Transcript Dynamics Problems - The Burns Home Page

Dynamics Problems
Warm Up
The system below is in equilibrium. If the scale is calibrated in N, what does it read?
Since the tension is distributed over
the entire (mass-less) rope, the scale
will read (9.8N/kg)(5kg) = 49N
5kg
5kg
Warm Up
Determine the Normal force in each of the following:
N w
F

N  w cos    F
N  W1  W2
F
F
1
2
N 0
N  F sin    mg
F
F
2
1
m
N  w cos  
N  mg cos 2   F sin 1 
N  F sin 1   F sin 2   mg
F
N  F sin  
Warm Up
Write F=ma for each scenario:
F
F
2
1
m
a
 F  ma
F cos 1   F cos  2     F sin 1   F sin  2   mg   ma
F
 F  ma

 w
w sin      w cos    F     a
g
F
1
 F  ma
2
mg sin  2   F cos 1     mg cos  2   F sin 1    ma
Example



A pumpkin of unknown mass is suspended by a cord attached to the ceiling and
pushed away from vertical. When a 24.0 N force is applied to the pumpkin at an
angle of 18.00 to horizontal, the pumpkin will remain in equilibrium when the cord
makes an angle of 32.00 with the vertical.
(A) What is the tension in the cord when the pumpkin is in equilibrium ?
(B) What is the mass of the pumpkin ?
Diagram
Free body Diagram
FT
Fap
24.0N
58
18
Fg
(A) What is the tension in the cord when the pumpkin is in
equilibrium ?
FT
Fap
24.0N
Since the pumpkin is in static equilibrium, we
need only look at the horizontal components
 Fx  0
FTx  Fapx  0
T cos  58   24.0 N cos 18   0
T
24.0 N cos 18 
cos  58 
 43.073N
43.1N
58
18
Fg
(B) What is the mass of the pumpkin ?
FT
Fap
24.0N
Since the pumpkin is in static equilibrium, we
need only look at the vertical components
F
y
18
58
0
Fg
FTy  Fap y  Fg  0
T sin  58   Fapp sin 18   mg  0
m

T sin  58   Fapp sin 18 
g
 43.073N  sin  58   24.0 N sin 18 
9.81
 4.4795kg
4.48kg
m
s2
Example


The tension in the horizontal rope is 30N
A) Determine the weight of the object
Diagram
Free body Diagram
500
FT 2
400
30N
FT 2 sin 50
500
FT 2 cos 50
FT 1  Fw
FT 3  30 N
A) Determine the weight of the object
FT 2
The weight is in static equilibrium, so the
appropriate net component forces must be
zero.
FT 2 sin 50
500
FT 2 cos 50
FT 1  Fw
Horizontal Component
F
x
0
F T 2x  F T 3  0
 FT 2 cos  50   FT 3  0
FT 2 cos  50   FT 3
FT 3
FT 2 
cos  50 

30 N
cos  50 
 46.672 N
Vertical Component
F
y
0
F T 2 y  F T1  0
FT 2 sin  50   Fw  0
Fw  FT 2 sin  50 
  46.672 N  sin  50 
 36 N
The weight of the mass is 36N
FT 3  30 N
Weight on a Wire


A rope extends between two poles. A 80N weight hangs from it as per the diagram.
A) Determine the tension in both parts of the rope.
Diagram
T1
100
150
T2
80N
Free body Diagram
T1
T1 sin 15
T2
15
10
T1 cos 15
T2 cos 10
FG
T2 sin 10
A) Determine the tension in both parts of the rope.
T1
T1 sin 15
The weight is in static equilibrium, so the
appropriate net component forces must be
zero.
T2
15
T2 sin 10
10
T1 cos 15
T2 cos 10
FG
Horizontal Component
F
x
Vertical Component
F
0
F T 1x  F T 2 x  0
T2 cos 10 
cos 15 
182.8 N cos 10 
T1 
cos 15 
 186.4 N
 186 N
0
F T1y  F T 2 y  F G  0
T1 cos 15   T2 cos 10   0
T1 
y
T1 sin 15   T2 sin 10   FG  0
T2 cos 10 
cos 15 
sin 15   T2 sin 10   80 N  0
80 N
0.437527
 182.8
T2 
 183 N
Example

