Download Project 2: Newton`s Laws

Project 2: Newton’s Laws
Student Objectives:
•
•
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Draw free body diagrams by recognizing which forces are present on an object and how these
forces act.
Develop the intuition to break out of the standard “horizontal/vertical” coordinate system in
order to work problems in tilted or otherwise convenient coordinate systems
Use Newton’s Laws to solve for unknown variables such as a missing force or acceleration.
Concepts:
Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting
upon an object in a given situation. A free-body diagram is a special example of the vector diagrams.
These diagrams will be used throughout our study of physics. The size of the arrow in a free-body
diagram reflects the magnitude of the force. The direction of the arrow shows the direction that the
force is acting. Each force arrow in the diagram is labeled to indicate the exact type of force. It is
generally customary in a free-body diagram to represent the object by a point (at least until our
discussion of extended bodies) and to draw the force arrow from the center of the point outward in the
direction that the force is acting. In addition, if the student wishes, the coordinate system for the
problem can also be drawn on the diagram. An example of a free body diagram for a book resting on a
table is shown below,
Ftable
mg
For a book resting on a table, there is no motion and so no net force should be acting on the book. The
two forces that do act on the book are the Earth’s gravitational force represented by mg and the force
from the table supporting the book, labeled Ftable.
Newton's second law of motion can be formally stated as follows:
The acceleration of an object as produced by a net force is directly proportional to the magnitude of the
net force, in the same direction as the net force, and inversely proportional to the mass of the object.
r
r Fnet
This verbal statement can be expressed in equation form as a =
m
Newton's acceleration equation is often rearranged to a more familiar form as shown below. The net
force is equated to the product of the mass times the acceleration.
r
r
Fnet = ma
In this entire discussion, the emphasis has been on the net force. The acceleration is directly
proportional to the net force; the net force equals mass times acceleration; the acceleration in the same
direction as the net force; an acceleration is produced by a net force. The NET FORCE. It is important to
remember this distinction. Do not use the value of merely any single force in the above equation. It is
the net force that is related to acceleration. The net force is the vector sum of all the forces.
r
r
r
Fnet = ∑ F = ma
If all the individual forces acting upon an object are known, then the net force can be determined.
Worked Examples:
1. A 5.0-kg and a 10.0-kg box are touching each other. A 45.0-N horizontal force is applied to the 5.0-kg
box in order to accelerate both boxes across the floor. Ignore friction forces and determine the
acceleration of the boxes and the force acting between the boxes.
The first approach to this problem involves the dual combination of a system analysis and an
individual object analysis. As mentioned, the system analysis is used to determine the
acceleration and the individual object analysis is used to determine the forces acting between
the objects. In the system analysis, the two objects are considered to be a single object. The
dividing line that separates the objects is ignored. The mass of the system of two objects is 15.0
kg. The free-body diagram for the system is shown below.
There are three forces acting upon the system - the gravity force (the Earth pulls down on the
15.0 kg of mass), the normal force (the floor pushes up on the system to support its weight),
and the applied force (the hand is pushing on the back part of the system). The force acting
between the 5.0-kg box and the 10.0-kg box is not considered in the system analysis since it is
an internal force. Just as the forces holding atoms together within an object are not included in
a free-body diagram, so the forces holding together the parts of a system are ignored. These
are considered internal forces; only external forces are considered when drawing free-body
diagrams. The magnitude of the force of gravity is m•g or 147 N. The magnitude of the normal
force is also 147 N since it must support the weight (147 N) of the system. The applied force is
stated to be 45.0 N. Newton's second law (a = Fnet/m) can be used to determine the
acceleration. Using 45.0 N for Fnet and 15.0 kg for m, the acceleration is 3.0 m/s2.
Now that the acceleration has been determined, an individual object analysis can be performed
on either object in order to determine the force acting between them. It does not matter which
object is chosen; the result will be the same in either case. Here the individual object analysis is
conducted on the 10.0 kg object (only because there is one less force acting on it). The freebody diagram for the 10.0-kg object is shown below.
