Physics 207 – Lecture 10
Lecture 10
l
Today:
v Review session
Assignment: For Monday, Read through Chapter 8
There will be a reading quiz posted at Mastering Physics.
Exam Thursday, Oct. 6th from 7:15-8:45 PM Chapters 1-6,7
One 8½ X 11 hand written note sheet and a calculator (for trig.)
Physics 207: Lecture 9, Pg 1
Textbook Chapters
l
l
l
l
l
l
l
Chapter 1 Concept of Motion
Chapter 2 1D Kinematics
Chapter 3 Vector and Coordinate Systems
Chapter 4 Dynamics I, Two-dimensional motion
Chapter 5 Forces and Free Body Diagrams
Chapter 6 Force and Newton’s 1st and 2nd Laws
Chapter 7 Newton’s 3rd Law
Exam will reflect most key points (but not all)
25-30% of the exam will be more conceptual
70-75% of the exam is problem solving
Physics 207: Lecture 9, Pg 2
Page 1
Physics 207 – Lecture 10
Example with pulley
l
•
•
T4
A mass M is held in place by a
force F. Find the tension in each
segment of the massless ropes
and the magnitude of F.
v Assume the pulleys are
massless and frictionless.
The action of a massless
frictionless pulley is to change the
direction of a tension.
This is an example of
static equilibrium.
T1
T3
T2
F
T5
M
Physics 207: Lecture 9, Pg 3
Example with pulley
l
•
•
•
•
•
A mass M is held in place by a force F.
Find the tension in each segment of the
rope and the magnitude of F.
v Assume the pulleys are massless
and frictionless.
v Assume the rope is massless.
The action of a massless frictionless
pulley is to change the direction of a
tension.
Here F = T1 = T2 = T3 = T
T4
T1
T3
T2
F
Equilibrium means Σ F = 0 for x, y & z
For example: y-dir ma = 0 = T2 + T3 – T5
and ma = 0 = T5 – Mg
So T5 = Mg = T2 + T3 = 2 F
T = Mg/2
T5
M
Physics 207: Lecture 9, Pg 4
Page 2
Physics 207 – Lecture 10
Another example with a pulley
Three blocks are connected on the table as shown. The
table is frictionless & the masses are m1 = 4.0 kg, m2 =
1.0 kg and m3 = 2.0 kg.
N
m2
T1
T1
m1g
T3
m1
m2g
m3
m3g
(A) 3 Free Body Diagrams
Physics 207: Lecture 9, Pg 5
Another example with a pulley
Three blocks are connected on the table as shown. The
table is frictionless & the masses are m1 = 4.0 kg, m2 =
1.0 kg and m3 = 2.0 kg.
N
T
T
1
(1)
(2)
(3)
m1 a1y= -m1g + T1
m2 a2yx= -T1 + T3
m3 a3y= -m3g + T3
3
m2
T1
m1g
m1
T1
T3
m2g
m3
m3g
Let a = a1y= a12y= - a3y or m3 a= m3g - T3
Add (1) & (2) (m1 +m2)a = -m1g + T3
Now add (3)
(m1 +m2+m3)a = -m1g + m3 g
a = (-m1g + m3 g)/(m1 +m2+m3)= -20 / 7 m/s2
Physics 207: Lecture 9, Pg 6
Page 3
Physics 207 – Lecture 10
Problem recast as 1D motion
Three blocks are connected on the table as shown. The
center table has a coefficient of kinetic friction of µK=0.40,
the masses are m1 = 4.0 kg, m2 = 1.0 kg and m3 = 2.0 kg.
m1g
N
T1
m1
m3g
m3
T3
m2
ff
frictionless
frictionless
m2g
m1g > m3g and m1g > (µkm2g + m3g)
and friction opposes motion (starting with v = 0)
so ff is to the right and a is to the left (negative)
Physics 207: Lecture 9, Pg 7
Another example with a pulley
Three blocks are connected on the table as shown. The
table has a coefficient of kinetic friction of µK=0.40, the
masses are m1 = 4.0 kg, m2 = 1.0 kg and m3 = 2.0 kg.
N
m2
T1
T1
m1g
T3
m1
m2g
m3
m3g
(A) FBD (except for friction)
(B) So what about friction ?
