Chapter 7
Forces and Motion
In
Two Dimensions
Equilibrium
An object is in equilibrium when the
Net Force on the object is zero.
FNet = 0
Acceleration = 0
Constant Velocity
Velocity = 0
Equilibrium in 2 Dimensions
F x = 0
F y = 0
Static Equilibrium
Objects are not moving.
FNet
=0
Acceleration = 0
Velocity = 0
20N
Static Equilibrium
50N
FNet = 0
Fx = 0
Fy = 0
50N
20N
60°
60°
2kg
F1
F2
2kg
Fg = mg
Fg = 20N
2
Fg = 2kgx9.8m/s
Fg = 20N
F1
F2
x direction
Fx = -F1x + F2x
60°
60°
0 = -F1x + F2x
F1x = F2x
F1cos60°= F2cos 60°
Fg = 20N
F1 = F2 = F
F1
60°
y direction
F2
Fy = F1y + F2y - Fg
0
=
F
+
F
F
1y
2y
g
60°
Fg = F1y + F2y
Fg = F1sin60°+ F2sin 60°
Fg = Fsin60°+ Fsin 60°
Fg = 20N
Fg = 2Fsin60°
Fg = 2Fsin60°
__Fg__
=F
2sin60°
_20N__
=F
2sin60°
11.5N = F
Homework
Finish Work Sheet
Ft
FP
Fg
50kg
Ft
FP
Fg
50kg
Ftx
Ft
Fty
30°
Fg = mg
2
Fg = 50kg·9.8m/s
Fg = 500N
FP
Fg
X - Components
Y- Components
Fx = Fpx - Ftx
0 = Fp - Ftx
Fpx = Ftx
Fy = Fpy - Fg
0 = Fpy - Fg
Fty = Fg
Fp = Ftcos30°
Ftsin30° = Fg
Fg
Ft =
sin30°
X - Components
Y- Components
Fp = Ftcos30°
Fg
Ft =
sin30°
Fp = 1000cos30°
Fp = 866N
500N
Ft =
sin30°
Ft = 1000N
Homework
Finish Work Sheet
Incline Plane
FN=Fg
FN
Fg
Incline Plane
Ff
θ
Fg
θ
Fg
Fg
FN
θ
Fg
Ff
Fgx
Fg
FN
Fgy
Problem
In a block/inclined plane system,
the inclined plane makes an angle
of 60° with the ground. The
coefficient of friction is 0.5. If the
block has a mass of 1.02kg, what
is the net force on the block? What
is the blocks acceleration?
Incline Plane
Ff
θ=60°
Fg
FN
θ
Fg
Ff
FN
Fgx
Fgy
Fg
Projectile Motion
Projectile Motion
Projectile Motion
Horizontal
Component
Vertical
Component
Projectile Motion
Projectile motion is the combination
of two independent motions, the
motion in the x direction and the
motion in the y direction. These two
motions are usually independent of
each other.
y component
Projectile Motion
x component
Problem Solving Strategy
y component
1. Break up the problem into two
interconnected one-dimensional
problems.
x component
Problem Solving Strategy
2. Vertical motion (y
component) is exactly
that of an object being
dropped or thrown
straight up or down.
(g - gravity!!!!!)
Problem Solving Strategy
3. Horizontal motion (x component)
is the same as solving constant
velocity problems.
Problem Solving Strategy
4. Vertical (y) and horizontal (y)
components are connected by
the variable time (t). Solving for
time in one dimension, x or y,
automatically gives you the time
for the other dimension.
*Basic Equations*
v = v0 + at
d = d0 +1/2(v+v0)t
d = d0 +v0t
2
v
2
2
+ ½at
= v0 +2a(d-d0)
Problem:
A ball is kicked horizontally, with a
velocity of 25m/s, off a 122.5m high
cliff. How far from the cliff did the
ball land?
