but no velocity?
Kinematics
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Topic 1: Newtonian Mechanics
¨ What are the three different ways an object can accelerate? Speed up, slow down, & turn
¨ If an object is accelerating, is its. . . ¤ speed changing? maybe ¤ velocity changing? Yes
¨ Can an object have a velocity but no acceleration?
¨ Can an object have acceleration but no velocity?
Kinematics
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Topic 1: Newtonian Mechanics
¨ What are the three different ways an object can accelerate? Speed up, slow down, & turn
¨ If an object is accelerating, is its. . . ¤ speed changing? maybe ¤ velocity changing? Yes
¨ Can an object have a velocity but no acceleration? Yes, if it’s traveling at constant velocity (cruise control)
¨ Can an object have acceleration but no velocity?
Kinematics
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Topic 1: Newtonian Mechanics
¨ What are the three different ways an object can accelerate? Speed up, slow down, & turn
¨ If an object is accelerating, is its. . . ¤ speed changing? maybe ¤ velocity changing? Yes
¨ Can an object have a velocity but no acceleration? Yes, if it’s traveling at constant velocity (cruise control)
¨ Can an object have acceleration but no velocity? Yes, at the instant its motion starts or a vertical projection at
is apex (changing direction)
Kinematics
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Topic 1: Newtonian Mechanics
¨ Is a negative acceleration the same thing as deceleration?
Kinematics
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Topic 1: Newtonian Mechanics
¨ Is a negative acceleration the same thing as deceleration? No / Sometimes – deceleration means slowing down but negative
acceleration means accelerating in the negative direction. (If an object is moving to right and slowing down it is decelerating & has negative
acceleration)
General Rule ¨ If velocity and acceleration are in the same direction, than the object
is speeding up. ¨ If velocity and acceleration are in opposite directions, than the
object is slowing down.
Kinematics
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Topic 1: Newtonian Mechanics
Kinematics
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Velocity vs. Time Graphs
Forwards Backwards
Slope is above the x-axis
Slope is below the x-axis
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Topic 1: Newtonian Mechanics
Kinematics
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Topic 1: Newtonian Mechanics
Kinematics AB BC
CD DE
AB DE
BD
AB CD
BC DE
C
A E
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Topic 1: Newtonian Mechanics
Kinematics
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Topic 1: Newtonian Mechanics
Kinematics AB DE
BD
NONE
NONE
NONE
NONE
B D
A C E What would a position time graph of an accelerating object look like?
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Topic 1: Newtonian Mechanics
Kinematics
!
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Topic 1: Newtonian Mechanics
Kinematics
!
!
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Topic 1: Newtonian Mechanics
Kinematics
The graph above shows the velocity versus time for an object moving in a straight line. At what time after t = 0 does the object again pass through its initial position?
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Topic 1: Newtonian Mechanics
Kinematics
Negative displacement
The graph above shows the velocity versus time for an object moving in a straight line. At what time after t = 0 does the object again pass through its initial position?
Positive displacement of equal area
Answer: Between 1 and 2 seconds
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Topic 1: Newtonian Mechanics
¨ The Law of Falling Bodies - in the absence of air resistance, all bodies near the surface of the Earth fall with the same constant acceleration (g). ¤ g = 9.8 m/s2 near the surface of the Earth ¤ direction for vector g = - depends on frame of
reference ¤ Approximate g as10 m/s2 on MC section of exam
Kinematics
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Topic 1: Newtonian Mechanics
Kinematics
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Topic 1: Newtonian Mechanics
Kinematics
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Topic 1: Newtonian Mechanics
¨ A stone is dropped from rest from the top of a tall building. After 3.00 s of free-fall, what is the displacement y of the stone? What is its velocity?
Kinematics
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Topic 1: Newtonian Mechanics
¨ A stone is dropped from rest from the top of a tall building. After 3.00 s of free-fall, what is the displacement y of the stone? What is its velocity?
Kinematics
y = y0 + v0t + ½ at2
y = ½ (-9.8 m/s2)(3.0 s)2
y = - 44.1 m = 44.1 m down v = v0 + at
v = (-9.8 m/s2)(3.0 s)
v = -29.4 m/s
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Topic 1: Newtonian Mechanics
Kinematics
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Topic 1: Newtonian Mechanics
¨ Examining half of flight ¤ ttop = ½ total time ¤ vf = 0 m/s
¨ Examining entire flight ¤ Use ttotal ¤ vf = - vi
Kinematics
Always treat g as being
negative for these
problems!!!
