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Lecture 07
Dynamics and Relative Velocity

Problem Solving: Dynamics Examples

Relative Velocity
Dynamics Problem Solving
 Draw
a Free Body Diagram
(FBD) for each Object
down F = ma for each
Object in each Direction
 Write
 Solve
for the Unknown(s)
Dynamics Example 1

Two boxes of mass 5 kg and 10 kg are
pushed across a floor (with coefficient of
friction of  = 0.2) by a force of 50 N. What is
the force each block exerts on the other and
the acceleration of each block?
Box 2 FBD:
Box 1 FBD:
FN
FN
y
F21
FP
Ff
Fg
F12
Ff
x
Fg
Dynamics Example 1

Two boxes of mass 5 kg and 10 kg are
pushed across a floor (with coefficient of
friction of  = 0.2) by a force of 60 N. What is
the force each block exerts on the other and
the acceleration of each block?
Box 1 F=ma
Box 2 F=ma
x-direction:
x-direction:
FP – Ff – F = m1a1
F – Ff = m2a2
y
y-direction:
FN – Fg = 0
x
y-direction:
FN – Fg = 0
Dynamics Example 1

Two boxes of mass 5 kg and 10 kg are
pushed across a floor (with coefficient of
friction of  = 0.2) by a force of 60 N. What is
the force each block exerts on the other and
the acceleration of each block?
Box 1
Box 2
FP – Ff – F = m1a
F – Ff = m2a
FP – FN – F = m1a
F – FN = m2a
FP – m1g – F = m1a
F – m2g = m2a
Dynamics Example 1

Two boxes of mass 5 kg and 10 kg are
pushed across a floor (with coefficient of
friction of  = 0.2) by a force of 50 N. What is
the force each block exerts on the other and
the acceleration of each block?

Solve each equation for acceleration
and set them equal:

Solve for F (the force each block
exerts on the other):

Solve for a (the acceleration of the
blocks):
FP  m1 g  F F  m2 g

m1
m2
FP m2
F

m1  m2
33.3 N
FP  m1 g  m2 g
a
 1.4 m/s2
m1  m2
Dynamics Example 2

A box of mass 3 kg is pulled on a smooth
(frictionless) surface by a second block of
mass 2 kg hanging over a pulley. What is the
acceleration of each block and tension in the
string connecting them?
Box 2 FBD:
Box 1 FBD:
FT
FN
y
FT
Fg
x
Fg
note: the tension is the same everywhere in the string
Dynamics Example 2

A box of mass 3 kg is pulled on a smooth
(frictionless) surface by a second block of
mass 2 kg hanging over a pulley. What is the
acceleration of each block and tension in the
string connecting them?
Box 1 F=ma
Box 2 F=ma
x-direction:
FT = m1a1
y-direction:
FN – Fg = 0
y-direction:
y
FT – Fg = m2a2
x
FT – m2g = m2a2
Dynamics Example 2


A box of mass 3 kg is pulled on a smooth
(frictionless) surface by a second block of
mass 2 kg hanging over a pulley. What is the
acceleration of each block and tension in the
string connecting them?
Substitute the expression for tension
from Box 1 into the Box 2 equation:

Solve for a (the acceleration of the
blocks):

Solve for FT (the the tension in the
string):
m1a  m2 g  m2 a
m2 g
a
 3.92 m/s2
m1  m2
m1m2 g
FT 
 11.76 N
m1  m2
Relative Velocity

Sometimes your velocity is known relative to a
reference frame that is moving relative to the
earth.
Example 1: a boat moving relative to water, which is
then moving relative to the ground.
Example 2: a plane moving relative to air, which is then
moving relative to the ground.

These velocities are related by vector addition:



vac  vab  v bc
» vac is the velocity of the object relative to
the ground
» vab is the velocity of the object relative to
a moving reference frame
» vbc is the velocity of the moving reference
frame relative to the ground
Relative Motion Example

You are in a plane traveling (relative to still air) at
200 mph at an angle of 30º east of north. There is
a wind blowing 50 mph due east (relative to the
ground). What is the velocity of the plane with
respect to the ground?
200 mph
50 mph
y
x
30º
Relative Motion Example

You are in a plane traveling (relative to still air) at 200 mph
at an angle of 30º east of north. There is a wind blowing 50
mph due east (relative to the ground). What is the velocity of
the plane with respect to the ground?
 To find the velocity with respect to the ground we just need to add the
velocity of the plane with respect to the air to the velocity of the air
with respect to the ground:



v pg  v pa  v ag
vag
vpa
vpg
Relative Motion Example

You are in a plane traveling (relative to still air) at 200 mph
at an angle of 30º east of north. There is a wind blowing 50
mph due east (relative to the ground). What is the velocity of
the plane with respect to the ground?
 Break the plane’s velocity (with respect to the air) into components.
 Add the respective components together to attain the x- and y-
components of the net velocity we are looking for!
x-direction:
vpgx = vpax + vagx
y-direction:
vpgy = vpay + vagy
vpgx = (200 mph)sin30º + (50 mph)
vpgy = (200 mph)cos30º + (0 mph)
vpgx = 150 mph
vpgy = 173 mph
Relative Motion Example

You are in a plane traveling (relative to still air) at 200 mph
at an angle of 30º east of north. There is a wind blowing 50
mph due east (relative to the ground). What is the velocity of
the plane with respect to the ground?
 With the components of the net velocity known, find the magnitude
and direction of the velocity!
150 mph
173 mph

vpg
Vpg = 229 mph
 = 41º
Summary of Concepts

X and Y directions are Independent!

F = ma applies in both x and y direction

Relative Motion (add vector components)
vag
200 mph
50 mph
vpg
30º
50 mph
173 mph
vpg
vpa