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Physics 207: Lecture 7, Pg 1
"Professor Goddard does not know the relation between action and reaction and the need to have something better than a vacuum against which to react. He seems to lack the basic knowledge ladled out daily in high schools."
New York Times editorial, 1921,
about Robert Goddard's revolutionary
rocket work.
"Correction: It is now definitely
established that a rocket can
function in a vacuum.
The 'Times' regrets the error." New York Times editorial, July 1969.
Lecture 7
Physics 207: Lecture 7, Pg 2
Lecture 7
Goals:Goals: Solve 1D and 2D problems with forces in equilibrium and non-equilibrium (i.e., acceleration) using Newton’ 1st and 2nd laws.
Distinguish static and kinetic coefficients of friction
Differentiate between Newton’s 1st, 2nd and 3rd Laws
Assignment: HW4, (Chapters 6 & 7, due 2/18, 9 am, Wednesday)
Read Chapter 7
1st Exam Wednesday, Feb. 18 from 7:15-8:45 PM Chapters 1-7
Physics 207: Lecture 7, Pg 3
Exercise, Newton’s 2nd Law
A. P + C < W
B. P + C > W
C. P = C
D. P + C = W
A woman is straining to lift a large crate, without success. It is too heavy. We denote the forces on the crate as follows: P is the upward force being exerted on the crate by the personC is the contact or normal force on the crate by the floor, and W is the weight (force of the earth on the crate). Which of following relationships between these forces is true, while the person is trying unsuccessfully to lift the crate? (Note: force up is positive & down is negative)
Physics 207: Lecture 7, Pg 4
Mass We have an idea of what mass is from everyday life. In physics:
Mass (in Phys 207) is a quantity that specifies how much inertia an object has
(i.e. a scalar that relates force to acceleration)
(Newton’s Second Law) Mass is an inherent property of an object. Mass and weight are different quantities; weight is
usually the magnitude of a gravitational (non-contact) force.
“Pound” (lb) is a definition of weight (i.e., a force), not a mass!
Physics 207: Lecture 7, Pg 5
Inertia and Mass The tendency of an object to resist any attempt to
change its velocity is called Inertia Mass is that property of an object that specifies how
much resistance an object exhibits to changes in its velocity (acceleration)
If mass is constant then
If force constant
Mass is an inherent property of an object Mass is independent of the object’s surroundings Mass is independent of the method used to measure it Mass is a scalar quantity The SI unit of mass is kg
netFa
ma 1||
|a|
m
Physics 207: Lecture 7, Pg 6
ExerciseNewton’s 2nd Law
A. increasingB. decreasingC. constant in timeD. Not enough information to decide
An object is moving to the right, and experiencing a net force that is directed to the right. The magnitude of the force is decreasing with time (read this text carefully).
The speed of the object is
Physics 207: Lecture 7, Pg 7
Exercise Newton’s 2nd Law
A. B
B. C
C. D
D. F
E. G
A 10 kg mass undergoes motion along a line with a velocities as given in the figure below. In regards to the stated letters for each region, in which is the magnitude of the force on the mass at its greatest?
Physics 207: Lecture 7, Pg 13
Moving forces around
Massless strings: Translate forces and reverse their direction but do not change their magnitude
(we really need Newton’s 3rd of action/reaction to justify)
Massless, frictionless pulleys: Reorient force direction but do not change their magnitude
string
T1 -T1
T1 -T1
T2
-T2| T1 | = | -T1 | = | T2 | = | T2 |
Physics 207: Lecture 7, Pg 14
Scale Problem
You are given a 1.0 kg mass and you hang it directly on a fish scale and it reads 10 N (g is 10 m/s2).
Now you use this mass in a second experiment in which the 1.0 kg mass hangs from a massless string passing over a massless, frictionless pulley and is anchored to the floor. The pulley is attached to the fish scale.
What force does the fish scale now read?
1.0 kg
10 N
?
1.0 kg
Physics 207: Lecture 7, Pg 15
Scale Problem
Step 1: Identify the system(s).
In this case it is probably best to treat each object as a distinct element and draw three force body diagrams. One around the scale One around the massless pulley (even
though massless we can treat is as an “object”)
One around the hanging mass
Step 2: Draw the three FBGs. (Because this is a now a one-dimensional problem we need only consider forces in the y-direction.)
?
1.0 kg
Physics 207: Lecture 7, Pg 16
Scale Problem
Fy = 0 in all cases
1: 0 = -2T + T ’
2: 0 = T – mg T = mg
3: 0 = T” – W – T ’ (not useful here) Substituting 2 into 1 yields T ’ = 2mg = 20 N
(We start with 10 N but end with 20 N)
?