The system in the diagram is just on the verge of slipping. If a 7.0N weight is hanging
from a ring, what is the coefficient of static friction between the block and the
tabletop?.
Free body Diagram
Block
Ring
F Tc
FN
FTc sin 30
30
35N
Ff
7.0N
F Tb
30
F Tb
FTc cos  30
F Gb
FG
The system in the diagram is just on the verge of slipping. If a 7.0N weight is hanging
from a ring, what is the coefficient of static friction between the block and the tabletop?.
The weight is just in static equilibrium, so the
appropriate net component forces must be
zero.
Ring
Block
F Tc
FN
Ff
30
F Tb F Tb
F Gb
From Block:
FTb  F f
  f FN
  f mg
F
u f  Tb
mg
FTb

35 N
From ring:
FTb  FTc cos  30 

FG  FTc sin  30 
FTc 
7N
sin  30 
FG
Combining:
uf 


FTb
35 N
FTc cos  30 
f 
35 N
7N
cos  30 
sin  30 
35 N
7 N cos  30 
35 N sin  30 
 0.35
Atwood’s Machine Example:
Masses m1 = 10 kg and m2 = 20kg are attached to an ideal massless string
and hung as shown around an ideal massless pulley.
a)
b)
What are the tensions in the
string T1 and T2 ?
Find the accelerations, a1 and a2,
of the masses.
Fixed Pulley
T1
a1
T2
m1
m2
a2


Draw free body diagrams for each object
Applying Newton’s Second Law:
T1 - m1g = m1a1
(a)

T2 - m2g = -m2a2
=> -T2 + m2g = +m2a2 (b)

Free Body Diagrams
But T1 = T2 = T
since pulley is ideal
and a1 = -a2 =a
since the masses are
connected by the string
T1
a1
T2
a2
m1g
m2g
Solve for Acceleration
-m1g + T = m1 a
-T + m2g = m2 a
(a)
(b)
Two equations and two unknowns
 we can solve for both unknowns (T and a).

Free Body Diagrams
Add (b) + (a):
 g(m2 – m1 ) = a(m1+ m2 )

a
( m 2  m1 )
g
( m1  m 2 )
(20 kg  10 kg ) 
m
9.8


(10 kg  20 kg ) 
s2 
m
 3.27 2
s
a
T1
a1
T2
a2
m1g
m2g
-m1g + T = m1 a
-T + m2g = m2 a
(a)
(b)
Solve for T
Plug a into (b) and Solve for T
T  m2 g  m2
(m2  m1 )
g
(m1  m2 )
Free Body Diagrams
(m2  m1 )
T  m2 g  m2
g
(m1  m2 )
 (m2  m1 ) 
 m2 g 1 

(
m

m
)

1
2 
2m1
 m2 g
m1  m1\2

2m1m2
g
m1  m2
2 10kg  20kg  
m

9.8
10kg  20kg   s 2 
 130.7 N
T1
a1
T2
a2
m1g
m2g
Atwood Machine Review

So we find:
( m 2  m1 )
a
g
( m1  m 2 )
T
T
2 m1 m 2
T =
g
( m1 + m 2 )
a
m1
m2
a
Example
A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E],
what is the acceleration of the box?
Pulling a Box (Part 1)
A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E],
what is the acceleration of the box?
Free body Diagram
FN
+y
FA
+x
FG
a
Pulling a Box (Part 1)
A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E],
what is the acceleration of the box?
FN
+y
a
FA
+x
FG
Vertical Forces
F
y
0
F N  FG  0
N  mg  0
N  mg
Horizontal Forces
F
x
 ma x
F A  ma x
FA  ma
FA
m
F
 A
m
a
Solving
FA
m
95.0 N

18.0kg
m
 5.28 2
s
a
Example
Two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg are
pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg
box, what is the acceleration of each box, and what is the tension T in the rope
connecting both boxes?
Pulling a Box (Part 2)
A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is
pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg
box, what is the acceleration of each box, and what is the tension T in the rope
connecting both boxes?
Free body Diagram
FN
FN
FT
FA
FT
a
+y
+x
FG
FG
Pulling a Box (Part 2)
A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is
pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg
box, what is the acceleration of each box, and what is the tension T in the rope
connecting both boxes?
FN
a
FN
+y
FA
FT FT
+x
Forces
4.00 kg Box
F
x
 m1 a x
F T  m1 a
T  m1a
FG
FG
6.00 kg Box
F
x
 m2 ax
F T  F A  m2 a
T  FA  m2 a
Adding to eliminate T and find a
T
 m1a
T  FA  m2 a
+
FA  m1a  m2 a
FA  a  m1  m2 
a
FA
m1  m2
Pulling a Box (Part 2)
A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is
pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg
box, what is the acceleration of each box, and what is the tension T at either end
of the rope connecting both boxes?
FN
a
FN
+y
FT
FA
FT
+x
Solve for Acceleration
FA
a
m1  m2
20.0 N

