Wappingers Central School District

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```					       FN

Ff                  FA

Fg

An Unbalanced Force

Fnet
a
m
What is an unbalanced force?
• According to Newton’s Second Law of
Motion:
– an unbalanced force is one that causes an
object to accelerate.
• Positively or negatively (increasing or decreasing
velocity)
or
• Change direction. (a force that causes an object to
move in a circular path)
Strategies for solving force related problems.
1.       Draw a free-body diagram that summarizes all the forces
that act on the object in the vertical (y) and horizontal (x)
directions.        F    N

FA = applied force
Ff = frictional force
Ff                          FA   FN = normal force
Fg = force due to gravity
or weight
Fg

2.       Write a mathematical expression that summarizes these
forces in each direction.
–      Fnet(x) = FA – Ff
•   Since the length of the vector for FA is longer than the one for Ff, the
block should be accelerating on the positive x-direction.
–      Fnet(y) = FN – Fg
•   Since the length of the vectors for FN and Fg are the same length,
the net force in the vertical direction should be zero as we would
expect. This also means that FN = Fg.
Strategies for solving force related problems.
3.       For horizontal surfaces:
•      FN = Fg = mg.
4.       For inclined surfaces, the force due to gravity (weight)
must be broken into two vectors; one that is
perpendicular to the incline, and the other that is parallel
to the incline.                      F  N

–      FN = Fg() = Fgcos
–      Fg(||) = Fgsin
Ff

Fg(||)



Fg( | )

Fg
Forces Acting on Objects in the Horizontal Direction
Does the applied
Is there an                                                v <> 0
Yes           force cause the                         Yes
applied force?
object to move?
v=0

No                                                            Does the object
No       v = constant
accelerate?
a=0

Fnet = 0
Is the object in
Yes         v = 0 m/s: FA = Ff(static)               Yes
motion?
v = constant: FA = Ff(kinetic)
a≠0                                                  a≠0

No           Fnet = Ff                                          Fnet = FA - Ff
a=0

If no friction
Fnet = 0               Fnet = ma                                     Fnet = FA
No Forces in the Horizontal Direction
FN

Fg

If there is no horizontally applied force, then the object will be:
• stationary (v = 0 m/s)
• or in motion, sliding along a frictionless surface at a constant velocity
(v = constant).
•Under both circumstances, Fnet = 0 N since there is no acceleration.

Friction in the Absence of an Applied Force
FN

v
Ff

Fg

Under the circumstance where the object is in motion and being acted
upon by a only a frictional force, the object will experience an
unbalanced force due to friction (Fnet = Ff).
•   For example, assume the block has a mass of 10 kg, and undergoes
an acceleration of -5.0 m/s2 as it slides from the left to the right.
1. Determine the frictional force.
2. Determine the normal force.
3. Determine the coefficient of friction.
Friction in the Absence of an Applied Force
FN

v
Ff

Fg
1. As per our free-body diagram, the only force acting in the x-direction is friction.
Therefore:
Fnet = -Ff
Ff = ma
Ff = (10kg)(-5m/s2) = -50N
2. On a horizontal surface, the normal force equals the weight.
FN = Fg
FN = mg
FN = (10kg)(9.81m/s2) = 98.1N
F   50 N
3. Since Ff = FN,   f           0.51.
An Applied Force with Friction

FN

Ff                                    FA

Fg

When multiple forces act on an object, they need to be summed up to
determine the net force (Fnet = FA - Ff).
•   Assume that there is an applied force of 120N acting on a 10kg block
to the right that causes it to go through an acceleration of 4m/s2.
1. Determine the net force.
2. Determine the frictional force.
3. Determine the coefficient of friction.
An Applied Force with Friction
FN

Ff                                             FA

Fg
1. The net force is simply determined by multiplying the object’s mass by its
acceleration.
Fnet = ma
Fnet = (10kg)(4m/s2) = 40N
2. From the free-body diagram, the net force is the sum of the forces acting in the x-
direction. We will then solve the relationship for the frictional force.
Fnet = FA - Ff
Ff = FA - Fnet
Ff = 120N – 40N = 80N
Ff        80N
3. Since Ff = FN,                    0.82.
An Applied Force with No Friction
FN

FA

Fg

When the surfaces are considered frictionless, the applied force will
equal the net force (Fnet = FA).
•   For example: Assume that there is an applied force of 120N acting on
a 10kg block to the right.
1. Determine the acceleration of the block.
• Since Fnet = FA, a = FA/m
Fapplied       120 N
a                        12 m / s 2
Stationary or Moving at a Constant Velocity
FN

Ff                                              FA

Fg

When objects move at a constant velocity (same speed in a linear direction)
then the net force will be zero. Similarly, if the object is not moving, then the
net force must be zero as well. Why?
•   In both cases, the acceleration is zero, hence (Fnet = ma = 0)
•   And if the net force is zero, then the applied force must equal the frictional force
(FA = Ff).
•   Example: A wooden 10kg block is placed on a wooden floor.
 Case #1: the block is stationary.
 Case #2: the block is moving at a constant velocity.
Stationary or Moving at a Constant Velocity
FN

Ff                                             FA

Fg
 Case #1: Determine the maximum force that can be applied to the block without
causing it to move.
 Case #2: Determine the applied force required to cause the block to move at a
constant velocity.
 In both cases, you will need to refer to your reference table to find the appropriate
values for the coefficient of friction for wood on wood.

Case #1: Stationary                                Case #2: Constant Velocity
s = 0.42                                          k = 0.30
FA = Ff                                            FA = Ff
FA = FN                                           FA = FN
FA = mg                                           FA = mg
FA = (0.42)(10kg)(9.81m/s2) = 41.2N                FA = (0.30)(10kg)(9.81m/s2) = 29.4N
Special Case – The Elevator Question
1.       When talking about elevators, they
spend a short period of time at the                 FN
beginning and end of their ascent
accelerating and decelerating,
respectively.
2.       All motion and forces occur in the
vertical direction.
3.       Write a mathematical expression that
summarizes these forces.
–     Fnet(y) = FN – Fg
•   Note: FN will be greater than Fg if the
acceleration is in the upward direction and
vice-versa if the acceleration is in the
downward direction.
Special Case – The Elevator Question

Note: You are weightless when in free-fall!
Unbalanced Forces and Uniform Circular Motion

• Whenever an object moves in a circular path, it
experiences an unbalanced force.
• The unbalanced force always acts perpendicular
to the direction of motion and towards the center
of the circular path.
• A centripetal force is an unbalanced force, which
is also a net force except that the motion is
mv 2
Fnet    Fc 
r
Uniform Circular Motion – A Horizontal Path(FBD’s)
Car on Road          Ball on String           The Rotor

Ff       FT
Ff
v
v

FN
FT
Ff

Fg

Fc = Ff               Fc = FT                 Fc = FN
Uniform Circular Motion – A Vertical Path(FBD’s)
Roller Coaster        Ball on String        Ferris Wheel
v
FT                                  FN
Top
FN                          Fg
Fg
Fg

Fc = F N + F g        Fc = FT + F g         Fc = FN - Fg

FN

Bottom           FN                     FT
Fg
v
Fg
Fg
Fc = FN - Fg          F c = FT - F g        Fc = FN - F g

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