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Chapter 4
Dynamics: Newton’s Laws of
Motion
• Force
• Newton’s First Law of Motion
• Mass
• Newton’s Second Law of Motion
• Newton’s Third Law of Motion
• Weight – the Force of Gravity; and the Normal Force
• Applications Involving Friction, Inclines
Assignment 4
Textbook (Giancoli, 6th edition).
Due on Oct. 15th.
http://ilc2.phys.uregina.ca/~barbi/academic/phys109/2009/phys109.html
1. On page 99 of Giancoli, problem 25.
2. On page 101 of Giancoli, problem 34.
3. On page 103 of Giancoli, problem 75.
4. A mountain climber, in the process of crossing between two cliffs by a rope,
pauses to rest. She weighs 535 N. As the drawing shows, she is closer to the left cliff
than to the right cliff, with the result that the tensions in the left and the right sides of
the rope are not the same. Find the tensions in the rope to the left and to the right of
the mountain climber. (Hint: Notice that the climber's and the rope's accelerations are
zero! Carefully draw a free body diagram for the point on the rope above the climber
and apply Newton's second law. You will need to consider both the x and the y axis
to solve for both tensions.)
Recalling Last Lecture
Force
In general we can state that if a system of two or more forces are applied on a object,
the net force on this object will be given by:
(4.1)
Example: Determine the net force
;
;
;
; and
acting on an object subject five different forces:
y
x
Newton’s Laws of Motion
Newton’s Second Law of Motion:
Newton’s Laws of Motion
Newton’s Laws of Motion
Newton’s Third Law of Motion:
The Newton’s third law can be mathematically represented as:
(4.3)
Where
is the force applied by A on B
and
is the force resulting of the reaction of by B on A
Weight – the Force of Gravity and the Normal Force
The force that produces the gravitational acceleration is called gravitational
force,
. We will discuss gravitational force in chapter 5.
points down toward to the center of the Earth, and so does
The relationship between
and
.
is given by Newton’s second law:
(4.4)
The magnitude of
on a object is called the object’s weight.
Weight – the Force of Gravity and the Normal Force
Normal Force:
The force exerted perpendicular to a surface is called the normal force. It is exactly
as large as needed to balance the force from the object
(if the required force gets too big, something breaks!)
Normal force is a contact force.
Newton’s Laws of Motion
So, the net force applied on an object by other objects is the reason for giving
acceleration to this object (changing its state of motion).
Important note: When drawing a diagram of forces (free-body diagram), consider
only the forces exerted on the body under study:
y
In translational motion,
you can consider all
forces as acting on the
center of the object
x
We are interested in the motion of the box. The forces we have to consider are the
ones applied to the box. For example:
is the force exerted by an object A on the
box; and
is the force exerted an object B on the box:
The resultant is:
Solving Problems
1. Draw a sketch.
2. For one object, draw a free-body diagram, showing all
the forces acting on the object. Make the magnitudes
and directions as accurate as you can.
3. Label each force.
4. If there are multiple objects, draw a separate diagram
for each one.
5. Resolve vectors into components.
6. Apply Newton’s second law to each component.
7. Solve.
Newton’s Laws of Motion
Newton’s Second Law of Motion:
Problem 4.33 (textbook: Three blocks on a frictionless horizontal surface are in
contact with each other, as shown in Fig. 4–51. A force
is applied to block 1 (mass
m1). (a) Draw a free-body diagram for each block. Determine (b) the acceleration of
the system (in terms of m1, m2 and m3), (c) the net force on each block, and (d) the
force of contact that each block exerts on its neighbor. (e) If m1 = m2 = m3 = 12.0 Kg
and F = 96.0 N , give numerical answers to (b), (c), and (d).
Problem 4.33 (textbook):
a) In the free-body diagrams below,
r
F1 2
r
F21
r
F23
r
F32
= force on block 1 exerted by block 2,
= force on block 2 exerted by block 1,
= force on block 2 exerted by block 3, and
= force on block 3 exerted by block 2.
By Newton’s 3rd law:
r
r
The magnitudes of F21 and
r F12 are
r equal and they point in opposite directions.
The magnitudes of F23 and F32 are equal and they point in opposite directions.
r
r
r
FN2
FN3
r
r
r
F
r
N1
r
F1 2
F23
F3 2
F21
F
r
m1g
r
m2g
r
m3g
Problem 4.33 (textbook):
b) All of the vertical forces on each block
add up to zero, since there is no
acceleration in the vertical direction.
Thus for each block,
FN = m g
For the horizontal direction, we have
∑F = F −F
12
r
F
+ F21 − F23 + F32 = F = ( m1 + m2 + m3 ) a →
r
FN1
r
m1g
r
F1 2
r
F21
r
FN2
r
m2g
r
F23
a=
F
m1 + m2 + m3
r
F3 2
r
FN3
r
m3g
Problem 4.33 (textbook):
c) For each block, the net force must be F = ma by Newton’s 2nd law. Each block
has the same acceleration since they are in contact with each other.