4.00kg  6.00kg
m
 2.00 2
s
FG
Now for Tension
T  m1a

N
  4.00kg   2.00 
kg 

 8.00 N
FG
We could have used
the other tension
formula from Box 2
and obtained the same
answer
Example
A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is
connected by a 1kg rope and they are pulled along the floor. If a horizontal force is
20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and
what is the tension T at either end of the rope connecting both boxes?
1.00kg
Pulling a Box (Part 3)
A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is
pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg
box, what is the acceleration of each box, and what is the tension T at either end
of the rope connecting both boxes?
Free body Diagram
1.00 kg
FN
F T1
FG
Because the
rope has
mass, the
two ends will
experience
different
tensions
FN
FA
FT 2
a
+y
+x
FG
Pulling a Box (Part 3)
A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is
pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg
box, what is the acceleration of each box, and what is the tension T at either end
of the rope connecting both boxes?
a
FN
FN
+y
FT
FA
FT
+x
Forces
4.00 kg Box
 F x  m1 a x
F T  m1 a
T1  m1a
FG
6.00 kg Box
F
x
 m2 ax
F T  F A  m2 a
T2  FA  m2 a
T2  FA  m2 a
FG
Using F=ma for the system to find a
 F  ma
F  m  m
A
1
2
 mrope  a
FA
a
m1  m2  mrope
Pulling a Box (Part 3)
A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is
pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg
box, what is the acceleration of each box, and what is the tension T in the rope
connecting both boxes?
FN
FN
a
+y
FT
FA
FT
+x
FG
FG
Solve for Acceleration
a
FA
m1  m2  mrope
20.0 N
4.00kg  1.00kg  6.00kg
m
 1.82 2
s

Now for T1
T1  m1a

N
  4.00kg  1.82 
kg 

 7.27 N
Now for T2
T2  FA  m2 a

N
 20.0 N   6.00kg  1.82 
kg 

 9.09 N
Example
A worker drags a 38.0 kg box along the floor by pulling on a rope attached to
the box. The coefficient of friction between the floor and the box is us=0.450
and uk=0.410.
a) What are the force of friction and acceleration when the worker applies a
horizontal force of 150N?
b) What are the force of friction and acceleration when the worker applies a
horizontal force of 190N?
38kg
Solution (Free Body Diagram)
a) What are the force of friction and acceleration of the worker applies a
horizontal force of 150N?
The Normal force up
Friction
to the
left
FN
The applied force of
tension to the right
38kg
FT
Ff
FG
The force of gravity
down
+y
+x
Solution (Vector Components)
What are the force of friction and acceleration of the worker applies a horizontal
force of 150N?
FN
+y
38kg
FT
Ff
+x
FG
To determine if the box will move, we must find the maximum static friction and
compare it to the applied force.
F
y
 ma y  0
F N  FG  0
FN  mg  0
FN  mg

N
  38kg   9.80 
kg 

 372.4 N
Ff  S FN
  0.450  372.4 N 
 168 N
Since the applied
force by the worker
is only 150N, the
box will not move
Solution (Vector Components)
b) What are the force of friction and acceleration of the worker applies a horizontal
force of 190N?
FN
+y
38kg
FT
Ff
+x
FG
x  ma
x
Since
applied
force is greater than 168N from part a), we will have an
 Fthe
nd
acceleration
F f  F T  ma x in the x direction. So we will apply Newton’s 2 Law in the
horizontal direction.
F =(0.410)(372.4N)=153N
 F  F  ma
f
T
x
a
K
FT  Ff
m
190 N   K FN

m
190 N   0.410  372.4 N 

38kg
m
 0.982 2
s
The acceleration of the box
is 0.982 m/s2 [E]
Example
A train of three masses is pulled along a frictionless surface. Calculate the tensions in
the ropes.
T1
8 kg
30 N
T2
5 kg
13 kg
We can find the acceleration of the train by treating the three masses as one unit.
F  ma
30 N   8kg  5kg  13kg  a
a  1.15
 F  ma
m
[left ]
2
s
Tension in rope T1
T1
Tension in rope T2
F
 F  ma
FA  T1  m1a
T1  FA  m1a
m

