Transcript Topic 2_2_Ext B__Applications of Newton`s second law

Topic 2.2 Extended
B – Applications of Newton’s second
Topic 2.2 Extended
B – Applications of Newton’s second
Newton’s 2nd law is all about forces acting on masses.
In this section you’ll be introduced to some of the
basic forces of mechanics.
If you drop a mass m
y
The fbd
=
=
=
=
ma
ma
mg
-mgy
W
W
W
y
∑F
W
W
W
x
x
very close to the
surface of the earth it
will fall, satisfying
the 2nd law:
vector form of weight
|W|= W = mg scalar form of weight
W = mg
weight
Weight always
points to the
center of Earth.
The fbd
N
N
The fbd
Topic 2.2 Extended
B – Applications of Newton’s second
W
W
Another force you are familiar with is the normal
force N.
This is the surface contact force that you feel when
you push on a solid, such as a block of wood, a wall,
or a floor.
The normal force always acts perpendicular to the
contact surface, and points away from it.
Note that the weight always points to the
center of Earth, but the normal force points
W
perpendicular to the contact surface.
N
Note that the normal
force only exists if the
mass is in contact with
a surface.
But the weight is always
present.
W
N
Topic 2.2 Extended
B – Applications of Newton’s second
Yet another force you are familiar with is the
friction force f.
If you attempt to slide an object to the right it will
resist you with friction.
The friction always acts in a direction opposite to
the pull and parallel to the contact surface.
Thus the two surface contact forces are friction and
the normal. They are perpendicular to each other.
N
The fbdW
N
f
W
f
the friction
force
W
your pull to
the right
Topic 2.2 Extended
B – Applications of Newton’s second
Topic 2.2 Extended
B – Applications of Newton’s second
A fourth force you are familiar with is the “tug of
war” force (aka the “pulling” force.)
A pull along the length of a wire or a rope or a pipe
is called a tension T.
Note that the tension appears only when the rope is
tightened. Tension points in the direction of the line
of the rope, away from the object being pulled.
You cannot create a pushing force (compression) with a
rope.
The fbd
W
N
N
N
f
T
T
Wthe tension
f
W
W
Topic 2.2 Extended
B – Applications of Newton’s second
We say that
Suppose a 5-kg crate is to be dragged across "T
a has been
virtually frictionless ice surface by a rope inclined
resolved."
at 30°. The pull is 40-n.
(a) What is the acceleration of the crate?
∑Fx =
T cos 30° =
40 cos 30° =
ax =
max
5 ax
5 ax
6.93 m/s2
y
N
T
30°
T cos 30°
N
x
The fbd
30°
W
W
T sin 30°
Topic 2.2 Extended
B – Applications of Newton’s second
Suppose a 5-kg crate is to be dragged across a
virtually frictionless ice surface by a rope inclined
at 30°. The pull is 40-n.
(b) What is value of the normal force?
∑Fy
N + T sin 30° + -W
N + 40 sin 30° + -W
N
N
N
N
=
=
=
=
=
=
may
5 ay
5  0
mg - 20
5(10) - 20
30 n
30°
W
y
N
T
30°
T cos 30°
The weight of the
crate is 50 n. Why
isn't the normal
force equal to the
weight?
T sin 30°
x
The fbd
W
Officially, accelerations are
Topic
not drawn from the
body. 2.2 Extended
If included
as a reference
B – Applications
of Newton’s
in the fbd, have the
acceleration off to the
Suppose a side.
5-kg crate is suspended by a
spring scale from the top of an elevator
which is accelerating upward at 2 m/s2.
What is the “apparent weight” of the
crate (as read by the spring scale)?
∑Fy = may
T - W = 5 ay
T - mg = 5 ay
T = 5 ay + mg
T = 5(2) + 5(10)
T = 60 n
The spring scale reads the tension.
Since the “real weight” of the crate is
50 n, it appears to have “gained” 10 n.
An observer in the elevator (a noninertial frame) would not know of the
elevator’s acceleration and would think
that 60-n was the actual weight.
second
a
T
W
y
a
T
W
The fbd
Topic 2.2 Extended
B – Applications of Newton’s second
Many problems in physics involve more
than one body.
As an example, consider the system
shown here:
When released from rest, the two masses
will begin to accelerate.
If the string doesn’t change length,
their accelerations will be the same.