F1 n et =
m1 F
m1 + m 2 + m 3
F2 net =
F3 net =
m2 F
m1 + m 2 + m 3
m3 F
m1 + m 2 + m 3
Problem 4.33 (textbook):
d) From the free-body diagram, we see that for m3
F32 = F3 net =
And by Newton’s 3rd law
m3 F
m1 + m 2 + m 3
F32 = F23 = F3 net =
m3 F
m1 + m 2 + m 3
r
r
Of course, F23 and F32 are in opposite directions.
Also from the free-body diagram, we see that for m1:
F − F12 = F1 net =
m1 F
m1 + m2 + m3
→ F12 = F −
By Newton’s 3rd law:
F12 = F21 =
m1 F
m1 + m2 + m3
( m2 + m3 ) F
m1 + m 2 + m 3
→
F12 =
( m2 + m3 ) F
m1 + m2 + m3
Problem 4.33 (textbook):
e) Using the given values:
a=
F
m1 + m 2 + m 3
=
96.0 N
36.0 kg
= 2.67 m s 2
Since all three masses are the same value, the net force on each mass is
(
)
Fnet = ma = (12.0 kg ) 2.67 m s 2 = 32.0 N
This is also the value of F32 and F23. The value of F12 and F21 is
(
)
F12 = F21 = ( m 2 + m3 ) a = ( 24 kg ) 2.67 m s 2 = 64.0 N
To summarize:
Fnet 1 = Fnet 2 = Fnet 3 = 32.0 N
F12 = F21 = 64.0 N
F23 = F32 = 32.0 N
The values make sense in that in order of magnitude, we should have F > F21 > F3 2 ,
since F is the net force pushing the entire set of blocks, F12 is the net force pushing
the right two blocks, and F23 is the net force pushing the right block only.
Solving problems
The examples in the textbook are good source of information on how to solve
problems using diagrams of forces and using the methods of components for vector
addition.
Please, study them !!!!!!!!!!!!!
Tension in a Flexible Cord
When a cord, string or rope pulls
on an object, it is said to be under
tension, and the force it exerts is
called a tension force.
The figure shows you pulling a SYSTEM consisting of two boxes connected by a
cord.
The force you apply on the system is
.
Note: If you consider each individual component of the system, the force
is actually only applied on box A and NOT on box B.
The force will be transmitted to box B via the tension in the cord
.
Tension in a Flexible Cord
In the figures:
Fig. (b) is a diagram of all forces applied on box A.
Note that box A exerts a force on the cord connecting
the two boxes. According to Newton’s third law,
the cord responds with a force (tension force),
,
of same magnitude but opposite direction.
Fig. (c) is a diagram of forces applied on B. Now the cord
pulls box B.
If we neglect the mass of the cord, this pull will
correspond to a force equal to the tension
found on the other end of the cord (connected to A).
Note that this is NOT true if a mass is assigned to the
cord.
Tension in a Flexible Cord
Problem 4.35 (textbook: Suppose two boxes on a frictionless table are connected
by a heavy cord of mass 1.0 kg. Calculate the acceleration of each box and the
tension at each end of the cord, using the free-body diagrams shown in Fig. 4–53.
Assume FP = 40.0 N, and ignore sagging of the cord. Compare your results to
Example 4–12 and Fig. 4–22.
Problem 4.35 (textbook):
Initially, treat the two boxes
and the rope as a single
system. (But note that the rope has
a mass, so the tension is not the same throughout the rope)
r
The only accelerating force on the system is FP . The mass of the system is
23.0 kg, and so using Newton’s 2nd law, the acceleration of the system is
a =
FP
m
=
40.0 N
23.0 kg
= 1.74 m s 2
This is the acceleration of each piece of the system.
r
Now consider the left box alone. rThe only force on it is F B T , and it has the
acceleration found above. Thus F B T can be found from Newton’s 2nd law.
(
)
FBT = m B a = (12.0 kg ) 1.74 m s 2 = 20.9 N
r
r
Now consider the rope alone. The net force on it is FTA − FTB , and it also has the
acceleration found above. Thus FTA can be found from Newton’s 2nd law.
(
)
FTA − FTB = mC a → FTA = FTB + mC a = 20.9 N + (1.0 kg ) 1.74 m s 2 = 22.6 N
Tension in a Flexible Cord
Problem 4.78 (textbook: (a) What minimum force F is needed to lift the piano
(mass M) using the pulley apparatus shown in Fig. 4–60? (b) Determine the tension
in each section of rope: FT1 , FT2 , FT3 , and FT4 ,
Problem 4.78 (textbook):
(a)
To find the minimum force, assume that the piano is moving with
a constant velocity. Since the piano is not accelerating,
FT 4 = M g
For the lower pulley, since the tension in a rope is the same
throughout, and since the pulley is not accelerating, it is seen that
FT 1 + FT 2 = 2 FT1 = M g
→
FT1 = FT 2 = M g 2
Also, since F = F
, then
T2
r
FT1
r
FT2
F = Mg 2
(b)
Draw a free-body diagram for the upper pulley.
From that diagram, we see that
FT 3 = FT 1 + FT 2 + F =
3M g
FT 1 = FT 2 = M g 2
FT 3 = 3 M g 2
r
FT4
Lower
Pulley
Upper
Pulley
r
FT3
r
FT2
r
F
r
FT1
2
FT 4 = M g