There are only three forces acting upon it - the force of gravity on the 10.0-kg, the support
force (from the floor pushing upward) and the rightward contact force (Fcontact). As the 5.0-kg
object accelerates to the right, it will be pushing rightward upon the 10.0-kg object; this is
known as a contact force (or a normal force or an applied force or …). The vertical forces
balance each other since there is no vertical acceleration. The only unbalanced force on the
10.0-kg object is the Fcontact. This force is the net force and is equal to m•a where m is equal
to 10.0 kg (since this analysis is for the 10.0-kg object) and a was already determined to be 3.0
m/s2. The net force is equal to 30.0 N. This net force is the force of the 5.0-kg object pushing
the 10.0-kg object to the right; it has a magnitude of 30.0 N. So the answers to the two
unknowns for this problem are 3.0 m/s2 and 30.0 N.
2. A block with a mass of 5 kg starts from rest and slides down an inclined plane that is inclined at an
angle of 35o with respect to the horizontal. There is no friction between the block and the plane.
Find the acceleration of the block.
We can visualize this problem as pictured below,
5 kg
a
35o
There are two forces acting on the block in this diagram. The force of gravity is acting straight
down, pulling the block towards the center of the Earth. The second force is the normal force
from the incline on the block. This force acts perpendicular to the surface of the incline. The
free body diagram can be drawn as,
F
m
g
The dashed line represents the surface of the incline. The toughest part of this problem is
realizing that rather than working this problem is the usual x as horizontal and y as vertical
coordinate system, there is another easier coordinate system to choose. In fact, working the
problem in this old, standard coordinate system would be extremely difficult. As shown in the
first figure, we expect the block to accelerate down the incline. Let us choose a coordinate
system where all of this acceleration is occurring in one direction (and we can assign this
direction as the x direction). As always, our second axis must be perpendicular to our first. Our
free body diagram can now be drawn with this coordinate system on it.
y
F
m
35
o
x
g
The 35o angle of our incline is the angle made between the gravitational force and the −y axis. This can
be seen since the tilted x axis has been rotated 35o from the horizontal axis. Thus the gravitational
force, which is pointed straight down (or −y in the standard coordinate system) must be rotated 35o as
well.
Since all of our acceleration is down the incline, this means that our only acceleration is in the x
direction. However, we can list Newton’s Laws from both directions in this case.
∑F
x
= ma = mg sin (35°)
∑F
y
= 0 = FN − mg cos(35° )
Only the first equation is needed to find the acceleration of our block,
ma = mg sin (35°)
And,
a = g sin (35°) = 5.62m/s 2 .
3. Two blocks are setup with one hanging from the edge of an incline while the other is sitting on the
incline. The blocks are connected by a massless string and frictionless pulley. The incline of the
plane is 300 and the hanging mass has a value of 5 kilograms.
(a) If the system is at rest, find m2.
(b) Find the normal force between m2 and the plane.
∑F = m a
1
1 1
= T − m1g
∑F
∑F
a)
2x
= m2 a2 x = T − m2 g sin θ
2y
= m2 a2 y = Fn − m2 g cos θ
Note if string is tight then
a 1 = a 2x .
m1a = T − m1g
m2a = T − m2 g sin θ
(have a=0):
0 = T − m1g
0 = T − m2 g sinθ
m1g = m2 g sin θ
m2 =
m1
5
=
= 10 kg
sin θ sin 30
b) FN = m2 g cos θ
= (10 )(9.8 ) cos 30°
= 84.9 N
(have a=0)
Homework:
1. A 4kg block is held up on a frictionless incline by an external force of 15 N. Find the angle of the
incline.
2. A man pushes an 80 N crate upward along a frictionless slope at an angle of 30o relative to the
ground. If the speed of the crate is constant, find the force he exerts parallel to the slope.
3. A block with a mass of 3 kg is placed on a table. A second block with a mass of 5 kg is placed on
top of the first block.
a) the normal force of the table on the system.
b) the normal force of the 3 kg block on the 5 kg block.
c) the normal force of the 5 kg block on the 3 kg block.