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Topic 1: Newtonian Mechanics
y = y0 + v0t + 1/2at2
Kinematics
Final displacement is negative (downward)
Initial position is zero
g is negative
t = - (v0) ± (v0)
2 - 4 12
a⎛⎝⎜
⎞⎠⎟
(y)
2 12
a⎛⎝⎜
⎞⎠⎟
= - (v0) ± (v0)
2 - 2(a)(y)(a)
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Topic 1: Newtonian Mechanics
Kinematics Analyze the horizontal and vertical components of the projectile.
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Topic 1: Newtonian Mechanics
Kinematics Analyze the horizontal and vertical components of the projectile.
v =
dt
y = y0 +v0t + 1/2 at2
&v = v0 +at
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Topic 1: Newtonian Mechanics
Kinematics What is the initial vertical velocity of a projectile launched horizontally?
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Topic 1: Newtonian Mechanics
Kinematics
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Topic 1: Newtonian Mechanics
Kinematics
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Topic 1: Newtonian Mechanics
Kinematics
vix = v cos θ
viy = v sin θ
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Topic 1: Newtonian Mechanics
¨ Maximum height occurs at ¤ _____ total time. ¤ _____ degrees.
¨ Maximum range occurs at ¤ _____ total time. ¤ _____ degrees.
¨ Maximum hang time occurs at ¤ _____ degrees.
Kinematics
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Topic 1: Newtonian Mechanics
¨ Maximum height occurs at ¤ 1/2 total time. ¤ 90 degrees.
¨ Maximum range occurs at ¤ total time. ¤ 45 degrees.
¨ Maximum hang time occurs at ¤ 90 degrees.
Kinematics
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Topic 1: Newtonian Mechanics
Statics & Dynamics
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Topic 1: Newtonian Mechanics
Newton’s Laws of Motion ¨ An object continues in uniform motion in a straight line or remains at
rest unless a resultant (net) external force acts on it. ¨ When unbalanced forces act on an object, the object will accelerate
in the direction of the resultant (net) force. The acceleration will be directly proportional to the net force and inversely proportional to the object’s mass.
¨ When two bodies A and B interact, the force that A exerts on B is
equal and opposite to the force that B exerts on A.
Statics & Dynamics
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Topic 1: Newtonian Mechanics
Statics & Dynamics
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Topic 1: Newtonian Mechanics
Statics & Dynamics
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Topic 1: Newtonian Mechanics
Statics & Dynamics
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Topic 1: Newtonian Mechanics
Statics & Dynamics
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Topic 1: Newtonian Mechanics
¨ Mass (m) – the amount of matter that comprises an object and determines its resistance to a change in its motion (inertia) ¤ Units – kg
¨ Weight (W or Fg) – the magnitude of the force of gravity acting upon an object. ¤ Units - N
Statics & Dynamics
g =
!Fgm
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¨ Gravitational mass – property of mass that produces a gravitational field. ¤ measured with the use of a double-pan or triple-beam
balance
¨ Inertial mass – an object’s resistance to acceleration ¤ measured with the use of an inertial balance, or spring-
loaded pan
a = ∑
!F
m = !Fnetm
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Topic 1: Newtonian Mechanics
¨ Density (ρ) - a measurement of how tightly matter is compacted.
¤ Units – kg/m3 ρ = m
V
V = ℓwh V = πr2ℓ V = 4
3πr3
Rectangular Solid Cylinder Sphere
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Topic 1: Newtonian Mechanics
¨ Normal Force (FN)– an electromagnetic repulsive contact force between two objects, acting perpendicular to their surface of contact.
n Tensional Force (FT) – the magnitude of the pulling force exerted by a string, cable, chain, or similar object on another object.
Statics & Dynamics
What’s the normal force acting on these block?
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Topic 1: Newtonian Mechanics
Statics & Dynamics
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Newton’s Laws of Motion
¨ Frictional Force (Ff) – an oppositional force due to an electromagnetic force of attraction between surfaces.
!Ff ≤ u
!Fn
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Newtonian Mechanics
¨ Net Force (ΣF, Fnet) – the sum of all forces acting on an object (resultant force)
a = ∑
!F
m = !Fnetm
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General Strategy: ¤ One body problem or two body
problem? ¤ Sketch and label a free-body
diagram. ¤ Decide if the object is in
equilibrium or not. ¤ Break forces into perpendicular
components, if appropriate. ¤ Write a net force equation for
each dimension and solve.
