Transcript F g cos

Refresher:
*Acceleration is only
caused by an
F = ma
unbalanced net force
acting on an object. F = F1 + F2 + …
*The weight of an object
is referred to as a
Gravitational Force
Pg. 1
Fg = mg
Refresher:
*Friction is a force that always
opposes motion. “” is the
coefficient of friction which
represents the roughness or
smoothness of the surfaces in
contact.
Ff = FN
*The Normal Force is a supporting
force caused by a surface. The
normal force is always perpendicular
to the surface. Sometimes called the
“apparent weight” of an object.
The equation
is typically
found in a
“net force”
equation.
Pg. 2
4.8 The Normal Force
FN  11 N  15 N  0
FN  26 N
FN  11 N  15 N  0
FN  4 N
Pg. 3
4.8 The Normal Force
Apparent Weight
The apparent weight of an object is the reading of the scale.
It is equal to the normal force the scale exerts on the man.
Pg. 4
4.8 The Normal Force
F
y
  FN  mg  ma
FN  mg  ma
apparent
weight
Pg. 5
true
weight
Refresher:
*An Applied Force is
any push or pull
*A Tension Force (FT) is
a supporting force
caused by a string or
cable.
Pg. 6
FA
The equation is
typically found
in a “net force”
equation.
Grocery Bag
FT
Fg
FN
Fg
Tension and Normal forces
help “support” the weight
Pg. 7
The coefficient of kinetic friction
between the red block and the floor is
0.3. In order to accelerate this system at
a rate of 2 m/s2, what force must be
applied? Assume the top block does not
slip.
Use g =
10 m/s2
4 kg
8 kg
Pg. 8
F
a.) Consider the frictionless pulley
that has two masses hanging over
each side. What will happen to
the apparatus if the blocks are
released from rest?
Which Block will
accelerate?
Pg. 9
a.) Consider the frictionless pulley
that has two masses hanging over
each side. What will happen to the
apparatus if the blocks are released
from rest?
If the green block is 3 kg and
the blue block is 1 kg,
calculate that acceleration.
Pg. 10
Use g =
10 m/s2
3 kg
1 kg
a.) Consider the frictionless pulley
that has two masses hanging over
each side. What will happen to the
apparatus if the blocks are released
from rest?
Hint: set up a Net Force
problem. Label the direction
3 kg
of your forces.
1 kg
Are these forces acting with
or against each other?
Pg. 11
A 20.0 kg mass sits on a tabletop. The
coefficient of kinetic friction between the
tabletop and the mass is 0.25.
20.0
It is attached
kg
to a 12.5 kg
mass that is
12.5
hung over a
kg
massless pulley
Use g =
10 m/s by a string.
What is the acceleration of the system?
2
Pg. 12
Two children push on a 5 kg toy box
at the same time. One pushes with
a force of 125 N West and the other
pushes 165 N South.
What is the magnitude and direction
of the Net Force (Resultant)?
R2 = x2 + y2
125 N
Ө
165 N
Pg. 13
207 N
Ө= Tan-1(o/a)
Ө = 53º S of W
Recall Pythagorean Thm.
Two children push on a 5 kg toy box
at the same time. One pushes with
a force of 125 N West and the other
pushes 165 N South.
What is the magnitude and direction
of the Acceleration?
F = ma
125 N
Ө
165 N
Pg. 14
207 N
207 N = (5kg)a
41.4
2
m/s
=a
Ө = 53º S of W
Two children push on a 5 kg toy box
at the same time. One pushes with
a force of 125 N West and the other
pushes 165 N South.
If a third child were to apply a force to this
box to suddenly make it maintain a
constant velocity (stop accelerating), how
much force would be needed and in what
direction? 125 N 207 N Ө = 53º N of E
Ө
165 N
Pg. 15
207 N
A 40.0 kg crate is pulled across the ice with a rope. A force of
100.0 N is applied at an angle of 30 above horizontal. Neglect
friction. A) Calculate the acceleration of the crate.
F
Fy = F sin
m
Ff = 0

Fx = F cos
Fx = 100N cos30 = 86.6 N = ma
Fx = 86.6 N = (40kg)a
a = 2.17 m/s2
Pg. 16
A 40.0 kg crate is pulled across the ice with a rope. A force of 100.0 N is applied at
B) Calculate the
upward force the ice exerts on the crate as
F
it is pulled (FN).
an angle of 30 above horizontal. Neglect friction.

m
mg
Fnorm
Fy = Fy + Fnorm – mg = 0
Fy + Fnorm = mg
Pg. 17
Fy = F sin
= 50 N
Since the box is not
accelerating upward or
downward, the sum of the
vertical forces = 0
(Upward = Downward)
4.8 The Normal Force
Pg. 18
A 40.0 kg crate is pulled across the ice with a rope. A force of 100.0 N
is applied at an angle of 30 above horizontal. Neglect friction. B)
Calculate the upward force the ice exerts on the crate as it is pulled.
F

m
mg
Pg. 19
Fy = F sin
Fy + Fnorm = mg
= 50 N
Fnorm
50 N + Fnorm = (40kg)(9.8 m/s2)
50 N + Fnorm = 392 N
Fnorm = 392 N – 50 N
= 342 N
Inclined Surface

Fnorm
Fg
Fg = Fnorm
Fnorm = Fg cos
Pg. 20
This is called the perpendicular component
Inclined Surface
F

Fnorm
Fg
F = Fg sin
Pg. 21
There is also a parallel component of Fg
Fg
What is the coefficient of friction?
Ff
Fg = Ff
Fg sin = (Fg cos)

Suppose this person
is not moving…
Label the forces
Pg. 22
mg sin = (mg cos)
sin = 
cos
tan = 
Suppose  = 30°
What is the coefficient of friction?
Ff
Fg = Ff
Fg

Suppose this person
slides downhill at a
constant velocity
Pg. 23
Same answer as
before…except
this would
represent the
Kinetic Friction
Pg. 24
What is the acceleration?
Fg = F
Fg sin = ma
Fg
mg sin = ma

Suppose this person
is accelerating
downhill because
there is zero friction.
g sin = a
Suppose  = 30°
Remember this for the lab
Pg. 25
Fa
Fg
Ff

Fa = Ff + mg
Because it slides at const velocity
Fa = (Fg cos) + Fg sin
Suppose this block gets pushed uphill at
a constant velocity…label the forces
Pg. 26
FT
mg
Ff
Pg. 27

FT
mg