If the pulley is light enough, we don’t
have to take into account its presence
in the system as a third accelerating
body.
All the pulley does is redirect the
tension.
Thus, we consider this system as a twobody system, with each body having the
same acceleration.
a
mm1m1 1mm1 1
m22
m
m2
m2
m
2
a
Topic 2.2 Extended
B – Applications of Newton’s second
Since there are two bodies, there are
two free body diagrams.
Each body has a weight.
a
mm1m1 1mm1 1
Body 1 has a normal force, because it
is in contact with a solid surface.
Each body has a tension.
Tension has the property that it is
the same everywhere in a cord (if it
is light). Therefore, only one
symbol T is needed.
N1
Body 1 also has friction,
a
acting in opposition to the
T
rightward pull of the tension.
f
And for book-keeping purposes
we can put the accelerations in
W1
the FBDs. Note only one symbol
FBD, m1
is needed, since they are the
same.
m22
m
m2
m2
m
a
2
T
a
W2
FBD, m2
Topic 2.2 Extended
B – Applications of Newton’s second
a
Suppose we want to find the values of
m1
T and a:
Each body will satisfy the 2nd law:
Body 1:
∑Fy = m1ay
∑Fx = m1ax
T – f = m1a
Body 2:
∑Fy = m2ay
T – W2 = m2(-a)
T = m2g - m2a
a
N1 – W1 = m1(0)
T = m1a + f
T – m2g = -m2a
m2
N1 = W1
expected
…and of no
use in this
problem…
T
N1
a
a
T
f
W1
FBD, m1
W2
FBD, m2
Topic 2.2 Extended
B – Applications of Newton’s second
a
The final equations for each body
were:
Body 1:
m1
Body 2:
m2
T = m1a + f
T = m2g - m2a
To find out what a is in terms of the
other quantities, set the two equations
equal (eliminating the T):
m1a + f = T = m2g - m2a
m1a + f = m2g - m2a
m1a + m2a = m2g - f
m2g – f
a =
m1 + m2
Note that we need values for m1,
m2, and f in order to find a:
T
N1
a
(m1 + m2)a = m2g - f
a
a
T
f
W1
FBD, m1
W2
FBD, m2
Topic 2.2 Extended
B – Applications of Newton’s second
a
The final equations for each body were:
Body 1:
Body 2:
T = m1a + f
T = m2g - m2a
Suppose m1 is 2 kg and m2 is 3 kg and
there is no friction. What are a and
T?
m2g – f
(3)(10) – 0
= 6 m/s2
a =
=
m1 + m2
2+ 3
You can use either equation for
T:
T = m1a + f = (2)(6) + (0) = 12 n
What is the maximum friction
force you can have before the
masses do not accelerate?
(3)(10) – f
 f = 30 n
0 =
2+ 3
m1
m2
a
T
N1
a
a
T
f
W1
FBD, m1
W2
FBD, m2
Topic 2.2 Extended
B – Applications of Newton’s second
Sometimes problems will involve bodies
30°
45°
T2
that are not even moving, and might not
T1
knot
be masses in the conventional sense:
T3
Suppose three cords support a crate of
mass m as shown.
A knot is at the intersection of the
three cords.
If we know the geometry of the system (the
angles), we can find all of the tensions.
Since the three ropes all do
T3
different things, their tensions
T
2
T1
will all be different.
45°
30°
Not only can we make a free body
diagram of the crate, but we can
make one for the knot.
T3
mg
No book-keeping acceleration is
needed, since nothing is moving.
FBD, m
FBD, knot
Topic 2.2 Extended
B – Applications of Newton’s second
From the free body diagram for the crate
of mass m:
T3 - mg = 0
T3 = mg
From the free body diagram for the knot:
∑Fx = 0
T2 cos 45° - T1 cos 30° = 0
0.707T2 - 0.866T1 = 0
T2 = 1.225T1
30°
T1
∑Fy = 0
T2 sin 45° + T1 sin 30° - T3 = 0
0.707T2 + 0.5T1 = mg
T2
T1
30°
45°
T2
knot
T3
T3
45°
0.707(1.225T1) + 0.5T1 = mg
T1 = mg/1.366
T3
mg
T2 = 1.225(mg/1.366)
T2 = 0.897mg
FBD, knot
FBD, m
Topic 2.2 Extended
B – Applications of Newton’s second
Our results:
T1 = mg/1.366
30°
T2 = 0.897mg
T1
T3 = mg
If the weight of the crate is 20 n, what
45°
T2
knot
T3
are the three tensions?