¨ Equilibrium:
ΣF = 0
¨ Non-equilibrium:
ΣF = m a
Newton’s Laws of Motion
How?
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Topic 1: Newtonian Mechanics
¨ 2. A 12.0-kg lantern is suspended from the ceiling by two wires, each at an angle of 300 from the vertical. What is the tension in each wire?
Statics & Dynamics
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Topic 1: Newtonian Mechanics
¨ 2. A 12.0-kg lantern is suspended from the ceiling by two wires, each at an angle of 300 from the vertical. What is the tension in each wire?
Statics & Dynamics
Σ F = 0 N(FT cos θ + FT cos θ ) + -Fg = 0 N2FT cos θ = mg
FT = (12 kg)(9.8 m/s2)
2 i cos 30°FT = 68 N
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Topic 1: Newtonian Mechanics
¨ A sled is pulled with a force at an angle on a frictionless floor. Write an expression for the normal force
Statics & Dynamics
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Topic 1: Newtonian Mechanics
¨ A sled is pulled with a force at an angle on a frictionless floor. Write an expression for the normal force
Statics & Dynamics
FTy
Fg
FN
ΣFy = 0 N FTy + FN + -Fg = 0 N
FN = Fg - FT sinθ FN = mg - FT sinθ
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Topic 1: Newtonian Mechanics
Elevators
¨ In each case, the scale will read the normal or reaction force, not the weight ¨ Apparent weightlessness: sensation due to lack of
normal force
Statics & Dynamics
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Topic 1: Newtonian Mechanics
¨ Calculate the acceleration of the man. Statics & Dynamics
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Topic 1: Newtonian Mechanics
¨ Calculate the acceleration of the man. Statics & Dynamics
Σ F = ma FN + -Fg = m a 1000N + (–700 N) = (70 kg) a a = +4.3 m/s2
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Topic 1: Newtonian Mechanics
Two Body Problems ¨ Two masses are hung as shown over a frictionless, massless
pulley. Determine their acceleration and the tension in the rope.
Statics & Dynamics
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Topic 1: Newtonian Mechanics
Two Body Problem
Statics & Dynamics
Both blocks accelerate together, so treat as one large block.
– +
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Topic 1: Newtonian Mechanics
Two Body Problem
Statics & Dynamics
Both blocks accelerate together, so treat as one large block.
– +
Σ F = (m1 + m2) aFg1 + -Fg2 = (m1+ m2) a
a = m1g + -(m2g)
(m1+ m2)
a = 7 kg(9.8 m/s2)⎡⎣ ⎤⎦ + - 9 kg(9.8 m/s
2)⎡⎣ ⎤⎦(7 kg + 9 kg)
a = -1.23 m/s2
Negative = left
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Topic 1: Newtonian Mechanics
Two Body Problem
Statics & Dynamics
Σ F = (m1 + m2) aFg1 + -Fg2 = (m1+ m2) a
a = m1g + -(m2g)
(m1+ m2)
a = 7 kg(9.8 m/s2)⎡⎣ ⎤⎦ + - 9 kg(9.8 m/s
2)⎡⎣ ⎤⎦(7 kg + 9 kg)
a = -1.23 m/s2
Negative = counterclockwise
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Topic 1: Newtonian Mechanics
Two Body Problem
Statics & Dynamics
To find tension pick ONE block!
Σ F = m1 aFT1 + -Fg = m1 a
FT1 = (m1 a) + Fg = (m1a) + (m1g)
FT1 = 7 kg (+1.23 m/s2)⎡⎣ ⎤⎦ + 7 kg (9.8 m/s
2)⎡⎣ ⎤⎦FT1 = 77 N = 77 N up
–
+
Block 1 accelerates up so a is +
Block 2 accelerates down so a is -
Σ F = m2 aFT2 + -Fg = m2 a
FT2 = (m2 a) + Fg = (m2a) + (m2g)
FT2 = 9 kg (-1.23 m/s2)⎡⎣ ⎤⎦ + 9 kg (9.8 m/s
2)⎡⎣ ⎤⎦FT2 = 77 N = 77 N up
FT = m (a + g)
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Topic 1: Newtonian Mechanics
Two Body Problem with Friction
¨ Find the acceleration of this system and the tension in the rope if the coefficient of kinetic friction between the box and the tabletop is 0.20.