 30 N   8kg  1.15 2 
s 

 20.8 N
T1
T1  T2  m2 a
or
T2  T1  m2 a
m

 20.8 N   5kg  1.15 2 
s 

 15.1N
T2
 F  ma
T2  ma
m

 13kg  1.15 2 
s 

 15.1N
Example
A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain
pulls the sled at and angle of 300 above the horizontal. How hard does the tractor
have to pull to keep the sled moving with a constant velocity if the sled and
firewood has a weight of 15,000 N and uk=0.40?
300
Solution (Free Body Diagram)
A farmer uses his tractor to pull a sled loaded with firewood. Suppose
the chain pulls the sled at and angle of 300 above the horizontal. How
hard does the tractor have to pull to keep the sled moving with a
constant velocity if the sled and firewood has a weight of 15,000 N
and uk=0.40?
The Normal force up
Friction
to the
left
FN
The applied force of
tension at 300
FT
FT sin 30
30
Ff
FT cos 30
300
Tension broken down into
components
FG
The force of gravity
down
Solution (Force Components)
A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain
pulls the sled at and angle of 300 above the horizontal. How hard does the tractor
have to pull to keep the sled moving with a constant velocity if the sled and
firewood has a weight of 15,000 N and uk=0.40?
FN
FT
+y
FT sin 30
30
Ff
FT cos 30
+x
300
FG
Vertical Components
Horizontal Components
Fy 0
 F x  ma x
F N  F G  F Ty  0
F f  F Tx  0
FN  mg  FT sin  30   0
 k FN  FT cos  30   0
FN  mg  FT sin  30 
Since we have a
constant velocity,
acceleration is 0
uk  mg  FT sin  30    FT cos  30   0
We will need FN, so
solve for FN
Solution (Force Components)
A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain
pulls the sled at and angle of 300 above the horizontal. How hard does the tractor
have to pull to keep the sled moving with a constant velocity if the sled and
firewood has a weight of 15,000 N and uk=0.40?
FN
FT
+y
FT sin 30
+x
30
Ff
FT cos 30
300
FG
Solve for FT
uk  mg  FT sin  30    FT cos  30   0
uk mg  uk FT sin  30   FT cos  30   0
uk FT sin  30   FT cos  30   uk mg
FT  uk sin  30   cos  30    uk mg
FT 
uk mg
uk sin  30   cos  30 
FT 
 0.40 15000 N 
 0.40  sin  30  cos  30
 5628.384 N
 5600 N
Example
A group of children toboggan down a hill with a 300 slope. Given that the
coefficient of kinetic friction is 0.10, calculate their acceleration and the speed
they will obtain after 6.0s.
Solution (Free Body Diagram)
A group of children toboggan down a hill with a 300 slope. Given that the coefficient
of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain
after 6.0s.
+y
Choose axis orientation to
match the direction of motion
Normal is
and the normal to the surface
perpendicular to
the surface
+x
FN
Ff
Object
Force of friction
F cos 30
opposes direction
of motion
G
FN
Ff
30
FG sin 30
Decompose
gravity into axis
components
FG cos 30 30
FG sin 30
FG
Force of gravity is
straight down
FG
Solution (Force Vectors)
A group of children toboggan down a hill with a 300 slope. Given that the
coefficient of kinetic friction is 0.10, calculate their acceleration and the speed
they will obtain after 6.0s.
FN
+y
Ff
FG cos  30
30
+x
FG sin  30
FG
y direction
F
y
0
F N  F Gy  0
FN  FG cos  30   0
FN  FG cos  30 
FN  mg cos  30 
x direction
F
 max
Remember
Ff to
FGxsolve
 max
for FN because we
uk F
 FG sin  30   max
will need
itNlater
uk mg cos  300   mg sin  30   max
x


ax  uk g cos  300   g sin  30 
Example 9:
Solution (Force Vectors)
A group of children toboggan down a hill with a 300 slope. Given that the
coefficient of kinetic friction is 0.10, calculate their acceleration and the speed
they will obtain after 6.0s.
FN
+y
Ff
FG cos  30
30
+x
FG sin  30
FG
Acceleration
ax  uk g cos  30   g sin  30 
0
m
m


