oppose

1
oppose
motion
surfaces
normal
Block not accelerating so net force 0 N the
normal force (push of the surface upward the wt)
FN wt mg FN (50.0 kg)(9.80 m/s2) FN
490. N
wt
FN
2

• Ex 2 A 45.0 kg block lies on a horizontal
  surface, and a person pushes down on the block
  with 200. N of force. Calculate the normal
  force.
• 2 Types of Friction acting on sliding objects
• 1. Static Friction the minimum frictional
  force which must be overcome to ________ an
  objects motion.
• 2. Kinetic Friction the frictional force
  acting on an object in ____________.
• Values of static friction are ____________ than
  kinetic friction due to the fact that once an
  object is set into motion it is easier to
  maintain its state of motion by Newtons first
  law.
• Example When you wish to push your couch
  across a level floor, you must first overcome the
  force of __________ friction to get the couch
  moving. Then you must supply a force equal to
  the ___________ friction to keep the couch in
  motion at a constant speed.

200. N
The surface receives the force from the persons
push as well as the wt of the block.
FN wt 200. N
FN mg 200. N
FN (45.0 kg)(9.80 m/s2) 200. N
FN 441.00 N 200. N
wt
FN 641 N
initiate
motion
larger
static
kinetic
3
Calculating Static Friction (FS) FS µS
FN _____ is the coefficient of static friction.
This variable is _____________ and depends upon
the types of materials in contact. A large value
of µS corresponds to a ____________ frictional
force while a small value corresponds to a
___________ frictional force. FN is the
____________ force. Ex A sled, with a mass of
15.0 kg, rests on a horizontal patch of snow.
The coefficient of static friction between the
sled and snow is 0.350. Calculate the force
required to barely set the sled in motion if a
boy with a mass of 55.0 kg sits on the sled.
µs
unitless
large
small
normal
Here, the surface feels the weight of the sled
and the boy
Force to set object in motion must overcome
static friction
Fs
wtboy
FN wt (boy sled)
wtsled
Fs µS FN
FN (mg)boy, sled
FN (55.0 15.0) kg (9.80 m/s2) FN (70.0
kg)(9.80 m/s2)
Fs (0.350) (686.00 N)
Fs 240. N
FN 686.00 N
4

• Calculating Kinetic Friction (FK)
• FK µK FN
• _____ is the coefficient of __________ friction.
  Coefficients of kinetic friction (µK) are
  __________ than coefficients of static friction
  (µS) for given surfaces in contact.
• Ex A sled, with a mass of 20.0 kg, rests on a
  horizontal patch of snow. The coefficient of
  kinetic friction between the sled and snow is
  0.050. Calculate the force required to push the
  sled at a constant velocity across the snow if a
  little girl, whos mass is 35.0 kg, sits on the
  sled.

smaller
µk
kinetic
For the sled to move at a constant velocity the
net force must be
0 N
Therefore the force in the direction of motion
must be equal to the force of
kinetic friction
FK
Force to push sled at constant velocity
FK µK FN
wt (girl sled)
FK 0.050 (539.00 N)
FK 27 N
FN wt (girl sled) FN mg (girl sled)
FN (20.0 kg 35.0 kg) (9.80 m/s2)
FN (55.0 kg) (9.80 m/s2)
FN 539.00 N
5

• More Frictional Notes
• Recall that the forces of static or kinetic
  friction were dependent upon 2 factors.
• 1. The type of surfaces in contact (___)
• 2. The normal force (_____).
• The forces are independent of the area of contact
  and the speed of the sliding motion if air drag
  can be neglected.
• Sample Problems
• Ex 1 A 20.0 kg block that rests on a horizontal
  surface. µs 0.500 and µk 0.215
• Calculate the force required to set the block
  into motion.
• b. If a force of 120. N acts on the block once
  it is set into motion, calculate the acceleration
  of the block after it is moving. Homework
  Problem 1