Physics 207: Lecture 9, Pg 8
Page 4
Physics 207 – Lecture 10
Problem recast as 1D motion
Three blocks are connected on the table as shown. The
center table has a coefficient of kinetic friction of µK=0.40,
the masses are m1 = 4.0 kg, m2 = 1.0 kg and m3 = 2.0 kg.
m1g
m1
T1
T1
N
m2
T3
T3
ff
frictionless
m3g
m3
frictionless
m2g
x-dir: 1.
Σ Fx = m2a = µk m2g
- T1 + T3
m3a = m3g - T3
m1a = − m1g + T1
Add all three: (m1 + m2 + m3) a = µk m2g+ m3g – m1g
Physics 207: Lecture 9, Pg 9
Analyzing motion plots
The graph is a plot of velocity versus time for an object. Which of the
following statements is correct?
A The acceleration of the object is zero.
B The acceleration of the object is constant.
C The acceleration of the object is positive and increasing in magnitude.
D The acceleration of the object is negative and decreasing in magnitude.
E The acceleration of the object is positive and decreasing in magnitude.
Velocity
l
Time
Page 5
Physics 207: Lecture 9, Pg 10
Physics 207 – Lecture 10
Chapter 2
Physics 207: Lecture 9, Pg 11
Chapter 2
Also average speed and average velocity
Physics 207: Lecture 9, Pg 12
Page 6
Physics 207 – Lecture 10
Chapter 3
Physics 207: Lecture 9, Pg 13
Chapter 3
Physics 207: Lecture 9, Pg 14
Page 7
Physics 207 – Lecture 10
Chapter 4
Physics 207: Lecture 9, Pg 15
Chapter 4
Physics 207: Lecture 9, Pg 16
Page 8
Physics 207 – Lecture 10
Chapter 5
Physics 207: Lecture 9, Pg 17
Chapter 5 & 6
Physics 207: Lecture 9, Pg 18
Page 9
Physics 207 – Lecture 10
Chapter 6
Note: Drag in air is proportional to v2
Physics 207: Lecture 9, Pg 19
Chapter 7
Physics 207: Lecture 9, Pg 20
Page 10
Physics 207 – Lecture 10
Conceptual Problem
The pictures below depict cannonballs of identical mass which
are launched upwards and forward. The cannonballs are
launched at various angles above the horizontal, and with
various velocities, but all have the same vertical component
of velocity.
Physics 207: Lecture 9, Pg 21
Graphing problem
The figure shows a plot of velocity vs. time for an object
moving along the x-axis. Which of the following statements
is true?
(A) The average acceleration
over the 11.0 second
interval is -0.36 m/s2
(B) The instantaneous
acceleration at t = 5.0 s is
-4.0 m/s2
(C) Both A and B are correct.
(D) Neither A nor B are
correct.
Note: ∆x ≠ ½ aavg ∆t2
Physics 207: Lecture 9, Pg 22
Page 11
Physics 207 – Lecture 10
Conceptual Problem
A block is pushed up a 20º ramp by a 15 N force which may
be applied either horizontally (P1) or parallel to the ramp (P2).
How does the magnitude of the normal force N depend on the
direction of P?
(A) N will be smaller if P is
horizontal than if it is parallel
the ramp.
(B) N will be larger if P is
horizontal than if it is parallel
to the ramp.
(C) N will be the same in both
cases.
(D) The answer will depend on
the coefficient of friction.
20°
Physics 207: Lecture 9, Pg 23
Conceptual Problem
A cart on a roller-coaster rolls down the track shown below.
As the cart rolls beyond the point shown, what happens to
its speed and acceleration in the direction of motion?
A. Both decrease.
B. The speed decreases, but
the acceleration increases.
C. Both remain constant.
D. The speed increases, but
acceleration decreases.
E. Both increase.
F. Other
Physics 207: Lecture 9, Pg 24
Page 12
Physics 207 – Lecture 10
Sample Problem
l
l
A 200 kg wood crate sits in the back of a truck. The
coefficients of friction between the crate and the truck are
s = 0.9 and k = 0.5.
The truck starts moving up a 20°slope. What is the
maximum acceleration the truck can have without the crate
slipping out the back?
