Chapter 5
Lecture 8:
Force and Motion: II
Particles Under a Net Force

If an object that can be modeled as a particle
experiences an acceleration, there must be a
nonzero net force acting on it



Model is particle under a net force model
Draw a free-body diagram
Apply Newton’s Second Law in component
form
Newton’s Second Law,
Example 1a

Forces acting on the
crate:



A tension, acting through
the rope, is the
magnitude of force T
The gravitational force, Fg
The normal force, n ,
exerted by the floor
Newton’s Second Law,
Example 1b

Apply Newton’s Second Law in
component form:
F
 Fy
x


 T  m ax
 n  Fg  0  n  Fg
Solve for the unknown(s)
If the tension is constant, then a is
constant and the kinematic equations
can be used to more fully describe the
motion of the crate
Note About the Normal Force


The normal force is not
always equal to the
gravitational force of the
object
For example, in this case
F
y
 n  Fg  F  0
a n d n  Fg  F

n may also be less than Fg
Inclined Planes

Forces acting on the object:



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The normal force acts
perpendicular to the plane
The gravitational force acts
straight down
Choose the coordinate system
with x along the incline and y
perpendicular to the incline
Replace the force of gravity with
its components
Multiple Objects


When two or more objects are connected or
in contact, Newton’s laws may be applied to
the system as a whole and/or to each
individual object
Whichever you use to solve the problem, the
other approach can be used as a check
5.9: Applying Newton’s Laws
Sample problem
Figure 5-12 shows a block S (the sliding block)
with mass M =3.3 kg. The block is free to move
along a horizontal frictionless surface and
connected, by a cord that wraps over a
frictionless pulley, to a second block H (the
hanging block), with mass m 2.1 kg. The cord
and pulley have negligible masses compared to
the blocks (they are “massless”). The hanging
block H falls as the sliding block S accelerates to
the right. Find (a) the acceleration of block S, (b)
the acceleration of block H, and (c) the tension
in the cord.
Key Ideas:
1. Forces, masses, and
accelerations are involved, and
they should suggest Newton’s

second law of motion: F  m a


2.
The expression F  m a is a
vector equation, so we can write
it as three component
equations.
3. Identify the forces acting on
each of the bodies and draw
free body diagrams.
5.9: Applying Newton’s Laws
For the sliding block, S, which does not
accelerate vertically.
Also, for S, in the x direction, there is only
one force component, which is T.
For the hanging block, because the
acceleration is along the y axis,
From the free body diagrams, write

Newton’s Second LawF  m a in the
vector form, assuming a direction of
acceleration for the whole system.
Identify the net forces for the sliding
and the hanging blocks:
We eliminate the pulley from
consideration by assuming its mass to be
negligible compared with the masses of
the two blocks.
With some algebra,
Forces of Friction

When an object is in motion on a surface or
through a viscous medium, there will be a
resistance to the motion


This is due to the interactions between the object
and its environment
This resistance is called the force of friction
Forces of Friction, cont.

Friction is proportional to the normal force

ƒs  µs n and ƒk= µk n
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μ is the coefficient of friction
These equations relate the magnitudes of the forces,
they are not vector equations
For static friction, the equals sign is valid only at
impeding motion, the surfaces are on the verge of
slipping
Use the inequality if the surfaces are not on the verge
of slipping
Forces of Friction, final


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The coefficient of friction depends on the
surfaces in contact
The force of static friction is generally greater
than the force of kinetic friction
The direction of the frictional force is opposite
the direction of motion and parallel to the
surfaces in contact
The coefficients of friction are nearly
independent of the area of contact
Static Friction




Static friction acts to keep the
object from moving
If F increases, so does ƒ s
If F decreases, so does ƒ s
ƒs  µs n

Remember, the equality holds
when the surfaces are on the
verge of slipping
Kinetic Friction



The force of kinetic
friction acts when the
object is in motion
Although µk can vary
with speed, we shall
neglect any such
variations
ƒk = µk n
Explore Forces of Friction


Vary the applied force
Note the value of the
frictional force


Compare the values
Note what happens
when the can starts to
move
6.3 Properties of friction
Property 1. If the body does not move, then the static frictional force and
the component of F that is parallel to the surface balance each other. They
are equal in magnitude, and is fs directed opposite that component of F.
Property 2. The magnitude of has a maximum value fs,max that is given by
where ms is the coefficient of static friction and FN is the magnitude of the
normal force on the body from the surface. If the magnitude of the component
of F that is parallel to the surface exceeds fs,max, then the body begins to
slide along the surface.
Property 3. If the body begins to slide along the surface, the magnitude of the
frictional force rapidly decreases to a value fk given by
where mk is the coefficient of kinetic friction. Thereafter, during the sliding, a
kinetic frictional force fk opposes the motion.
Friction in Newton’s Laws
Problems


Friction is a force, so it simply is included in
the  F in Newton’s Laws
The rules of friction allow you to determine
the direction and magnitude of the force of
friction
Friction Example



The block is sliding down the
plane, so friction acts up the
plane
This setup can be used to
experimentally determine the
coefficient of friction
µ = tan q


For µs, use the angle where
the block just slips
For µk, use the angle where
the block slides down at a
constant velocity
Friction Example, 2
Uniform Circular Motion,
Acceleration

A particle moves with a constant speed in a
circular path of radius r with an acceleration:
ac 


v
2
r
The centripetal acceleration, a c is directed toward
the center of the circle
The centripetal acceleration is always
perpendicular to the velocity
Uniform Circular Motion, Force



A force, Fr , is
associated with the
centripetal acceleration
The force is also
directed toward the
center of the circle
Applying Newton’s
Second Law along the
radial direction gives
F
 m ac  m
v
r
2
Uniform Circular Motion, cont
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

A force causing a centripetal
acceleration acts toward the
center of the circle
It causes a change in the
direction of the velocity vector
If the force vanishes, the
object would move in a
straight-line path tangent to
the circle

See various release points in
the active figure
Motion in a Horizontal Circle


The speed at which the object moves
depends on the mass of the object and the
tension in the cord
The centripetal force is supplied by the
tension
T sin θ = m ac
Conical Pendulum

The object is in
equilibrium in the
vertical direction and
undergoes uniform
circular motion in the
horizontal direction



∑Fy = 0 → T cos θ = mg
∑Fx = T sin θ = m ac
v is independent of m
v 
L g s in q ta n q
Horizontal (Flat) Curve


The force of static friction
supplies the centripetal
force
The maximum speed at
which the car can negotiate
the curve is
v 

 s gr
Note, this does not depend
on the mass of the car