Statics & Dynamics
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Topic 1: Newtonian Mechanics
Two Body Problem with Friction
Statics & Dynamics
Σ F = (mA + mB)a-FgB + FfA = (mA + mB)a
-FgB + µFNA = (mA + mB)a
-FgB + µ -FgA = (mA + mB)a
a = -FgB + µ(-mA i g)
(mA + mB)
a = -mBg + µ(-mA i g)
(mA + mB)
a = -19.6 N + 0.20(5.0 kg i 9.8 m/s2)
(5.0 kg + 2.0 kg)a = -1.4 m/s2
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Topic 1: Newtonian Mechanics
Two Body Problem with Friction
Statics & Dynamics
Σ F = maFT - Fg= m a
FT= m g + m a
FT= m (g + a)
FT= 2.0 kg (9.8 m/s2 + -1.4 m/s2)
FT= 16.8 N
Block B accelerates down so a is -
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Topic 1: Newtonian Mechanics
Inclined Planes Statics & Dynamics
Fg⊥
Fg II
Fg
Fg - weight
Fg = mg
Fg II - pulls object downhill
Fg II = Fg sin θ
Fg⊥ - pulls object into hill
Fg⊥ = Fg cos θ
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Topic 1: Newtonian Mechanics
¨ A girl accelerates down a frictionless slide inclined at 38°. Determine the acceleration of the girl.
Statics & Dynamics
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Topic 1: Newtonian Mechanics
¨ A girl accelerates down a frictionless slide inclined at 38°. Determine the acceleration of the girl.
Statics & Dynamics
Σ F = ma
a = Σ Fm
a = -FgIIm
= m i g i sinθ
ma = -(9.8 m/s2)(sin 38°)a = 6.0 m/s2
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Topic 1: Newtonian Mechanics
¨ Write an expression for the steepest angle at which a box can rest on a hill before sliding (the angle of repose).
Statics & Dynamics
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Topic 1: Newtonian Mechanics
¨ Write an expression for the steepest angle at which a box can rest on a hill before sliding (the angle of repose).
Statics & Dynamics
Ff = µ FNmg sinθ = µ mg cosθ
µ = sinθcosθ
µ = tanθθ = tan-1µs
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¨ Period ( T ) – time it takes to complete one cycle around a circle
Gravitation & Circular Motion
T =
time# revolutions
v =
dt
= circumference
period =
2rT
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¨ The direction of the object’s instantaneous velocity is always tangent to the circle in the direction of motion.
¨ Its instantaneous velocity is always perpendicular to a radius drawn to the point of tangency.
¨ Since the direction of the object’s motion is always changing, its velocity is always changing therefore the object is always accelerating.
¨ The force causing the change in direction is an inwardly directed (centripetal) force.
¨ According to Newton’s second law of motion, an object always accelerates in the direction of net force so the acceleration must also be inwardly directed (centripetal).
¨ If the object is constantly subject to a net force and always accelerating, then it can never be in equilibrium.
Gravitation & Circular Motion
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Gravitation & Circular Motion
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Fgrav
Fnorm Ffric
Ftens
Gravitation & Circular Motion
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¨ Determine the maximum speed at which a car can safely negotiate an unbanked turn.
Gravitation & Circular Motion
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¨ Determine the maximum speed at which a car can safely negotiate an unbanked turn.
Gravitation & Circular Motion
FF = Fc
µ iFN = mv2
r
v = µ img i r
m
v = µ i g i r
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¨ Write an expression for the speed of the chair in terms of g, L, and θ .
Gravitation & Circular Motion
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¨ Write an expression for the speed of the chair in terms of g, L, and θ .
Gravitation & Circular Motion
FTy = mg AND FTx = mv2
r
FT cos θ = mg AND FT sinθ = mv2
rmg
cosθ = FT =
mv2
r i sinθ
m i g i r i sinθm(cosθ )
= v
v = L i g i tanθ
v = (Lsinθ ) i g i tanθ
r = L sinθ
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Topic 1: Newtonian Mechanics
¨ Write an expression for the speed of the chair in terms of g, L, and θ .
Gravitation & Circular Motion
r = L sinθ This formula shows that the angle and the radius depend on the velocity alone and not the mass of the chairs. A good job too if you think about an actual ride - if this wasn't true the wires would all get tangled up if the people in the chairs had different masses.
v = (Lsinθ ) i g i tanθ
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¨ What is the minimum coefficient of static friction that must exist between a rider's back and the wall, if the rider is to remain in place when the floor drops away?
