Introduction to Mechanics
Applying Newton’s Laws
Pulleys and Force
Lana Sheridan
De Anza College
Mar 2, 2016
Last time
• friction examples
Overview
• finish last friction example
• pulleys
Friction Question
#5, page
Hopping into your Porsche, you floor it and accelerate at 12 m/s2
without spinning the tires. Determine the minimum coefficient of
static friction between the tires and the road needed to make this
possible.
0
Walker, “Physics”.
Friction Question
#5, page
Hopping into your Porsche, you floor it and accelerate at 12 m/s2
without spinning the tires. Determine the minimum coefficient of
static friction between the tires and the road needed to make this
possible.
Sketch a free body diagram for the car. (What force causes the
car’s forward acceleration?)
0
Walker, “Physics”.
Friction Question
#5, page
Hopping into your Porsche, you floor it and accelerate at 12 m/s2
without spinning the tires. Determine the minimum coefficient of
static friction between the tires and the road needed to make this
possible.
Sketch a free body diagram for the car. (What force causes the
car’s forward acceleration?)
Hypothesis: coefficients of friction are usually between 0 and 1.
Car tires are designed not to slip on asphalt. µs should be high,
but we are looking for the minimum it could be. Guess: 0.5.
0
Walker, “Physics”.
Friction Question
Hopping into your Porsche, you floor it and accelerate at 12 m/s2
without spinning the tires. Determine the minimum coefficient of
static friction between the tires and the road needed to make this
possible.
Strategy: Newton’s 2nd law.
Fnet = ma
Fnet = fs . If we want to find the minimum coefficient of static
friction, assume that we are getting the max possible force from
that coefficient: fs = fs,max = µs n.
µs mg
µs
µs
= ma
a
=
g
=
12 m/s2
9.81 m/s2
= 1.2
Friction Question
Hopping into your Porsche, you floor it and accelerate at 12 m/s2
without spinning the tires. Determine the minimum coefficient of
static friction between the tires and the road needed to make this
possible.
µs
= 1.2
Reasonable?: Woah! This is not only much bigger than my guess,
it is bigger than 1!
This would mean that it requires less force to pick up the entire
Porsche and move it to one side than it does to push it along the
ground starting from rest.
Friction Question
Hopping into your Porsche, you floor it and accelerate at 12 m/s2
without spinning the tires. Determine the minimum coefficient of
static friction between the tires and the road needed to make this
possible.
µs
= 1.2
Reasonable?: Woah! This is not only much bigger than my guess,
it is bigger than 1!
This would mean that it requires less force to pick up the entire
Porsche and move it to one side than it does to push it along the
ground starting from rest.
Research → Apparently, yes! This is the sort of number you can
get for high-performance racing tires. Cool.
Pulleys
M
Pulleys “turn tensions around a corner”.
ly over a frictionless pulley of
t is called an Atwood machine.
termine the value of g. Deterobjects and the tension in the
S
T
+
m1
igure 5.14a in action: as one
wnward. Because the objects
elerations must be of equal
S
m 1g
m2
+
For the moment, we are just considering massless, frictionless
S
m 2g
pulleys. What does that mean?
e subject to the gravitational
onnected to them. Therefore,
particles under a net force.
ts are shown in Figure 5.14b.
T exerted by the string and
S
m1
m2
S
T
a
b
Figure 5.14 (Example 5.9) The
Atwood machine. (a) Two objects
connected by a massless inextensible
string over a frictionless pulley.
(b) The free-body diagrams for the
Pulleys
M
Pulleys “turn tensions around a corner”.
ly over a frictionless pulley of
t is called an Atwood machine.
termine the value of g. Deterobjects and the tension in the
S
T
+
m1
igure 5.14a in action: as one
wnward. Because the objects
elerations must be of equal
m1
m2
S
m 1g
S
T
m2
+
For the moment, we are just considering massless, frictionless
S
m 2g
pulleys. What does that mean?
e subject to the gravitational
a
b
we do not have to worry about force needed to
onnected •to Massless:
them. Therefore,
accelerate
each atom Figure
in the 5.14
pulley(Example 5.9) The
particles under
a net force.
Atwood machine. (a) Two objects
• Frictionless: the axle of
the pulley
has no inextensible
friction to resist the
connected
by a massless
ts are shown in Figure 5.14b.
S
wheel
turning
string over a frictionless pulley.