µ
FN
FS
FK
F to put block in motion
Must overcome Static Friction Fs µsFN Here the
normal force is equal to the weight of the
block Fs µsmg Fs 0.500 (20.0 kg)(9.80 m/s2)
Fs 98.0 N
120. N
wt
FN
a Fnet / m a 77.86 N 20.0 kg a 3.9
m/s2
Fnet F? - F? Fnet 120. N Fk Fnet 120. N
42.140 N Fnet 77.86 N
Fk µKFN Fk µk mg Fk (0.215)(20.0 kg)(9.80
m/s2) FK 42.140 N
6
Ex 2 A 125 N force is acting on a 15.0 kg block
that slides across a horizontal surface at a
constant velocity. Determine the coefficient of
kinetic friction between the block and the
surface.
µK FK / FN
µ k 125 N /
147.00 N
125 N
FK
FN wt FN mg
µ k 0.850
wt
FN
FN (15.0 kg)(9.80 m/s2) FN 147.00 N
Fnet
0 N
Fk
125 N
Ex 3 A 350. N force acts on a 50.0 kg block
that is accelerating across a horizontal surface
at 1.50 m/s2. Calculate the coefficient of
kinetic friction between the block and the
surface.
Fnet F? - Fk Fnet ma ma F? - Fk
µK FK / FN
350. N
FK
µK 275.00 N 490.00 N
wt
Fk F? - ma Fk 350. N (50.0 kg)(1.50m/s2)
FN
µK 0.561
Fk 350. N 75.000 N
FN wt FN mg
Fk 275.00 N
FN (50.0 kg)(9.80 m/s2) FN 490.00 N
7
Ex 4 A boy on his sled is coasting at 4.00 m/s
along a horizontal stretch of snow. The sled has
a mass of 13.0 kg and the boy has a mass of 25.0
kg. The coefficient of kinetic friction is
0.0500 between the sled and the snow. How far
will the sled travel before coming to a halt?
vo 4.00 m/s vf 0 m/s d (vf2 vo2 )/ 2a a
?
0 N
Fk
a Fnet / m
a 18.620 N / 38.0 kg
wtb,s
FN
a 0.49000 m/s2
Fk µkFN
Fk (0.0500) 372.40 N
Fnet F? - F?
Fk 18.620 N
Fnet 0 N - Fk
Fnet 0 N 18.620 N
Fnet 18.620 N
FN (mg) b,s FN (25.0 13.0)kg (9.80 m/s2) FN
38.0 kg (9.80 m/s2) FN 372.40 N
d (vf2 vo2) / 2a
d (4.00 m/s)2 2(-0.49000 m/s2)
d 16.3 m
8

• Force at Angles Notes
• A box with a mass of 25.0 kg is pulled across a
  horizontal surface at a constant velocity when
  pulled by 125 N at a 35.0o angle to the
  horizontal. (Notice the force is acting upward)
• Calculate the normal force.
• b. Calculate the force of kinetic friction.

125 N
Fy
wt mg wt (25.0 kg)(9.80 m/s2) wt 245.00 N
FN wt Fy
FN 245.00 N 71.697 N
Fx
FN 173 N
wt
FN wt - Fy
sin 35.0o Fy / 125 N
Fy sin 35.0o (125 N)
Fy 71.697 N
Since the block moves at a constant velocity, the
net force
0 N
Fk
Fx
cos 35.0o Fx / 125 N
c. Calculate µk
Fx Fk cos 35.0o (125 N)
µk Fk / FN
Fk 102 N
µk 102.39 N /173.30 N
µk 0.591
9
Ex 2 A 5.00 kg block is pulled at a constant
velocity when a force of 20.0 N at a 55.0o angle
acts upon it. Calculate the coefficient of
kinetic friction.
µk Fk / FN
20.0 N
Fy
µk 11.472 N 32.617 N
Fx
Fk
µk 0.352
wt
Fk Fx
cos 55.0o Fx / 20.0 N
Fk cos 55.0o (20.0 N)
Fk 11.472 N
FN wt - Fy
FN mg sin 55.0o(20.0 N)
FN (5.00 kg)(9.80 m/s2) sin 55.0o(20.0 N)
FN 49.000 N 16.383 N
FN 32.617 N
10
3. A force of 45.0 N at a 45.0o angle to the
horizontal is required to set a 10.0 kg block in
motion on a horizontal surface. Calculate the
coefficient of static friction. (Notice the
force acts downward)
45.0 N
Fy
µs Fs / FN
Fx
Fs
µs 31.820 N / 129.820 N
µs 0.245
wt
Fx Fs
Fs cos 45.0o(45.0 N)
FN wt Fy
Fs 31.820 N
FN mg sin 45.0o(45.0 N)
FN 10.0 kg (9.80 m/s2) 31.820 N
FN 98.000 N 31.820 N
FN 129.820 N
11
4. A downward force of 50.0 N at an angle of
41.0o to the horizontal acts on a wood block on a
horizontal surface. The block accelerates at a
rate of 1.50 m/s2. Calculate the coefficient of
kinetic friction.
µk Fk / FN
50.0 N
Fy
µk 30.235 N 81.803 N
Fx
Fk
µk 0.370
wt
FN wt Fy
Fnet Fx - Fk
FN mg sin 41.0o(50.0 N)
ma Fx - Fk
FN 5.00kg (9.80 m/s2) 32.803 N
Fk Fx - ma
Fk cos 41.0o (50.0 N) (5.00 kg)(1.50 m/s2)
FN 49.000 N 32.803 N
Fk 37.735 N 7.5000 N
FN 81.803 N
Fk 30.235 N
12