Gravitation & Circular Motion
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¨ What is the minimum coefficient of static friction that must exist between a rider's back and the wall, if the rider is to remain in place when the floor drops away?
Gravitation & Circular Motion
FFY = Fg AND Fcx = FN
FFY = mg AND FN = mv2
rµ iFN = mg
µ i mv2
r⎛⎝⎜
⎞⎠⎟
= mg
µ = mgrmv2
µ = r i gv2
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Gravitation & Circular Motion
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Gravitation & Circular Motion
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Gravitation & Circular Motion
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Gravitation & Circular Motion
Treat inward as + (positive)
∑F = mv2
r
Fn - Fg = mv2
r
Fn = mv2
r + Fg
∑F = mv2
r
Fn + Fg = mv2
r
Fn = mv2
r - Fg
Bottom: Top:
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¨ Write an expression for the minimum speed the cyclist needs in order to complete the loop.
Gravitation & Circular Motion
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¨ Write an expression for the minimum speed the cyclist needs in order to complete the loop.
Calculate at top, which is the critical point for making the loop.
Gravitation & Circular Motion
∑F = FN + FgFc = FN + Fgmv2
r = FN + mg
FN = mv2
r - mg
At minimum or critical speed FN = 0
0 = mv2
r - mg
mv2
r = mg
v = r i g
Roller coasters need to EXCEED the critical velocity, whereas dryers need a lower velocity
to allow the cloths to fall and >luff.
You can use the above equation in reverse to see how fast you should go to loose contact with the road when driving over a hill to feel weightless.
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Gravitation & Circular Motion
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Gravitation & Circular Motion
¨ Newton’s Law of Universal Gravitation - Every point mass attracts every single other point mass by a force pointing along the line intersecting both points.
!Fg =
Gm1m2r2
G = 6.67 x 10-11 m3/kg s2
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Gravitation & Circular Motion
¨ A force of 400 N is measured between two objects. Determine the new force of attraction when:
¤ both masses double.
¤ the separation distance doubles.
¤ the separation is decreased to 1/3rd of its original value
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Gravitation & Circular Motion
¨ A force of 400 N is measured between two objects. Determine the new force of attraction when:
¤ both masses double.
¤ the separation distance doubles.
¤ the separation is decreased to 1/3rd of its original value
Fgrav=
Gm1m2r2
= 1 ⋅2 ⋅212
= 4 = 1600 N
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Gravitation & Circular Motion
¨ A force of 400 N is measured between two objects. Determine the new force of attraction when:
¤ both masses double.
¤ the separation distance doubles.
¤ the separation is decreased to 1/3rd of its original value
Fgrav=
Gm1m2r2
= 1 ⋅2 ⋅212
= 4 = 1600 N
Fgrav=
Gm1m2r2
= 1 ⋅1 ⋅122
= 14
= 100 N
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Gravitation & Circular Motion
¨ A force of 400 N is measured between two objects. Determine the new force of attraction when:
¤ both masses double.
¤ the separation distance doubles.
¤ the separation is decreased to 1/3rd of its original value
Fgrav=
Gm1m2r2
= 1 ⋅2 ⋅212
= 4 = 1600 N
Fgrav=
Gm1m2r2
= 1 ⋅1 ⋅122
= 14
= 100 N
Fgrav=
Gm1m2r2
= 1 ⋅1 ⋅1(1 / 3)2
= 9 = 3600 N
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g - acceleration due to gravity or gravitational field strength
Gravitation & Circular Motion
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Gravitation & Circular Motion
The gravitational potential energy at an infinite distance away is ZERO. The gravitational potential energy near a planet is then
negative, since gravity does positive work as the mass approaches. This negative potential is indicative of a "bound state"; once a mass is near a large body, it is trapped until something can provide enough energy to allow it to escape.
Ug = -
Gm1m2r
ΔUg = mgΔy
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Topic 1: Newtonian Mechanics
¨ Orbital Speed of a Satellite ¤ Write an expression for this “orbital speed.”
Gravitation & Circular Motion
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Topic 1: Newtonian Mechanics
¨ Orbital Speed of a Satellite ¤ Write an expression for this “orbital speed.”
Gravitation & Circular Motion
FC = FgmSv
2
r =
GmEmSr2
v = GmE
r
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Topic 1: Newtonian Mechanics
Gravitation & Circular Motion
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Topic 1: Newtonian Mechanics
¨ How fast would this 20. meter diameter space station have to spin to simulate earth’s gravity?