   0.10   9.80 2  cos  30    9.80 2  sin  30 
s 
s 


m
 4.05 2
s
m
 4.1 2
s
Speed
v f  vi  at
m
m
 4.05 2  6 s 
s
s
m
 24.3
s
m
 24
s
0
Example
Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient
of kinetic friction between all surfaces is 0.20. Find the magnitude of the horizontal force,
F, necessary to drag block B to the left at a constant speed if A and B are connected by a
light, flexible cord passing around a fixed, frictionless pulley.
Solution (Free Body Diagram)
Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient
of kinetic friction between all surfaces is 0.20. Find the magnitude of the horizontal force,
F, necessary to drag block B to the left at a constant speed if A and B are connected by a
light, flexible cord passing around a fixed, frictionless pulley.
+y
Box B
FN
a
+x
FN
Normal Force
Friction from
Friction from
Object B A
Table
F f (box A)
F f (box A)
F f (table) of Gravity
Force
F T A and B
from
FA
Applied Force
FG
F f (table)
FA
FT
Tension
FG
Solution (Free Body Diagram)
Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient
of kinetic friction between all surfaces is 0.20. Find the magnitude of the horizontal force,
F, necessary to drag block B to the left at a constant speed if A and B are connected by a
light, flexible cord passing around a fixed, frictionless pulley.
FN
+y
a
+x
Normal Force
Box A
F f (box A)
F f (table)
FA
FT
FN
FG
FN
Friction
Object
Tension
FT
Ff
FT
Ff
F GA
+y
a
+x
Force of Gravity
from A only
F GA
Solution (Force Vectors)
Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is
0.20. Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are
connected by a light, flexible cord passing around a fixed, frictionless pulley.
A
B
FN
FN
a
+y
F f (box A)
FT
Ff
+x
F GA
x-direction for Box A
F
x
0
F f  FT  0
 F f  FT  0
  k FN  FT  0
  k mA g  FT  0
FT  k mA g
F f (table)
FA
FT
FG
x-direction for Box B
F
x
0
FA  F f (box A)  F f (table )  F T  0
 FA   k FN ( A)  k FN (table )  FT  0
FA  k mA g  k  mA  mB  g  k mA g
 3k mA g  uk mB g
 FA  k mA g  k  mA  mB  g  FT  0
FA  k mA g  k  mA  mB  g  FT
This was determined
using Box A
Solution (Force Vectors)
Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is
0.20. Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are
connected by a light, flexible cord passing around a fixed, frictionless pulley.
A
B
FN
FN
a
+y
F f (box A)
FT
Ff
+x
F GA
FA  3k mA g  uk mB g
 3  0.20 1.50 N    0.20  4.30 N 
 1.76 N
F f (table)
FA
FT
FG
Example
How much force is needed to give the blocks an acceleration of 3.0 m/s 2 if the coefficient
of kinetic friction between the blocks and the floor is 0.20 ? How much force does the
1.50 kg block exert on the 2.0 kg block?
2.0
1.5 kg
kg
1.0
kg
Example
(free body diagram)
How much force is needed to give the blocks an acceleration of 3.0 m/s 2 if the coefficient
of kinetic friction between the blocks and the floor is 0.20 ? How much force does the
1.50 kg block exert on the 2.0 kg block?
3 Blocks taken as a Single Unit
Normal Force
FN
Friction
Object
Applied
FA
Ff
FG
+y
a
+x
Force of Gravity
2.0
1.5 kg
kg
1.0
kg
Example
(force vectors)
How much force is needed to give the blocks an acceleration of 3.0 m/s 2 if the coefficient
of kinetic friction between the blocks and the floor is 0.20 ? How much force does the
1.50 kg block exert on the 2.0 kg block?
FN
+y
1.5
kg
a
FA
Ff
2.0
kg
1.0
kg
+x
FG
F
x
 max
Ff  FA  ma
 Ff  FA  ma
 k FN  FA  ma
  mg  FA  ma
FA  ma   mg