• Inclined Planes Notesheet
• A 50.0 kg block slides down an inclined plane,
  that has an angle of 30.0o to the horizontal.
• The block slides with a constant velocity.
• Calculate the weight of the block and locate this
  force on the diagram.
• wt mg
• wt (50.0 kg)(9.80 m/s2)
• wt 490. N
• b. Calculate the normal force and locate this
  force on the diagram.
• cos 30.0o FN / wt
• FN cos 30.0o (490.00 N)
• FN 424.35 N 424 N
• Calculate the force pulling the block down the
  incline and locate this force on the diagram.
• sin 30.0o Fd / wt
• Fd sin 30.0o (490.00 N)
• Fd 245.00 N 245 N
• d. What is the net force acting on the block and
  what is the frictional force acting on the block?
  
• Fnet Fd FK 0 N
• Fd Fk

FN
wt
FK
Fd
13
2. A 10.0 kg block slides down an incline with a
25.0o angle to the horizontal at a constant
velocity. Calculate the coefficient of kinetic
friction between the block and inclined
plane. µK FK / FN µK 41.417 N /
88.818 N µK 0.466 wt mg wt
(10.0 kg)(9.80 m/s2) wt 98.000 N cos
25.0o FN / wt FN cos 25.0o (98.000 N) FN
88.818 N sin 25.0o Fd / wt Fd sin
25.0o (98.000 N) Fd 41.417 N Fnet Fd
FK 0 N Fd Fk Fk 41.417 N
FN
wt
FK
Fd
14
3. A 50.0 kg block slides down an incline with
an angle of 35.0o to the horizontal. Calculate
the acceleration of the block if the coefficient
of kinetic friction between the block and
incline is 0.550. a Fnet / m a 60.29
N / 50.0 kg a 1.2 m/s2 Fnet Fd
FK Fnet 281.05 N - 220.76 N Fnet
60.29 N cos 35.0o FN / wt FN cos
35.0o (490.00 N) FN 401.38 N sin 35.0o
Fd / wt Fd sin 35.0o (490.00 N) Fd
281.05 N
FN
wt
FK
Fd
wt mg wt (50.0 kg)(9.80 m/s2) wt 490.00
N
Fk µK FN Fk (0.550)(401.38 N) Fk 220.76 N
15
4. A 25.0 kg block slides up an incline plane
with an angle of 40.0o to the horizontal at a
constant velocity. The block is being pulled by
a 20.0 kg block hanging from a pulley. Calculate
the coefficient of kinetic friction between the
block and incline.
FK
Since the block moves at a constant velocity, the
net force is zero so the sum of the forces up the
inclined plane will be equal to the sum of the
forces down the inclined plane.
Fp
Fp Fd FK
FN
Fk Fp Fd
wt
Fd
Fd
Fk
196.00 N
Fd
- 157.48 N
Fk 38.52 N
wt pulls the block on inclined plane up the
Plane (Fp)
wt mg
wt (25.0 kg)(9.80 m/s2)
µk Fk / FN
wt 245.00 N
Fp mg
µk 38.52 N / 187.68 N
Fp (20.0 kg)(9.80 m/s2)
sin 40.0o Fd / 245.00 N
µk 0.21
Fd (sin 40.0o) 245.00 N
Fp 196.00 N
Fd 157.48 N
cos 40.0o FN / 245.00 N
FN (cos 40.0o) 245.00 N
FN 187.68 N