T exerted by the string and
(b) The free-body diagrams for the
Because of the weight of the rope, the
hand with a force T, the string
tension is noticeably different at points 1,
tension T in the string. To be m
2, and 3. As the rope becomes lighter, howsome point, the tension T is
ever, the difference in tension decreases. In
(massless)
and
themass,
pulley
is massless
and—that is, T is the for
Figure 6–5
the limit of a rope
of zero
the tension
is the same throughout the rope.
to hold the cut ends togethe
Pulleys and Tension
If the rope is light
frictionless, the tension in the rope on both sides of the pulley is
equally to the right and to the
the same.
As an example, consider a
T
T
▲ FIGURE 6–7 A pulley changes the
0
of a tension
Figure from Walker,direction
“Physics”.
to a box with a weight of 105 N
tion, suppose the rope is unifo
is the tension in the rope (i) w
(iii) where it attaches to the ce
First, the rope holds the bo
to the box is simply the weig
rope, the tension supports th
rope. Thus, T2 = 105 N + 1212
sion supports the box plus all
that the tension pulls down o
From this discussion, we c
from top to bottom because of
difference in tension between
rope’s mass were to be vanish
as well. In this text, we will a
practically massless—unless s
tension is the same throughou
Pulleys are often used to r
Figure 6–7. In the ideal case, a
Thus, an ideal pulley simply ch
changing its magnitude. If a sys
Pulleys
Free-body diagrams:
ctionless pulley of
n Atwood machine.
e value of g. Deterthe tension in the
S
T
+
m1
in action: as one
cause the objects
must be of equal
the gravitational
them. Therefore,
der a net force.
m1
m2
S
m 1g
S
T
m2
+
S
m 2g
a
b
Figure 5.14 (Example 5.9) The
the y direction as u 5 40.0° west of north. (c) Compare
your solutions to parts (a) and (b). Do the results agree?
Is one method significantly easier?
28. The systems shown in Figure P5.28 are in equilibrium.
W If the spring scales are calibrated in newtons, what do
they read? Ignore the masses of the pulleys and strings
and assume the pulleys and the incline in Figure
P5.28d are frictionless.
Tension and Pulleys
5.00 kg
5.00 kg
b
5.00 kg
5.00 kg
c
1
30.0!
5.00 kg
Figure
Serway & Jewett, 9th ed, page
141.P5.28
d
ure P5.33. Two
angles u1 5 60
with the hor
the system i
find the tens
in the wires.
34. A bag of cem
S is Fg hangs in
three wires a
P5.33. Two o
angles u1 and
tem is in equi
hand wire is
5.00 kg
a
33. A bag of ceme
AMT hangs in e
W three wires a
35. Two people
ropes attache
If they pull
acceleration
opposite dire
0.518 m/s2 to
force each pe
other horizon
Pulleys and the Atwood Machine
The Atwood Machine can be used to make careful determinations
of g , as well as explore the behavior of forces and accelerations.
onless pulley of
Atwood machine.
alue of g. Detere tension in the
S
T
+
m1
n action: as one
use the objects
ust be of equal
m1
m2
S
m 1g
S
T
m2
+
S
m 2g
he gravitational
hem. Therefore,
1
a
b
http://en.wikipedia.org/wiki/Atwood machine
Pulleys and the Atwood Machine
We can consider the motion for each mass separately:
Fnet,1y = T − m1 g = m1 a
(1)
Fnet,2y = T − m2 g = −m2 a
(2)
Pulleys and the Atwood Machine
We can consider the motion for each mass separately:
Fnet,1y = T − m1 g = m1 a
(1)
Fnet,2y = T − m2 g = −m2 a
(2)
Be careful about the signs! Both masses must accelerate
together - one up, one down.
Two equations, two unknowns. Solve as you like!
Pulleys and the Atwood Machine
Fnet,1y = T − m1 g = m1 a
(3)
Fnet,2y = T − m2 g = −m2 a
(4)
Take eq (1) − eq (2):
m2 g − m1 g
= m1 a + m2 a
(m2 − m1 )g
a =
m1 + m2
Pulleys and the Atwood Machine
Fnet,1y = T − m1 g = m1 a
(3)
Fnet,2y = T − m2 g = −m2 a
(4)
Take eq (1) − eq (2):
m2 g − m1 g
= m1 a + m2 a
(m2 − m1 )g
a =
m1 + m2
T =
2m1 m2 g
m1 + m2
Summary
• pulleys
• pulleys and tension
• example problems
Homework
Walker Physics:
• PREV: Ch 6. Questions: 3, 15; Problems: 1, 3, 11, 13, 15, 87
• Ch 6, onward from page 177. Questions: Problems: 7, 33, 37,
49