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Chapter 5: Laws of Motion including friction
5.1 The Concept of Force
(external force, equilibrium)
5.2 Newton's First Law and Inertial Frames
5.3 Mass
5.4 Newton's Second Law
5.5 The Gravitational Force and Weight
5.6 Newton's Third Law
5.7 Some Applications of Newton's Laws
(tension, equilibrium)
5.8 Forces of Friction
5.1 The Concept of Force
Net force is the summation of all forces on an object
If net force is zero…then acceleration is zero.
If mass increases…and net force is zero, then acceleration decreases.
Gottfried Leibniz (calculus) was born on 1646 on July 1st, published his
finding on calculus in 1684.
Sir Isaac Newton developed calculus in 1665, 19 years earlier
than Leibniz, but didn’t publish results until 1687.
Source www.prenhall.com/physics
Physics 3/E daily calendar by James S. Walker
5.2 Newton's First Law and Inertial Frames
Newton's 1st Law
In absence of external forces an object at rest remains at rest and an object in motion continue in
motion with a constant velocity, when viewed from an inertial reference frame.
If an object doesn't interact with other objects, it's possible to identify a reference frame in which
the object has zero acceleration
Inertial Frames
Have you ever been at a red light and thought the driver next to you was moving very slowly
backwards?
And it turned out your foot had moved ever so slightly off the break.
Since it was such a small acceleration your body notice, thus didn’t signal your brain that you were
moving forward, thus you thought the car (or sign, etc) was moving backwards.
Example 1: A boy
in the back seat of
a moving SUV
throws a ball
straight up. For
the observer who is
at a bus stop, which
the SUV passes,
the ball appears to
travel in a ballistic
path.
Explanation
Once again you throw a ball up (vi = 4 m/s) in the moving car (vcar = 20 m/s)
To you the ball travels directly up and then returns directly down.
As you drive past your friend standing at his/her house...they see this ball
moving along with your car.
So to them it appears the path the ball follows a ballistic path with both with
vx = 20 m/s
vy = 4 m/s.
One interesting note.
The y-component of the velocity did have a force upon it...GRAVITY!
Gravity only affect components in the vertical direction
The x-component of the velocity was constant (no forces acting upon it).
* If the vx has no net forces acting upon it, vx will be constant.
This holds true for low velocities for most objects when air friction is neglible.
Does this hold true for a feather?
Obviously not, as you will see when we discuss air friction which is dependant upon
* Exposed surface area
* shape
* ρair (density)
*  v dv = ½v²
Important to remember: must use SI units (mks) (or be consistant; i.e. cgs, english, etc).
Demo: Newtonian Table Cloth: ME-F-NT
Demo: Hoop & Tack: ME-F-HT
5.3 Mass
Mass is a property of the object, just like density or volume.
What a coincidence…there is even a formula linking these
three properties together.
=m/V
Since mass is a property of a material and changes as easily as the volume (thus not easy), mass is
essentially a constant over a limited amount of time (at non-relativistic speeds)
So…you weigh 100 kg.
You jump on the space shuttle…and in orbit you are weightless, so you weigh 0 kg.
Hmmm…interesting.
How can 100 kg = 0 kg?
So if you ever tell someone you weight 100 kg…then you are saying that you can substitute 0 in any
equation where there is a value of 100.
Result…can ANYONE weight 100 kg?
Ans: NO… kilograms is a unit of mass, which is a property of the body.
Gravity gives an object weight. So weight can only be measured in Newtons, pounds, dynes, or other
force units.
5.4 Newton's Second Law
Newton's 2nd Law
The acceleration of an object is directly proportional to the net force acting on it and inversely
proportional to its mass
ΣF = ma
(when viewed from a inertial reference frame)
Demo: Rope and 3 Students: ME-J-RS
5.5 The Gravitational Force and Weight
F = ma
FW = m a g
FW = mg
So weight is mg
(g is the local acceleration due to gravity, about 9.8
m/s2 for earth…about 1.6 m/s2 for the moon)
5.6 Newton's Third Law
Newton's 3rd Law
If two objects interact, the force exerted by the first object (on the second object) is equal (in
magnitude) and opposite (in direction) to the force exerted by the second object (on the first
objected.)
· No isolated force can exist. Forces Always occur in pairs.