W = 20 n = mg
T1 = 20/1.366 = 14.64 n
T2 = 0.897(20) = 17.94 n
T3 = 20 n
Why don’t T1 and T2 add up to T3?
T2
T1
30°
T3
45°
Is it possible to have a
geometry such that T1 and T2
each support exactly half the
crate’s weight.
T3
mg
FBD, knot
FBD, m
Topic 2.2 Extended
B – Applications of Newton’s second
Sometimes you have to do a
little analysis to decide
which way the masses will
accelerate.
If you choose an arbitrary
a
direction, consistent with all
bodies, you can still solve the
problem.
Either of these choices is
consistent.
Suppose W1 = 10 n, W2 = 30 n, and
that there is no friction.
I will assume the acceleration
of the system is as shown.
The FBDs are therefore:
Thiscs
csrequires
requiresyou
you to
This
tobreak
break
downonly
3 1
Here
is ourdown
m2g N.
vectors:vector,
a, T, and
reference
angle:
a
m2
m1
30°
a
y
y
T
N
T
a
x
30°
x
m2g
FBD, m2
m1g
FBD, m1
Topic 2.2 Extended
B – Applications of Newton’s second
For m1:
W1 = m1g
10 = m1(10)
m1 = 1 kg
-m
T – m1g =
T – 10 = -1a
T = 10 +
For m2:
W2 = m2g
30 = m2(10)
m2 = 3 kg
a
1a
-1a
m2
a
m1
30°
a
y
T
x-direction
N
a
m2g sin 30° - T = -m2a
30 sin 30° - T =
-3a
x
T = 15 + 3a
10 +
-1a
= 15 + 3a
4a = -5
a = -1.25 m/s2
T
30°
m2g
FBD, m2
T = 10 + -1a
T = 10 + -1(-1.25)
T = 11.25 n
m1g
FBD, m1
Topic 2.2 Extended
B – Applications of Newton’s second
Interpreting the results:
a
Since our acceleration turned
out to be negative, we know
that we picked the wrong
direction.
Thus, m1 really goes upward,
and m2 goes down the incline.
The numeric values we got for
a and T during the analysis
are still valid.
m2
a
m1
30°
a
y
T
N
T
a
x
30°
m2g
FBD, m2
m1g
FBD, m1
Topic 2.2 Extended
B – Applications of Newton’s second
An alternate solution:
Since both masses are
m1
connected by the cord and move
as one, we can treat them as
one mass (of 1 + 3 kg):
a
The only forces we need are
the ones that are in the
direction of motion.
In the case of m1 that force
is m1g = 10 n.
In the case of m2 that force
is m2g sin 30° = 15 n.
15
10
m2
the winner
a
15 - 10 = (1 + 3)a
a = 1.25 m/s2
a
m2
m1
30°
a
y
T
N
T
a
x
30°
m2g
FBD, m2
m1g
FBD, m1
Topic 2.2 Extended
B – Applications of Newton’s second
Two masses M and m are connected by a
very light cord over a very light
pulley, as shown. Assume M > m.
Suppose we want to know what the
acceleration of the blocks is, once
released from rest.
For m
M
For M
T – mg = ma
T – Mg =
T = ma + mg
-Ma
T = Mg - Ma
ma + mg = Mg - Ma
ma + Ma = Mg - mg
(m + M)a = (M – m)g
a =
m
M – m
g
M + m
Question: If m = M, what does our equation
predict for a?
T
T
a
a
mg
Mg
FBD, m
FBD, M
Topic 2.2 Extended
B – Applications of Newton’s second
Finally, we can tie together Newton's second law
with the kinematic equations by the acceleration a:
Suppose a 750-kg car that is traveling at 20 m/s
slams on the brakes and comes to a halt after
traveling 125 m. What was the friction force
between the tires and the pavement during the
braking?
Since time is not given or required, use
v2 = vo2 + 2ad
02 = 202 + 2a(125)
a = -1.6 m/s2
F = ma
F = (750)(-1.6)
F = 1200 n
kinematics
dynamics