Download Document

General Physics I
Spring 2011
Applying Newton’s Laws
1
Equilibrium
• An object is in equilibrium if the net force acting on it is zero.
According to Newton’s first law, such an object will remain at
rest or moving at constant velocity. Thus, the acceleration of
the object is zero (consistent with Newton’s second law).
• Since the net force on the object is zero, we have
Fnet = F1 + F2 + F3 +... = 0.
• For two-dimensional situations, we resolve into components:
(Fnet )x = F1x + F2x + F3x +... = 0.
(Fnet ) y = F1y + F2 y + F3 y +... = 0.
• A shorthand way of writing the two equations above is to use
the summation symbol:
∑ Fx = 0 and ∑ Fy = 0.
(Equilibrium)
2
Example: Equilibrium in One and Two
Dimensions
Engine
Ring
3
Equilibrium Example (Continued)
• For the engine, we have
∑ Fy =T1 − w = 0. ---- (i)
Thus, T1 = w.
• For the ring, we have:
∑ Fx =T3 cos60°−T2 = 0. ---- (ii)
∑ Fy =T3sin60°−T1 = 0. ---- (iii)
Suppose the weight of the engine = 2000 N.
From Eq. (i), we have: T1 = w = 2000 N.
T1
2000 N
From Eq. (iii), we have: T3 =
=
= 2309 N.
sin 60° sin 60°
From Eq. (ii), we have: T2 = T3 cos 60° = (2309 N) cos 60° = 1155 N.
4
Example: Ramp or Incline
5
Workbook: Chapter 5, Question 6
6
Dynamics and Newton’s Second Law
• If the net force on an object is not zero, then the object will
accelerate, according to Newton’s second law. If we know
the forces acting, we can find the acceleration. We can then
use the kinematics equations to find velocity and position,
which give complete information about the motion of the
object. This is the subject of dynamics.
• As we have seen previously, Newton’s second law is:
Fnet = ma.
• Rewriting in component form gives:
∑ Fx = max and ∑ Fy = may .
7
Solving Dynamics Problems
• Identify all the forces acting on the object under
investigation.
• Choose a coordinate system such that the acceleration is
along one axis (e.g., x-axis)
• Draw a free-body diagram (very important!)
• Use Newton’s second law in component form to find the net
force along the x and y directions.
• Solve for acceleration if given all the forces. Solve for
unknown force if given acceleration and other forces.
8
Workbook: Chapter 5, Questions, 7, 8
9
Textbook: Chapter 5, Problem 11
10
Mass and Weight
• An object’s mass is a measure of its inertia. The mass
can be determined from Newton’s 2nd law: apply a net
force, measure the resulting acceleration and determine
the mass from their magnitudes:
m = Fanet .
• An object’s mass is independent of its location.
11
Mass and Weight
• An object’s weight is the gravitational
force acting on it due to a nearby
massive body (e.g., Earth, the Moon,
etc.). Thus, if the weight of an object
is w and the gravitational acceleration it produces is ag, then
w
ag = m , or, w = mag .
Now, ag = g, so in terms of magnitudes,
w = mg.
• Close to the Earth’s surface, the
gravitational (free-fall) acceleration
has a nearly constant magnitude of
9.8 m/s2 and points toward the
center of the Earth (i.e., locally
vertically downward).
12
Mass and Weight
• Note that g is independent of
mass.
• Note further that the weight is
proportional to the gravitational
acceleration. Thus, weight
depends on location since g
varies with location. It varies
slightly at different locations on
Earth. On the Moon, g = 1.6
m/s2. The mass,≈however, is
the same everywhere. It is
easier to lift things on the
Moon than on Earth, but just
as difficult to push something
horizontally.
13
You throw a ball straight up. Moments before it reaches its
maximum height, what force(s) are acting on the ball after
it has been released?
1.
2.
3.
4.
Weight (down)
Weight (down) and inertia (up)
Weight (down) and the force of the throw (up)
Inertia (up)
14
Apparent Weight
• When you stand on a scale at
rest, it does not directly read
your weight. Rather, it measures
the normal force that you exert
on it due to contact with your feet
(which causes springs to be
compressed). Since you are in
equilibrium, the upward normal
force that the scale exerts on
you is equal in magnitude to your
weight. Thus, the normal force
that you exert on the scale
(which is the reaction to the force
exerted by the scale on you) is
equal in magnitude to your
weight.
n
w
a = 0;
n= w
15
Apparent Weight
• If you stand on a scale that is
accelerating, its reading will
not be equal to your weight.
Its reading, i.e., the magnitude
of the normal force, is said to
be your apparent weight.
• Suppose that a person stands
on a scale in an elevator,
which has an upward
acceleration of magnitude a.
We can find her apparent
weight using Newton’s
second law:
∑ Fy = n − w = may = ma.
n = w+ ma = m( g + a). Thus,
n > w.
16
Apparent Weight
• If the elevator has a downward
acceleration, then
n – w = m(– a). So, n= m(g – a).
• If the elevator cable breaks, then a =
g, and so n = 0, Hence, the person
(and everything else in the elevator)
has an apparent weight of zero. This
is apparent weightlessness. This is
what astronauts in orbit (or people
on the "Vomit Comet") feel. This is
because the astronauts and the
space station are all in free fall
around Earth with exactly the same
acceleration, just like the person,
scale, and elevator are all in free fall
in the runaway elevator.
n
ay = – a
w
"Zero-g"
17
Workbook: Chapter 5, Question 19
18
Textbook: Chapter 5, Problem 52
(a) Choose positive y direction upward.
Rocket initially at rest so no air drag force.
∑ Fy = Fthrust − w = may.
Fthrust − w Fthrust − mg Fthrust
ay =
=
= m −g
m
m
5
= 3.0×10 N −9.8 m/s2 = 5.2 m/s2.
20,000 kg
Fthrust
w
(b) ∑ Fy = Fthrust − mg = may.
Fthrust = m( g + ay ), i.e.,
Fthrust
m=
. Assuming the same thrust, we have
(g + ay )
5N
3.0
×
10
4 kg.
m=
=
1.9
×
10
(9.8 m/s2 + 6.0 m/s2)
Thus, mass of fuel burned = 20,000 kg −19,000 kg =1000 kg.
19