Demo: Hopper Popper: ME-H-AR
5.7 Some Applications of Newton's Laws (tension, equilibrium)
Example: A Big Bird walking a tight rope
A bird likes to walk on a tight rope. This is a very
large bird, about 60 kg.
If he starts 5 meters from the left pole and the
total distance across is 20 meters, how far from
the other end will he get before the tight rope
breaks.
The tight rope can support a maximum of 500 N.
Assume vertical deflection at any time will be 0.8
meters below horizontal.
Since the tight rope stationary at any instance
between steps we know:
Fx = 0 and ΣFy = 0
Σ Fx
=0N
sinβ T1 - sinθ T2
=0N
T1 = sinθ T2 / sinβ
ΣFy
=0N
cosβ T1
+ cosθ T2 - 600N = 0
cosβ (sinθ T2 / sinβ) + cosθ T2 = 600 N
cotβ sinθ T2
+ cosθ T2 = 600 N
T2 = 600 N / (cotβ sinθ + cosθ)
Input the measured values for β and θ; solve for
tension of T2 and then T1
Atwoods Machine
In the Atwoods Machine two masses are attached on both ends of the end
of a string which is stretched over a pulley. If the pulley if Frictionless,
the tension in both sides MUST be EQUAL (it’s the same string and no
force is applied at the pulley).
Let’s calculate the tension caused by block A and by block B IF THE pulley
is held stationary.
Think About It
You enter an elevator. The elevator begins to rise.
You “feel” heavier…
Next…you enter an elevator…and it begins its descent.
During initial 1 to 2 seconds during descent you “feel”
lighter.
TA = m1 g
TB = m2 g
TA = 0.9kg * 10m/s²
TB = 1.1kg * 10m/s²
TA = 9 N
TB = 11 N
Next the now Frictionless pulley is released.
Thus the tension in both sides MUST BE
EQUAL.
What’s going on?
You are accelerating in the same direction of gravity
or you’re accelerating in a direction opposing gravity.
We also know the heavier side accelerates
down and the lighter side accelerates up.
 If “g” = 10 m/s2 and you accelerate in the
direction of gravity…at 1 m/s2 you will feel 10%
lighter
 And at 3 m/s2 you will feel 30% lighter
 If you accelerate down (in the direction of
gravity) at a rate of 10 m/s2 you will feel 100%
lighter, which is called “free fall”.
We’ll use this concept to calculate the tension in the
elevator cable for this problem.
So you noticed when the pulley was stopped
the tension in the lines was not equal, but
what happens when you release the pulley so
that the weights fall toward the side of Fnet.
It's labelled as T1 & T2 as if the tension in
the rope is different. Is this true?
NO!!! This is ONE string...a pulley redirect
ONLY.
If T1 ≠ T2 you have a problem. This is the
check.
Let's try it.
Above Fnet = 0.2 Newtons
KEY QUESTION...HOW MUCH MASS IS
MOVING?
Total mass = m1 + m2 = (0.9 + 1.1) kg
Newtons 2nd law
ΣF = mtotal a
Important note
Acceleration due to gravity is about 10 m/s²
Rarely I use agravity = 9.8 m/s²
Many books use "g" instead of agravity
As in Fweight = mg
Actually many books use W(eight) = mg
I prefer Fweight = mg
Check
Does T1 = T2? Yes, so you know the results are correct.
Did you notice something else?
The average Tension when at rest was 10 N ((9+11)/2).
The tension in the line while accelerating will always be
less than the average while at rest.
Side note
What is the basic units of a Newton?
ΣF = ma/
N = kg m/s²
You may be wondering why
T1 = m1 (g+a)
T2 = m2 (g-a)
ALWAYS use your personal experience as a guideline.
Everyone one of you knows the answer
Many years ago...one of my friend's had a little sister
that was afraid of elevators (she was 14 at the time)
0.2 + = 0.2 kg * a
a = 1 N/kg
a = 1 m/s²
Now calculate the tension for both sides
In the drawing,
T1 = m1 (g+a)
T1 = 0.9 (10+1)
T1 = 9.9 N and
T2 = m2 (g-a)
T2 = 1.1 (10-1)
T2 = 9.9 N
We loved going to the unversity center and sticking her
in an elevator and then we ran up the steps to meet her
at the top.
So even she knew that as soon as the elevator started
rising that she felt heavier (g+a)
This I believe is one of the reason for her fear.