Solving:
v Visualize the problem, Draw a picture if necessary
v Identify the system and make a Free Body Diagram
v Choose an appropriate coordinate system
v Apply Newton’s Laws with conditional constraints
(friction)
v Solve
Physics 207: Lecture 9, Pg 25
Sample Problem
l
A physics student on Planet Exidor throws a ball that
follows the parabolic trajectory shown. The ball’s position is
shown at one-second intervals until t = 3 s. At t = 1 s, the
ball’s velocity is v = (2 i + 2 j) m/s.
a. Determine the ball’s velocity at t = 0 s, 2 s, and 3 s.
b. What is the value of g on Planet Exidor?
Physics 207: Lecture 9, Pg 26
Page 13
Physics 207 – Lecture 10
Sample Problem
l
l
You have been hired to measure the coefficients of friction
for the newly discovered substance jelloium. Today you will
measure the coefficient of kinetic friction for jelloium sliding
on steel. To do so, you pull a 200 g chunk of jelloium across
a horizontal steel table with a constant string tension of 1.00
N. A motion detector records the motion and displays the
graph shown.
What is the value of k for jelloium on steel?
Physics 207: Lecture 9, Pg 27
Sample Problem
Σ Fx =ma = F - ff = F - µk N = F - µk mg
Σ Fy = 0 = N – mg
µk = (F - ma) / mg & x = ½ a t2
0.80 m = ½ a 4 s2
2
a = 0.40 m/s
µk = (1.00 - 0.20 · 0.40 ) / (0.20 ·10.) = 0.46
Physics 207: Lecture 9, Pg 28
Page 14
Physics 207 – Lecture 10
Exercise: Newton’s 2nd Law
A force of 2 Newtons acts on a cart that is initially at rest
on an air track with no air and pushed for 1 second.
Because there is friction (no air), the cart stops
immediately after I finish pushing.
It has traveled a distance, D.
Force
Cart
Air Track
Next, the force of 2 Newtons acts again but is
applied for 2 seconds.
A.
B.
The new distance the cart moves relative to D
is:
C.
D.
8 x as far
4 x as far
2 x as far
1/4 x as far
Physics 207: Lecture 9, Pg 29
Exercise: Solution
Force
Cart
Air Track
We know that under constant acceleration,
∆x = a (∆t)2 /2
(when v0=0)
Here ∆t2=2∆t1,
F2 = F1 ⇒ a2 = a1
1
a ∆ t 22 (
∆x2 2
2 ∆ t1 )2
=
=
=4
∆ x1 1 a ∆ t 2
∆ t12
1
2
(B) 4 x as long
Physics 207: Lecture 9, Pg 30
Page 15
Physics 207 – Lecture 10
Another question to ponder
How high will it go?
l One day you are sitting somewhat pensively in an
airplane seat and notice, looking out the window, one of
the jet engines running at full throttle. From the pitch of
the engine you estimate that the turbine is rotating at
3000 rpm and, give or take, the turbine blade has a radius
of 1.00 m. If the tip of the blade were to suddenly break
off (it occasionally does happen with negative
consequences) and fly directly upwards, then how high
would it go (assuming no air resistance and ignoring the
fact that it would have to penetrate the metal cowling of
the engine.)
Physics 207: Lecture 9, Pg 31
Another question to ponder
How high will it go?
v ω = 3000 rpm = (3000 x 2π / 60) rad/s = 314
rad/s
v r = 1.00 m
l vo = ωr = 314 m/s (~650 mph!)
l h = h0 + v0 t – ½ g t2
l vh = 0 = vo – g t
t = vo / g
So
l h = v0 t – ½ g t2 = ½ vo2 / g = 0.5 x 3142 / 9.8 = 5 km
or ~ 3 miles
Physics 207: Lecture 9, Pg 32
Page 16
Physics 207 – Lecture 10
Sample exam problem
An object is at first travelling due north, turns and finally
heads due west while increasing its speed. The average
acceleration for this maneuver is pointed
A directly west.
B somewhere between west and northwest.
C somewhere between west and southwest.
D somewhere between northwest and north.
E somewhere between southwest and south.