Sketch the Problem
y = 122.5 m
y component
v = 25 m/s
x component
y component (up is +)
d = d0 +v0t + ½at2
d = ½at2
-122.5m =
2(-122.5m)
2
(-9.8m/s )
2
2
½(-9.8m/s )t
=
2
t
y component (up is +)
√
2(-122.5m)
(-9.8m/s2)
= t
5.0s = t
Use this to solve for distance in the
x direction!!!
x component
d = d0 + v0t + ½at2
d = v0t
d = (25m/s)(5s)
d = 125m
Projectiles Launched at an Angle
θ
Range
Launch Problem
A football is thrown with a speed
15m/s at an angle of 60° with the
horizontal. How far is the football
thrown?
y component
Sketch the Problem
x component
1. Known
 x0
=0
 y0 = 0
 v0 = 15m/s
 θ0 = 60°
 a = -g =-9.8m/s2
2. Unknown
 vy
 vx
 dx
 dy
t
3. Find the x/y components of the velocity.
vx0 = (15m/s)cos60°
vx0 = 7.5m/s
vy0 = (15m/s)sin60°
vy0 = 13m/s
4. Break up the x/y components and
find time, then distance.
y - component
vy will be zero at the top
vy = v0 + at
0 = 13m/s +
2
(9.8m/s )t
2
(-9.8m/s )t
= 13m/s
13m/s
t=
2
9.8m/s
13m/s
t=
9.8m/s2
t = 1.3s
This is the time for half the trip!!
ttotal = 2.6s
x component
dx = d0 +vx0t
2
+ ½at
dx = vx0t
dx = (7.5m/s)(2.6s)
dx = 19.5m
Let’s find the height!!
y - component
dy = dy0 +vy0t + ½at2
dy = vy0t + ½at2
dy = 13m/s(1.3s) + ½(-9.8/s2)(1.3s)2
dy = 16.9m – 8.3m
dy = 8.6m
Circular Motion
d
v= t
d = 2πr
v
r
t=T
T is the Period: the time it takes
to make one revolution.
d
v= t
2πr
v= T
Centripetal Acceleration
2
v
ac = r
ac =
2
4π r
2
T
Centripetal Force
Centripetal Force is the
force toward the center of
the circle that keeps an
object moving in a circle.
Centripetal Force
Fc
Centripetal Force
Fc = mac
( )
Fc = m
2
4π r
2
T
Example
 A 15g
whistle is being swung
on a lanyard 0.30m long. If
one revolution takes 0.5s,
what is the centripetal force?
Example
Example
Given:
m = 15g = .015kg
T = 0.5s
Find:
Fc = ?
2
4π r
2
T
( )
4π
(.3m)
F = .015m(
)
(0.5s)
Fc = m
2
c
2
Fc= 0.7N
Homework
Worksheet
Due: 12/14/06
Universal Gravity
What does gravity depend on?
Mass
Distance
G
m1
m2
m
m
F∞ 1 2
r
r
1
F∞ 2
r
m1
m2
r
m1m2
F ∞ r2
m1m2
G
F = r2
Gravitational
G – Constant
G – 6.67 X
-11
2
2
10 Nm /kg
m
M
Mm
G
F=
2
r
What is the magnitude of the
gravitational force that acts on each
particle, m1 is 12kg and m2 is 25kg
and the two are 1.2m away.
m1
r
m2
Gm1m2
F=
2
r
F=
-11
(6.67x10 Nm2/kg2)(12kg)(25kg)
2
(1.2m)
F= 1.4 x
-8
10 N
Acceleration due the Gravity(g)
GMEarthm
F=
2
r
GMEarthm
mg =
2
r
Acceleration due the Gravity(g)
GMEarth
g=
2
r
r
GMEarth
g=
2
r
g=
-11
24
2
2
(6.67x10 Nm /kg )(5.97x10 kg)
6
2
(6.38x10 m)
g=
2
9.8m/s
Homework
Page: 194
Prob: 20,27,28,35,37
Due: 12/18/06
Test: 12/21/06
Satellites
1km/s
.6mi/s
Satellites
Satellites – 17500mi/hr
Lets find a satellite speed!!!
r
Fc = Fg
Fc = F g
GmEm
mac = r2
2
v
GmE
=
2
r
r
2
v
GmE
=
r
v=
√
GmE
r
Homework
Worksheets
Page: 194
Prob: 23,24,25,50,51
Due: 12/20/06
Test: 12/21/06
Homework
Review
Worksheet
Due: 12/21/06
Test: 12/21/06