Conversely, if you are at the top and the elevator
starts lowering you feel lighter (g-a)
(until it acheived constant velocity).
So apply this to the Atwood's machine.
Box Pulled by weight over Pulley
Apply the same logic as in Atwoods Machine example
mtotal = M + m
mtotal = 8kg + 2kg
mtotal = 10 kg.
Fnet = mg
Fnet = 2kg * 10 m/s²
Fnet = 20 N
ΣF = ma
20 N = 10 kg * a
a = 2 m/s²
Once again the Tension in the rope better be equal for all involved.
Big Box on top of table
ΣF = ma
ΣF = 8kg * 2 m/s²
ΣF = 16 N
Hanging Weight
ΣF = ma
ΣF = 2kg * (10 - 2) m/s²
ΣF = 16 N
Boxed Pulled by weight over Pulley w/Friction
Remember this...
Friction is F=μN
Ffriction = μ FNormal
Normal force = cos θ mg
So apply this to the above case.
8 kg block is resting on the surface
whre cos θ= cos(0°)
cos(0°) = 1
Ff = μ mg = 0.2 * 80 N = 16 N
Which direction is Friction?
ALWAYS opposes motion....thus the Ff vector must be drawn in to
the left as shown
Fnet = mg - μMg
Fnet = 20 - 16
Fnet = 4 N
How much mass is being moved?
ΣF = mtotala
4 N = 10 kg * a
a = 0.4 m/s²
Downhill ramp (w/Friction) while weight hangs over a Pulley
Coefficient of kinetic friction = 0.6
It's actually better to show the normal force originating at the
table top (ramp) and rising perpendicular to the surface of the
ramp. (instead of below as shown)
I just want to show how to Fweight is broken into its respective
components. Remember the weight must be the largest value
(hypotenuse) when you are going to calculate the Normal Force and
the Force in the direction of the motion (due to the angle of the
ramp.)
Two forces (component of weight & 2kg block) accelerate this
block system down the ramp, and one resistive force (friction, μ
FNormal) opposes this motion.
ΣF = Fin dir + Fhanging block- Ffriction
ΣF = sinθ Mg +
mg
- μcosθ Mg
Let θ = 30°
ΣF = 40 N + 20 N - 41.6 N
ΣF = 18.4 N
ΣF = 18.4 N = (8kg + 2kg) * a
a = 1.84 m/s²
Tension Between Block with Friction and Weight Over Pulley
One question: what is the tension in the line of a
block on an incline attached to a weight suspended
by a pulley
where Ff = μ N & N = cos30°(4kg)10m/s² = 34.6 N
I believe we said that m1 was 4 kg and m2 was 2 kg, thus
Fnet = mtotala
Fnet = m2g + sinθm1g Ff
=
mtotal a
Fnet = 20N + ½ 40N - 0.7*34.6N = (2kg +4kg)*a
Fnet = 15.8 N = 6 kg * a
a = 2.6 m/s²
And you were asked to calculate the tension in T1
This is where I simplify the problem.
T1 = T2 = m2*(g-a) = 2kg*(10-2.6) = 14.8 N
This is where estimating g = 10 m/s² may cause problems.
For instance if a = 9.6 m/s²...using g = 10 m/s² results in a 50% error!!!
(g-a) = 10 - 9.6 = 0.4 m/s²
(g-a) = 9.8 - 9.6 = 0.2 m/s²
So when ever you are subtracting you must be careful of your estimations.
A very similar situation arises with uncertainty in your lab...when you subtract uncertainties you can end up
with uncertainty associated with your result being larger than your result itself.
5.8 Forces of Friction
Static (not sliding) and Kinetic (moving/sliding) Friction
Demo: Static Equilibrium Balls: ME-K-SE
Demo: Static vs Kinetic Incline Plane: ME-K-SK
Think back to when you had to push (slide across a floor) something heavy (sofa, desk, etc).
Was it easier to start it moving or
was it easier to keep it moving once you started it to slide
In my experience...once it started moving in a small area you have to be careful not to hit a wall.
BECAUSE IT SLIDES EASIER ONCE IT STARTS TO MOVE!!!
This is called kinetic friction which is lower than static frictional Coefficient
We will also use this example to introduce Ch 7.
· The work you do is the force you apply to an object through a displacement
Note: simplified definition