Physics 1A
Lecture 4D
"The test of a first-rate intelligence is the ability to
hold two opposed ideas in mind at the same time and
still retain the ability to function.”
--F. Scott Fitzgerald
Quiz 2
Quiz 2 will include:
some of chapter 3:
problems like the boat traversing the river.
knowledge of general concepts
all of Chapter 4.
Quiz 2
My suggestion is to do the homework,
class examples, and book examples.
Don’t forget your quiz code number or a
Scantron on Monday !!!
There will be only seven questions, i.e.
you ought to have more time than in
Quiz 1.
Normal Force
Example
An applied 12N horizontal force
(Fapplied) pushes a block weighing
block
5.0N against a vertical wall. The
coefficient of static friction
Fapplied
between the wall and the block
is 0.60, and the coefficient of
kinetic friction is 0.40. Assume
that the block is not moving
initially. Will the block ever
move?
Answer
First, you must define a coordinate system.
Let’s choose up as positive y and the direction
of the applied force as positive x.
Normal Force
Answer
Next, draw a force diagram for the block.
Ffriction, wall on block
Fnormal, wall on block
Fgravity, Earth on block
block
Fapplied, you on block
No need to break the forces into components, so
we can turn to Newton’s Laws.
Which of Newton’s Laws do we apply?
We know it doesn’t accelerate in the x-direction,
but we are not sure about the y-direction.
Normal Force
Answer
We should apply Newton’s 1st Law in the x-direction.
ΣFx = 0
ΣFx = Fapplied - Fnormal = 0
Fnormal = Fapplied = 12N
If the block is not going to move then we can say that
the maximum static friction force must be greater than
than or equal to the force of gravity.
Ffriction,max > Fgravity = 5.0N
Ffriction,max = (μs) Fnormal = (0.60)12N
Ffriction,max = 7.2N
So, the block will not move, since Ffriction can be 5N!
Multiple Objects
Example
Two blocks are in contact on a frictionless table. You
apply a horizontal force to the larger block. If
m1 = 2.3kg, m2 = 1.2kg, and Fapplied = 3.2N, find the
magnitude of the force between the two blocks.
m1
Fapplied
m2
Answer
First, you must define a coordinate system.
Let’s choose up as positive y and the direction
of the applied force as positive x.
Multiple Objects
Answer
Next, draw a force diagram for each block separately.
Fnormal, table on m1
Fcontact, m2 on m1
m1
Fapplied, you on m1
Fgravity, Earth on m1
m2
Fnormal, table on m2
Fgravity, Earth on m2
Fcontact, m1 on m2
No need to break the forces into components, so
we can turn to Newton’s Laws.
Multiple Objects
Answer
We should apply Newton’s 2nd Law separately to
each object.
Since the blocks move together, ax = a1x = a2x.
For the m1 in the x-direction we have:
ΣFx = m1a1x
ΣFx = Fapplied - Fcontact = m1ax
Fapplied = Fcontact + m1ax
For the m2 in the x-direction we have:
ΣFx = m2a2x
ΣFx = Fcontact = m2ax
ax = (Fcontact)/m2
Multiple Objects
Answer
We can substitute ax into the previous equation.
More on multiple objects
As an exercise, assume that both objects are
subject to friction.
How does this change the answer?
Attached Masses
When masses are attached by ropes/strings then
tensions will be involved.
By Newton’s Third Law, the tension on the first mass
will be equal to the tension on the second mass (but
opposite in direction).
Also, assuming the rope cannot be stretched
(inextensible), then the acceleration in either mass
must be the same.
Pulleys
Pulleys will change the
direction of the tension
force in ropes.
This could mean that the
tension for two masses may
be in the same direction.
Or this also could mean
that a tension force on
one mass in the
horizontal direction will
have a Third Law Pair
with another mass in
the vertical direction.
Pulleys
Example
Two objects with masses 2.00kg
(m1) and 6.00kg (m2) are
connected by a light string that
passes over a frictionless pulley.
Determine the acceleration of
each mass and the tension in the
string.
Answer
First, you must define a
coordinate system.
Let’s choose up as positive y.
Pulleys
Answer
Next we should draw a force diagram for each
mass:
Ftension, string on 6kg mass
6kg mass
Fgravity, Earth on 6kg mass
Ftension, string on 2kg mass
2kg mass
Fgravity, Earth on 2kg mass
Forces are already broken up into components.
So we should apply Newton’s 2nd Law separately to
each object.
Pulleys
Answer
Since the string cannot be stretched, a = a1 = -a2.
Also the tensions will be the same magnitude: T
For the 2kg mass (m1) we have:
ΣFy = m1a1
ΣFy = T - m1g = m1a
T = m 1a + m 1g
For the 6kg mass (m2) we have:
ΣFy = m2a2
ΣFy = T - m2g = m2(-a)
T = m 2g - m 2a
Answer
Pulleys
Set T = T and solve for the resulting acceleration.
T =T
m1 a + m 1 g = m 2 g - m 2 a
m1 a + m 2 a = m 2 g - m 1 g
a(m2 + m1) = g(m2 - m1)
a = g(m2 - m1)/(m2 + m1)
Plugging in the values we get:
a = (9.80m/s2)(6kg - 2kg)/(6kg + 2kg)
a = (9.80m/s2)(4kg)/(8kg) = 4.90m/s2
Answer
Pulleys
Recall the equation that we have for tension from
Newton’s Second Law (either one will work):
T = m 2g - m 2a
T = m2(g - a)
T = 6kg(9.80m/s2 - 4.90m/s2)
T = 6kg(4.90m/s2) = 29.4N
In class Question
Let’s say that in the previous example the masses were
changed to m1 = 3kg and m2 = 9kg. How would this
affect the resulting motion of the Atwood’s Machine?
A) Acceleration of the two masses would decrease.
B) Acceleration of the two masses would increase.
C) Acceleration of the two masses would remain the
same.
For Next Time (FNT)
Finish the Chapter 4 HW.
Start Reading Chapter 5.