F None of these are correct
Physics 207: Lecture 9, Pg 33
Sample exam problem
An object is at first travelling due north, turns and finally
heads due west while increasing its speed. The average
acceleration for this maneuver is pointed
a = (vf – vi) / ∆ t
A directly west.
B somewhere between west and northwest.
C somewhere between west and southwest.
D somewhere between northwest and north.
E somewhere between southwest and south.
F None of these are correct
Physics 207: Lecture 9, Pg 34
Page 17
Physics 207 – Lecture 10
Sample exam problem
A small block moves along a frictionless
incline which is 45° from horizontal.
Gravity acts down at 10 m/s2. There
is a massless cord pulling on the
block. The cord runs parallel to the
incline over a pulley and then straight
down. There is tension, T1, in the
cord which accelerates the block at
2.0 m/s2 up the incline. The pulley is
suspended with a second cord with
tension, T2.
A. What is the tension magnitude, T1, in the 1st cord?
B. What is the tension magnitude,T2, in the 2nd cord?
(Assume T1 = 50. N if you don’t have an answer to part A.)
Physics 207: Lecture 9, Pg 35
Sample exam problem
a = 2.0 m/s2 up the incline.
What is the tension magnitude, T1, in
the 1st cord?
Use a FBD!
Along the block surface
Σ Fx = m ax = -mg sin θ + T
T = 5 x 2 N + 5 x 10 x 0.7071 N
= (10 + 35) N = 45 N
Physics 207: Lecture 9, Pg 36
Page 18
Physics 207 – Lecture 10
Sample exam problem
a = 0.0 m/s2 at the pulley.
What is the tension magnitude,T2, in
the 2nd cord?
Use a FBD!
Physics 207: Lecture 9, Pg 37
Conceptual Problem
l
A person initially at point P in the illustration stays there a
moment and then moves along the axis to Q and stays
there a moment. She then runs quickly to R, stays there a
moment, and then strolls slowly back to P. Which of the
position vs. time graphs below correctly represents this
motion?
Physics 207: Lecture 9, Pg 38
Page 19
Physics 207 – Lecture 10
Sample exam problem
l
You have a 2.0 kg block that moves on a linear path on a horizontal
surface. The coefficient of kinetic friction between the block and the
path is k. Attached to the block is a horizontally mounted massless
string as shown in the figure below. The block includes an
accelerometer which records acceleration vs. time. As you increase
the tension in the rope the block experiences an increasingly positive
acceleration. At some point in time the rope snaps and then the block
slides to a stop (at a time of 10 seconds). Gravity, with g = 10 m/s2,
acts downward.
Physics 207: Lecture 9, Pg 39
Sample exam problem
A. At what time does the string break and, in one sentence,
explain your reasoning?
B. What speed did the block have when the string broke?
C. What is the value of k?
D. Using k above (or a value of 0.25 if you don’t have one),
what was the tension in the string at t = 2 seconds?
Physics 207: Lecture 9, Pg 40
Page 20
Physics 207 – Lecture 10
Sample exam problem
B. What speed did the block have when the string broke?
Don’t know initial v (t=0) so can’t integrate area at t < 4 sec.
vf = 0 m/s and from t = 4 to 10 sec (6 second) a = - 2 m/s2
0 = vi + a t = vi – 2 x 6 m/s
vi = 12 m/s
Physics 207: Lecture 9, Pg 41
Sample exam problem
C. What is the value of k?
Use a FBD!
Σ Fx = m ax = - fk = - k N
Σ Fy = 0 = mg – N
N = mg
So m ax = - fk = - k mg
k = - ax / g = - (-2)/10 = 0.20
Physics 207: Lecture 9, Pg 42
Page 21
Physics 207 – Lecture 10
Sample exam problem
D. What was the tension in the string at t = 2 seconds?
Again a FBD!
Σ Fx = m ax = - fk + T
Σ Fy = 0 = mg – N
N = mg
T = fk + m ax = (0.20 x 2 x 10 + 2 x 3 ) N = 10 N
Physics 207: Lecture 9, Pg 43
Recap
Exam Thursday, Oct. 6th from 7:15-8:45 PM Chapters 1-6, 7 One
8½ X 11 hand written note sheet and a calculator (for trig.)
Physics 207: Lecture 9, Pg 44
Page 22