physics 207: lecture 7, pg 1 lecture 7 l goals: solve 1d and 2d problems with forces in equilibrium...
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Physics 207: Lecture 7, Pg 1
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 (1st part)due 2/17, 9 am,
Wednesday)
For Thursday, Read Chapter 7
1st Exam Thursday, Feb. 17 from 7:15-8:45 PM Chapters 1-7
Physics 207: Lecture 7, Pg 2
Identifying Forces: Non-contact
All objects having mass exhibit a mutually attractive force (i.e., gravity) that is distance dependent
At the Earth’s surface this variation is small so little “g” (the associated acceleration) is typically set to 9.80 or 10. m/s2
FB,G
Physics 207: Lecture 7, Pg 3
Contact (i.e., normal) Forces
Certain forces act to keep an object in place. These have what ever force needed to balance
all others (until a breaking point).
FB,T
Physics 207: Lecture 7, Pg 4
No net force No acceleration
FB,T Normal force is always to a surface
0
0
0net
y
x
F
F
amFF
FB,G
(Force vectors are not always drawn at contact points)
mgN
NmgFy
0
y
Physics 207: Lecture 7, Pg 5
No net force No acceleration
0net amFF
If zero velocity then “static equilibrium” If non-zero velocity then “dynamic equilibrium” This label, static vs. dynamic, is observer dependent
Forces are vectors
321net FFFamFF
Physics 207: Lecture 7, Pg 6
A special contact force: Friction
What does it do? It opposes motion (velocity, actual or that which
would occur if friction were absent!) How do we characterize this in terms we have learned?
Friction results in a force in a direction opposite to the direction of motion (actual or, if static, then “inferred”)!
maFFAPPLIED
ffFRICTION mgg
NN
ii
j j
Physics 207: Lecture 7, Pg 7
Friction...
Friction is caused by the “microscopic” interactions between the two surfaces:
Physics 207: Lecture 7, Pg 8
Friction...
Force of friction acts to oppose motion: Parallel to a surface Perpendicular to a NNormal force.
maFF
ffF mgg
NN
ii
j j
Physics 207: Lecture 7, Pg 9
Static Friction with a bicycle wheel
You are pedaling hard and the bicycle is speeding up.
What is the direction of the frictional force?
You are breaking and the bicycle is slowing down
What is the direction of the frictional force?
Physics 207: Lecture 7, Pg 10
Important notes Many contact forces are conditional and, more
importantly, they are not necessarily constant
We have a general notion of forces is from everyday life.
In physics the definition must be precise. A force is an action which causes a body to
accelerate.
(Newton’s Second Law)
On a microscopic level, all forces are non-contact
Physics 207: Lecture 7, Pg 11
Pushing and Pulling Forces
A rope is an example of something that can pull
You arm is an example of an object that can push or push
Physics 207: Lecture 7, Pg 14
Ropes provide tension (a pull)
In physics we often use a “massless” rope with opposing tensions of equal magnitude
Physics 207: Lecture 7, Pg 15
Forces at different angles
Case 1 Case 2
F
mg
N
Case1: Downward angled force with friction
Case 2: Upwards angled force with friction
Cases 3,4: Up against the wall
Questions: Does it slide?
What happens to the normal force?
What happens to the frictional force?
mg
Cases 3, 4
mg
NN
F
F
ff
ff
ff
Physics 207: Lecture 7, Pg 16
Free Body Diagram
A heavy sign is hung between two poles by a rope at each corner extending to the poles.
Eat at Bucky’s
A hanging sign is an example of static equilibrium (depends on observer)
What are the forces on the sign and how are they related if the sign is stationary (or moving with constant velocity) in an inertial reference frame ?
Physics 207: Lecture 7, Pg 17
Free Body Diagram
Step two: Sketch in force vectorsStep three: Apply Newton’s 2nd Law (Resolve vectors into appropriate components)
mg
T1
T2Eat at Bucky’s
T1
T2
mg
Step one: Define the system
Physics 207: Lecture 7, Pg 18
Free Body Diagram
Eat at Bucky’s
T1 T2
mg
Vertical : y-direction 0 = -mg + T1 sin1 + T2
sin2
Horizontal : x-direction 0 = -T1 cos1 + T2 cos2
Physics 207: Lecture 7, Pg 19
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 20
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 21
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 22
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 23
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 29
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 30
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 31
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 32
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 35
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 36
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 37
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 38
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 39
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 40
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 41
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 42
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
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