F) Physics 11
1) Dynamics – forces
could be 2 dimensions


F net  m a
Newton’s 2nd Law:
use a “free body diagram”
Forces:
Gravity:
F
G m1 m2
r2
F m g
Friction:
Springs:
anywhere
on the Earth
g=9.80 N/kg
F f   FN
F k x
could be kinetic (moving)
or static (stuck)
2) Kinematics – how things move
vi = initial velocity = vo
vf = final velocity = v
d = displacement
a = acceleration
t = time
v f  vi  a t
1 2
d  vi t  a t
2
v f  vi  2 a d
 vi  v f
d  
 2
2
v f  vi  2 a d
2
OR
2
2
OR
v ave

 t

 vi  v f
 
 2



eg) A 140g baseball traveling at 30 m/s is stopped by a
baseball mitt that exerts a force of 315 N.
How far will the mitt recoil?
free body diagram:
Fa = -315
Fnet = m a
-315 = 0.140 a
a = -2250 m/s2
vi
vf
d
a
t
=
=
=
=
=
30
0
?
-2250
x
vf2 – vi2 = 2 a d
02 – 302 = 2 (2250) d
-900 = 4500 d
d = 0.20 m
eg) If a construction crate accelerates a 1400 kg beam of
steel upward with an acceleration of 0.22g, what is the
tension in the cable?
a = (0.22) (9.8)
a = 2.156 m/s2
free body diagram:
Fnet
= m a
T – mg = m a
T – (1400)(9.8) = (1400)(2.156)
T – 13720 = 3018.4
T = 16 700 N
Interactions Between 2 Objects
eg)
Find the tension when the objects are released
4.0 kg
12 kg
eg)
Find the tension in the two ropes.
The acceleration is 2.2 m/s2 upward.
T1= ?
m1= 400 kg
T2= ?
m2=750 kg
eg)
Find the tension in the string when released.
16 kg
µk=0
24 kg
eg)
The tension in the string is 47.04 N.
Find the size of the unknown mass “m”
m
µk=0
24 kg
eg) Find the interaction force between the two boxes.
The elf applies a force of 320 N
20 kg
60 kg
3) About Friction - Two Types: Static & Kinetic
- One Equation:
eg)
m= 15 kg,
Ff = μ FN
μs = 0.40,
μk = 0.30
Static:
FN
Ff = μ FN
Ff = (0.40)(147)
Ff
Ff = 58.8 N
Fg
This is the max friction value
before the dog “snaps” free
It can vary from zero up
to this value
eg)
FN
m= 15 kg,
μs = 0.40,
μk = 0.30
Ff = 58.8 N
Ff
Fpull
Fg
if
if
Fpull = 0 N
then,
if
Fpull = 55 N
Fpull = 60 N
then,
Ff = 55 N
then,
Ff = fails
nothing happens
velocity
dog snaps free
velocity
eg)
m= 15 kg,
μs = 0.40,
μk = 0.30
Kinetic:
FN
Ff = μ FN
Ff = (0.30)(147)
Ff
Fpull
Ff = 44.1 N
Fg
The sliding friction force
will always have this value.
Even if the dog is:
Speeding up
Slowing down
Or cruising at consant “v”
eg)
FN
m= 15 kg,
μs = 0.40,
μk = 0.30
Ff = 44.1 N
Ff
Fpull
Fg
if Fpull = 14.1
if Fpull = 44.1
if Fpull = 89.1
Fnet = m a
Fnet = m a
Fnet = m a
14.1 – 44.1 = (15)a 44.1 – 44.1 = (15)a 89.1 – 44.1 = (15)a
-30 = (15)a
0 = (15)a
45 = (15)a
a = -2.0 m/s2
a = 0 m/s2
decreasing velocity constant velocity
a = +3.0 m/s2
increasing velocity
eg)
If μs = 0.85, what is the maximum acceleration
possible from the dragster?
eg) If the truck bed has μs = 0.70 , what is the maximum
deceleration possible without the crate sliding?
eg) Find the vf The block starts from rest and accelerates
for 3.0 m.
Fa=117 N
30o
μ=0.200
20 kg
Free Body Diagram:
20
Fnet = m a
101.32 – 27.5 = (20) a
73.82 = (20) a
a = 3.691 m/s2
vi = 0
vf = ?
d = +3.0
a = +3.691
t = x
vf2 – vi2 = 2 a d
vf2 – 02 = 2 (3.691)(3)
vf2 = 22.146
vf = +4.71 m/s
eg)
Squirels are added until the dog finally slides.
What will the acceleration of the dog be?
How tight is the rope during the slide?
m= 15 kg,
μs = 0.40,
μk = 0.30
Max static friction (break free)
Ff = μs m g
Ff = (0.40)(15)(9.8) = 58.8
Mass of the bucket
Fg = m g
58.8 = m (9.8)
m = 6.0 kg
All:
Dog:
Fnet = m a
58.8-44.1 = 21 a
14.7 = 21 a
Fnet = ma
T – 44.1 = (15)(7)