1.0 kg
? 1.0 kg
-mg
T
-T -T
-T ’
T”
W
1:2:3:T ’
Physics 207: Lecture 7, Pg 17
No Net Force, No acceleration…a demo exercise
In this demonstration we have a ball tied to a string undergoing horizontal UCM (i.e. the ball has only radial acceleration)
1 Assuming you are looking from above, draw the orbit with the tangential velocity and the radial acceleration vectors sketched out.
2 Suddenly the string brakes.
3 Now sketch the trajectory with the velocity and acceleration vectors drawn again.
Physics 207: Lecture 7, Pg 18
Static and Kinetic Friction Friction exists between objects and its behavior has been
modeled.
At Static Equilibrium: A block, mass m, with a horizontal force F applied,
Direction: A force vector to the normal force vector N N and the vector is opposite to the direction of acceleration if were 0.
Magnitude: f is proportional to the applied forces such that
fs ≤ s N
s called the “coefficient of static friction”
Physics 207: Lecture 7, Pg 19
Friction: Static frictionStatic equilibrium: A block with a horizontal force F applied,
As F increases so does fs
Fm
1
FBD
fs
N
mg
Fx = 0 = -F + fs fs = F
Fy = 0 = - N + mg N = mg
Physics 207: Lecture 7, Pg 20
Static friction, at maximum (just before slipping)
Equilibrium: A block, mass m, with a horizontal force F applied,
Direction: A force vector to the normal force vector N N and the vector is opposite to the direction of acceleration if were 0.
Magnitude: fS is proportional to the magnitude of N
fs = s N
Fm fs
N
mg
Physics 207: Lecture 7, Pg 21
Kinetic or Sliding friction (fk < fs)
Dynamic equilibrium, moving but acceleration is still zero
As F increases fk remains nearly constant
(but now there acceleration is acceleration)
Fm
1
FBD
fk
N
mg
Fx = 0 = -F + fk fk = F
Fy = 0 = - N + mg N = mg v
fk = k N
Physics 207: Lecture 7, Pg 22
Sliding Friction: Quantitatively
Direction: A force vector to the normal force vector N N and the vector is opposite to the velocity.
Magnitude: ffk is proportional to the magnitude of N N
ffk = k N N ( = Kmg g in the previous example)
The constant k is called the “coefficient of kinetic friction”
Logic dictates that S > K for any system
Physics 207: Lecture 7, Pg 23
Coefficients of Friction
Material on Material s = static friction k = kinetic friction
steel / steel 0.6 0.4
add grease to steel 0.1 0.05
metal / ice 0.022 0.02
brake lining / iron 0.4 0.3
tire / dry pavement 0.9 0.8
tire / wet pavement 0.8 0.7
Physics 207: Lecture 7, Pg 24
An experiment
Two blocks are connected on the table as shown. The
table has unknown static and kinetic friction coefficients.
Design an experiment to find S
Static equilibrium: Set m2 and add mass to m1 to reach the breaking point.
Requires two FBDsm1
m2
m2g
N
m1g
T
T
Mass 2
Fx = 0 = -T + fs = -T + S N
Fy = 0 = N – m2g
fS
Mass 1
Fy = 0 = T – m1g
T = m1g = S m2g S = m1/m2
Physics 207: Lecture 7, Pg 25
A 2nd experiment
Two blocks are connected on the table as shown. The
table has unknown static and kinetic friction coefficients.
Design an experiment to find K.
Dynamic equilibrium: Set m2 and adjust m1 to find place when
a = 0 and v ≠ 0
Requires two FBDs
m1
m2
m2g
N
m1g
T
T
Mass 2
Fx = 0 = -T + ff = -T + k N
Fy = 0 = N – m2g
fk
Mass 1
Fy = 0 = T – m1g
T = m1g = k m2g k = m1/m2
Physics 207: Lecture 7, Pg 26
An experiment (with a ≠ 0)
Two blocks are connected on the table as shown. The
table has unknown static and kinetic friction coefficients.
Design an experiment to find K.
Non-equilibrium: Set m2 and adjust m1 to find regime where a ≠ 0
Requires two FBDs
T
Mass 2
Fx = m2a = -T + fk = -T + k N
Fy = 0 = N – m2g
m1
m2
m2g
N
m1g
T
fk
Mass 1
Fy = m1a = T – m1g
T = m1g + m1a = k m2g – m2a k = (m1(g+a)+m2a)/m2g
Physics 207: Lecture 7, Pg 29
Inclined plane with “Normal” and Frictional Forces
Weight of block is mg
NormalForce
Friction ForceSliding Down
“Normal” meansperpendicular
Note: If frictional Force = Normal Force (coefficient of friction)Ffriction = Fnormal = mg sin then zero acceleration
1. At first the velocity is v up along the slide
2. Can we draw a velocity time plot?
3. What the acceleration versus time?
v
mg sin
fk Sliding
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