Download Advanced Placement Physics “B”

Type: Double
Date: ______________
Objective: Non-Equilibrium Applications, Friction and Newton’s
Laws III
Homework: Assignment (1-16)
Do PROBS #’s (40, 43, 66) Ch. +
Do AP 1988 #1 (handout)
AP Physics “B”
Mr. Mirro
Date: ________
Non-Equilibrium, Friction and Newton’s Laws III
Consider the following “two-body” system. Two boxes are connected by a cord running over a pulley.
If we ignore the mass of the cord and pulley, as well as the friction in the pulley,
… which means we can assume that a force applied to one end of the cord will have the same
magnitude at the other end…
then - we can analyze the behavior of the “system” as mass #2 falls downward causing mass #1 to move
across the table to the right.
m1
frictionless surface
m2
+y
Mass #1 does not move vertically, so the normal force
just balances the weight.
FN
∑F1y = m1 (a1y) = 0
+FN - m1 g = 0
FN = m1 g
T
m1
(if m1 = 1 kg)
FN = 1 kg (10m/s2) = 10 N
Fg = m1g
In the horizontal direction, there is one force on Mass #1 called the tension (T) in the cord, which we do not know.
∑F1x = m1 (ax1)
(Let a1x = a)
+T = m1 a
[Eq. #1]
+x
Next, consider mass #2 The force of gravity (Fg = mg) pulls downward and the cord pulls upward with a force or tension (T).
So we can write Newton’s second law for box #2 taking the
downward direction as positive in this case - since the direction
of motion is downward.
-y
T
∑F2y = m2 (a2y)
+m2 g - T = m2 a
(Let a2y = a)
[Eq. #2]
m2
Fg = m2g
+y
We now have two unknowns, a and T, and we also have two equations !
Therefore, we have to solve a system of two equations with two unknowns.
Thus, we can substitute Eq. #1, the tension (T) into Eq. #2 thereby reducing the number of variables to one.
+T = m1 a
[Eq. #1]
m2 g - [T] = m2 a
[Eq. #2]
m2 g -- [m1 a] = m2 a
[Eq #2 with tension (T) substitution]
m2 g - m1 a = m2 a
Distribute negative sign
m2 g = m1 a + m2 a
Transpose terms
m2 g = a (m1 + m2 )
Factor out the common acceleration (a)
a = m2 g
m1 + m2
For example, if mass #2 is m2 = 1 kg, we could effectively compute the acceleration (a) of system !
a = (1 kg)(10 m/s2) = 10 N = 10 kg m/s2 = 5 m/s
1 kg + 1 kg
2 kg
2 kg
2
AP Physics “B”
Mr. Mirro
Date: ________
Non-Equilibrium, Dynamics and Newton’s Laws I
Ex: Two boxes are connected by a massless cord running over a frictionless pulley. As box II moves
down, box I moves to the right. The coefficient of kinetic friction between box I and the table is 0.20.
[Giancoli4.16]
a. Find the acceleration (a), of the “system” which
will have the same magnitude for both boxes
assuming that the cord does not stretch.
5 kg
I
II
b. What is the tension (T) in the cord ?
2 kg
AP Physics “B”
Mr. Mirro
Date: ________
Non-Equilibrium, Dynamics and Newton’s Laws I
Newton’s 1st Law of Inertia states that if no “net” force is acting on a body, it remains either at rest, or moving
at a constant velocity – depending on the initial state of motion.
But,
What happens if a net force is exerted on the body ?
‰
“Speed increase”
‰
“Reduce speed”
‰
“Change direction”
Since a “change” in speed or velocity is an acceleration, we can say that a net or unbalanced force gives
rise to this acceleration !
a = Δv ⇒ ∑ F
Δt
m
Therefore,
What precisely is the relationship between acceleration and force ?
Everyday experience can answer this question.
Consider the force required to push a cart whose friction is minimal.
If you push with a gentle but constant force for a certain period of time you will make the cart
accelerate from rest up to some speed…say 3 m/s.
If you push with twice the force, you will find that the cart will reach 3 m/s in half the time.
That is, the acceleration will be “twice” as great.
If you double the force, the acceleration is doubled. If you triple the force, the acceleration is tripled. etc.
Thus, the acceleration of a body is directly proportional to the net applied force. But the acceleration also
depends upon the mass of the object as well.
ie. If you push an empty grocery cart with the same force as you push one that is filled with
groceries, you will find that the “full” cart accelerates more slowly.
So mass affects acceleration - The greater the mass, the less the acceleration.
a = FNet
M
Newton’s Second Law
We can rearrange Newton’s Second Law to obtain the familiar form of F = ma.
Considering the net force or the sum of the forces we get the expression:
In general,
<i> x-direction:
∑F = m a
∑Fx = m ax
(where a ≠ 0 necessarily)
<ii> y-direction:
combine all the
forces in the
x-direction
∑Fy = m ay
combine all the
forces in the
y-direction
Note: Be consistent when applying signs (+/-) and direction.
We can now solve problems where the system may not be in static equilibrium – these systems are called
dynamic systems, the study of which is called “dynamics.”
AP Physics “B”
Mr. Mirro
Date: ________
Non-Equilibrium, Friction and Newton’s Laws III
Ex 1: A flatbed truck is carrying a crate up a 10° hill as shown. The coefficient of static friction between
the truck bed and the crate is μs = 0.35. Find the maximum acceleration that the truck can attain
before the crate begins to slip backward relative to the truck. [Cutnell4.17]
Spacial-Diagram
Draw the Free-Body Diagram
Ex 2: Consider a model rocket weighing 1.1 nt including the engine and propellant. If the rocket engine
provides 11 nt of thrust, neglect the variation in mass as the engine burns fuel and determine the
following:
a. The net unbalanced force acting on the model rocket.
b. The acceleration of the rocket.
Ex 3: What net force is required to bring a 1500 kg car to rest from a speed of 28 m/s within a distance of
55 m ? [Giancoli4.2]