Gravitation & Circular Motion
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Topic 1: Newtonian Mechanics
¨ How fast would this 20. meter diameter space station have to spin to simulate earth’s gravity?
Gravitation & Circular Motion
ac = v2
rv = r i a
v = (10. m)(9.8 m/s2) v = 9.9 m/s
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Topic 1: Newtonian Mechanics
Torque Gravitation & Circular Motion
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Torque
Torque
Sign convention: CW -, CCW +
τ = r⊥F = rF sinθ
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Topic 1: Newtonian Mechanics
¨ A 20. kg person and a 5.0 kg dog want to balance on a 8.0 m uniform 12. kg board that is centered on a pivot. If the dog sits at one end of the see-saw, how far away does the person have to sit to balance it?
Gravitation & Circular Motion
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Topic 1: Newtonian Mechanics
¨ A 20. kg person and a 5.0 kg dog want to balance on a 8.0 m uniform 12. kg board that is centered on a pivot. If the dog sits at one end of the see-saw, how far away does the person have to sit to balance it?
Gravitation & Circular Motion
Forces ΣF = 0
Torques
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Topic 1: Newtonian Mechanics
Torque 1. forces through the axis of rotation are not torques
2. don’t neglect the weight of the board (pole) if it is not supported at its center of mass – force and torque
3. use the see-saw law if the only torques are due to weight and the object is in rotational equilibrium
Gravitation & Circular Motion
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Topic 1: Newtonian Mechanics
¨ How much force is the person applying to hold up the end of the 12 kg board that is 8.0 meters long? How much normal force is the fulcrum applying?
Gravitation & Circular Motion
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Topic 1: Newtonian Mechanics
¨ How much force is the person applying to hold up the end of the 12 kg board that is 8.0 meters long? How much normal force is the fulcrum applying?
Gravitation & Circular Motion
1.0 m
FN FA = τA
Fg = τg
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Topic 1: Newtonian Mechanics
¨ How much force is the person applying to hold up the end of the 12 kg board that is 8.0 meters long? How much normal force is the fulcrum applying?
Gravitation & Circular Motion
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Topic 1: Newtonian Mechanics
¨ How much force is the person applying to hold up the end of the 12 kg board that is 8.0 meters long? How much normal force is the fulcrum applying?
Forces Torques
Gravitation & Circular Motion
ΣF = 0 N FA + FN + -‐FG = 0
Solve torques >irst
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Topic 1: Newtonian Mechanics
¨ How much force is the person applying to hold up the end of the 12 kg board that is 8.0 meters long? How much normal force is the fulcrum applying?
Forces Torques
Gravitation & Circular Motion
ΣF = 0 N FA + FN + -‐FG = 0
Solve torques >irst
Στ = 0 Nm -‐FGr1 + FAr2 = 0 Nm
FGr1 = FAr2 m g r1 = FAr2
(12 kg)(9.8 m/s2)(1.0 m) = FA(5.0 m) FA = 24 N
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Topic 1: Newtonian Mechanics
¨ How much force is the person applying to hold up the end of the 12 kg board that is 8.0 meters long? How much normal force is the fulcrum applying?
Forces Torques
Gravitation & Circular Motion
ΣF = 0 N FA + FN + -‐FG = 0
Solve torques >irst
24 N + FN – 120 N = 0
FN = 96 N
Στ = 0 Nm -‐FGr1 + FAr2 = 0 Nm
FGr1 = FAr2 m g r1 = FAr2
(12 kg)(9.8 m/s2)(1.0 m) = FA(5.0 m) FA = 24 N
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Topic 1: Newtonian Mechanics
¨ A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of plank hanging over the right support (see the drawing). To what distance x can a person who weights 450 N walk on the overhanging part of the plank before it just begins to tip?
Gravitation & Circular Motion
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Topic 1: Newtonian Mechanics
¨ A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of plank hanging over the right support (see the drawing). To what distance x can a person who weights 450 N walk on the overhanging part of the plank before it just begins to tip?
Gravitation & Circular Motion
As the plank just begins to tip, a normal force will cease to exist at the far left end. Thus, we can treat this like a see-saw
Στ = 0 Nm Σmr = Σmr m1r1 = m2r2
(22.5 kg)(1.4 m)= (45 kg)(x) r = 0.70 m