chapter 4 dynamics: newton’s laws of...
TRANSCRIPT
• 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
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 .
Tension in a Flexible CordTension in a Flexible 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 CordTension 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), , 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.
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: and
Tension in a Flexible CordTension in a Flexible Cord
, , , T3T2T1 FFF ,T4F
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,
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
T 4F Mg=
T1 T 2 T1 T1 T 22 2F F F Mg F F Mg+ = = → = =
Also, since , then
(b)
Draw a free-body diagram for the upper pulley.
From that diagram, we see that
T 2F F=
2F Mg=
T 3 T1 T 2
3
2
MgF F F F= + + =
T1 T 2 T 3 T 42 3 2 F F Mg F Mg F Mg= = = =
Lower
Pulley
Upper
Pulley
Fr
T1Fr
T1Fr
T2Fr
T2Fr
T3Fr
T4Fr
On a microscopic scale, most surfaces are rough.
When we try to SLIDE an object across a table, the
roughness of both surfaces (object’s and table’s
surfaces) tends to oppose the object‘s motion. This
effect is called sliding friction, or kinetic friction.
FrictionFriction and and InclinedInclined PlanesPlanes
effect is called sliding friction, or kinetic friction.
The opposition to the body’s motion is due to a force
called friction force.
FrictionFriction and and InclinedInclined PlanesPlanes
The friction force acts in the opposite direction to the object’s motion.
In the figure, the forces applied on the box are
: the applied force;
: the normal force;
: the gravitational force;
: the friction force.: the friction force.
The reason for the friction force is not very well
understood, though it has been very well modeled.
Experiments have shown that the magnitude of the friction force is approximately
proportional to the magnitude of the normal force between two surfaces, and does
not depend on the total surface area of contact between the two surfaces.
(4.5)
FrictionFriction and and InclinedInclined PlanesPlanes
The term is called coefficient of kinetic friction, and its value depends on the
nature of the two contacting surfaces.
Sliding friction (or kinetic friction) is not present when the object is not SLIDING.
(4.5)
Sliding friction (or kinetic friction) is not present when the object is not SLIDING.
However, you might have experienced the fact that it takes different (magnitude)
forces to slide a same object across different surfaces.
This is due to the so called static friction. The force of static friction is parallel to the
two contacting surfaces.
Note: Let me stress again � there is no sliding force if the object moves (or is at
rest) without sliding.
FrictionFriction and and InclinedInclined PlanesPlanes
The force of static friction varies with the applied force on the object � The more
force you apply, the bigger is the force of static friction.
There will be a critical applied force that will ultimately set the object in motion. This
point will correspond to the maximum force of static friction:
(4.6)
The tem is called the coefficient of static friction.
Note that, at any point, it is true that force of static
friction will be such that:
(4.6)
FrictionFriction and and InclinedInclined PlanesPlanes
The magnitude of the friction forces depends on the nature of the two surfaces in
contact.
Problem 4-36 (textbook: If the coefficient of kinetic friction between a 35-kg crate
and the floor is 0.30, what horizontal force is required to move the crate at a steady
speed across the floor? What horizontal force is required if is zero?
Tension in a Flexible CordTension in a Flexible Cord
Problem 4.36 (textbook):
A free-body diagram for the crate is shown.
The crate does not accelerate vertically, and so
Also, the crate does not accelerate horizontally
(steady speed), and so
NF m g=
P frF F=
mgr
NFr
frFr
PFr
Putting this together, we have
If the coefficient of kinetic friction is zero, then the horizontal force required is 0 N,
since there is no friction to counteract. Of course, it would take a force to START the
crate moving, but once it was moving, no further horizontal force would be necessary
to maintain the motion.
P frF F=
( )( )( )2 2
P fr N0.30 35 kg 9.8m s 103 1.0 10 N
k kF F F mgµ µ= = = = = = ×
Problem 4-38 (textbook Suppose that you are standing on a train accelerating at
0.20 g. What minimum coefficient of static friction must exist between your feet and
the floor if you are not to slide?
Tension in a Flexible CordTension in a Flexible Cord
Problem 4.38 (textbook):
A free-body diagram for you as you stand on the train is shown.
You do not accelerate vertically, and so
NF mg=
mgr
NFr
frFr
The maximum static frictional force is , and that must be greater than or equal
to the force needed to accelerate you.
The static coefficient of friction must be at least 0.20 for you to not slide.
s NFµ
fr N 0.20 0.20
s s sF ma F ma mg ma a g g gµ µ µ≥ → ≥ → ≥ → ≥ = =
FrictionFriction and and InclinedInclined PlanesPlanes
The problem with inclined plans are no different from the problems we have
approached so far.
However, remember:
1- The normal force is ALWAYS perpendicular to the plane where the object lies on.
� So, in an inclined plane its is not going to be in the vertical direction.� So, in an inclined plane its is not going to be in the vertical direction.
(See next slide)
FrictionFriction and and InclinedInclined PlanesPlanes
Three forces can ALWAYS be identified acting on an object moving on an inclined
place:
• Gravity (vertical);
• Friction (along the surface);
• Normal (perpendicular to the surface).friend 1
friend 2
So, it is convenient to define your
coordinate system such that the x
direction coincides with the inclined
line, and y in the direction perpendicular
to the plane.
You can surely have other applied forces:
� for instance, you (or you and a group
of friends) can push or pull this object.
you
Problem 4-41 (textbook A 15.0-kg box is released on a 32º incline and accelerates
down the incline at 0.30 m/s2. Find the friction force impeding its motion. What is the
coefficient of kinetic friction?
Tension in a Flexible CordTension in a Flexible Cord
NFr
frFr
y
x
θ
θ
mgr
frF
Problem 4.41 (textbook):
A free-body diagram for the box is shown.
Write Newton’s 2nd law for each direction:
Notice that the sum in the y direction is 0, since
y
x
θ
θ
mgr
NFr
frFr
fr
N
sin
cos 0
x x
y y
F mg F ma
F F mg ma
θ
θ
= − =
= − = =
∑∑
Notice that the sum in the y direction is 0, since
there is no motion (and hence no acceleration) in
the y direction. Solve for the force of friction.
Now solve for the coefficient of kinetic friction. Note that the expression for the
normal force comes from the y direction force equation above.
mg
( ) ( )( )fr
2 o 2
fr
sin
sin 15.0 kg 9.80m s sin 32 0.30m s 73.40 N 73 N
x
x
mg F ma
F mg ma
θ
θ
− = →
= − = − = ≈
( )( )( )fr
fr N 2 o
73.40 Ncos 0.59
cos 15.0 kg 9.80m s cos 32k k k
FF F mg
mgµ µ θ µ
θ= = → = = =
Problem 4-65 (textbook A bicyclist of mass 65 kg (including the bicycle) can coast
down a 6.0º hill at a steady speed of 6.0 Km/h because of air resistance. How much
force must be applied to climb the hill at the same speed and same air resistance?
Tension in a Flexible CordTension in a Flexible Cord
Problem 4.41 (textbook):
Consider a free-body diagram for the cyclist coasting downhill at a constant speed.
Here we call Ffr the friction due to air resistance (and not sliding or static friction).
Since there is no acceleration, the net force in each direction must be zero. Write
Newton’s 2nd law for the x direction.
This establishes the size of the air friction force at
6.0 km/h, and so can be used in the next part.
fr frsin 0 sin
xF mg F F mgθ θ= − = → =∑ y
θ
NFr
frFr
θ6.0 km/h, and so can be used in the next part.
Now consider a free-body diagram for the cyclist
climbing the hill. Fp is the force pushing the cyclist
uphill. Again, write Newton’s 2nd law for the x direction,
with a net force of 0.
xθ
mgr
θ
y
x
θθ
mgr
NFr
frFr
PFr
fr Psin 0
xF F mg Fθ= + − = →∑
( )( )( )P fr
2 o 2
sin 2 sin
2 65 kg 9.8 m s sin 6.0 1.3 10 N
F F mg mgθ θ= + =
= = ×
AssignmentAssignment 55
Textbook (Giancoli, 6th edition), Chapters 4 and 6:
Due on Thursday, October 23, 2008
- Problem 52 - page 102 of the textbook
- Problem 87 - page 105 of the textbook
- Problems 6 and 8 - page 162 of the textbook
Tutorial next week, Oct. 15, CL-127 ���� solving an old midterm.
MidtermMidterm
Date: October 16, 1pm – 2:15pmRemind:
- There are four problems � Full mark will be considered only when all
problems have been completed
- You ARE allowed to use a calculator
- A equation sheet will be provided with some useful equations and identities.
- You are NOT allowed to:
- Use laptops;
- Have cell phones (please, turn it off if you have one);
- Use or consult any material other than those provided by me;
-Make sure you bring your photo ID with you
- Do not forget to write your name and UofR ID on each of the notebook used
to solve your problems..