m
N

 1.5kg  2.0kg  1.0kg   3.0 2    0.20 1.5kg  2.0kg  1.0kg   9.8 
s 
kg 


 22.32 N
 22 N
Example (free body diagram)
How much force is needed to give the blocks an acceleration of 3.0 m/s 2 if the coefficient
of kinetic friction between the blocks and the floor is 0.20 ? How much force does the
1.50 kg block exert on the 2.0 kg block?
2.0 kg Block and 1.0 kg taken as a Single Unit
Normal Force
Since we are considering 2.0kg and 1.0kg
block as a unit, then the Force is the push
of 1.5 kg block on the combined block
FN
Friction
Object
Applied
FA
Ff
Force of Gravity
F G2.01.0
+y
a
+x
2.0
1.5 kg
kg
1.0
kg
Example
(force vectors)
How much force is needed to give the blocks an acceleration of 3.0 m/s 2 if the coefficient
of kinetic friction between the blocks and the floor is 0.20 ? How much force does the
1.50 kg block exert on the 2.0 kg block?
+y
FN
+x
F block1
Ff
F G2.01.0
F
x
a
2.0
kg
1.0
kg
 ma x
F f  F block 1  ma
uFN  Fblock1  ma
umg  Fblock1  ma
Fblock1  ma   mg

m
N

  2.0kg  1.0kg   3.0 2    0.20  2.0kg  1.0kg   9.8 
s 
kg 


 14.88 N
 15 N
Example
A 5.00 kg object placed on a frictionless, horizontal table is connected to a string that passes
over a pulley and then is fastened to a hanging 9.00 kg object.
a) Find the acceleration of the two objects
b) Find the tension in the string.
Step 1: Free body Diagram
m1
FT
FT
+
+
FG=9 kg
The easiest way to choose the signs for the forces is to logical choose
what you believe will be correct direction and follow that direction from
one object to the other. If your final answer is negative, it just means
your initial direction choice was wrong.
Example (Solution)
A 5.00 kg object placed on a frictionless, horizontal table is connected to a string that passes
over a pulley and then is fastened to a hanging 9.00 kg object.
a) Find the acceleration of the two objects
m1
FT
FT
+
FT  m1a
FG=9 kg
+
these equations will
 FAdding

F
T
G  m2 a
Horizontal
Vertical
F  ma
1
FT  m1a
F m a
2
 FT  FG  m2 a
remove the force of tension,
a  m2 a us to
thus F
giving
G  m1allowing
solve
m for acceleration.
 9kg   9.8

  m1  m2  a
2 
s 

Since the answer
88.2 N  14kg  a is positive our
Note: don’t just say that a force of (9.8x9) is pulling on the 5 kg mass. This will give
an acceleration of 17.6, which is greater than the acceleration due to gravity of the
falling 9 kg mass, and you would not have tension in the rope. Ugly!
m
a  6.3 2
s
initial direction
choice was correct.
Example (Solution)
A 5.00 kg object placed on a frictionless, horizontal table is connected to a string that passes
over a pulley and then is fastened to a hanging 9.00 kg object.
b) Find the tension in the string.
m1
FT
FT
+
FG
+
Horizontal
Vertical
F  ma
1
FT  m1a
F m a
2
 FT  FG  m2 a
We need only substitute the
acceleration value into either
the horizontal or vertical
FT equation.
 m1a
m

  5kg   6.3 2 
s 

 31.5 N
Example
Blocks A, B, and C are placed as in the figure and connected by ropes of negligible
mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the
blocks and the surface is 0.35. Block C descends with constant velocity.
a) Determine the tension in the rope connecting Block A and B.
b) What is the weight of Block C?
c) If the rope connecting A and B were cut, what would be the acceleration of C?
Example
Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N
each, and the coefficient of kinetic friction between the blocks and the surface is 0.35. Block c descends with constant
velocity.
a) Determine the tension in the rope connecting Block A and B.
b) What is the weight of Block C?
c) If the rope connecting A and B were cut, what would be the acceleration of C?
FN
F T (C )
FN
Ff
F Tf
F T ( A)
Normal
Block A
FN
Object
Ff
Friction
FG
FG
Block B
Normal
FG
Tension from C
FN
Tension
FT
Friction
Tension from A
Ff
Gravity
F T ( A)
F T (C )
Object
36.9
FG
Gravity
Example (Force Vectors)
Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N
each, and the coefficient of kinetic friction between the blocks and the surface is 0.35. Block c descends with constant
velocity.
a) Determine the tension in the rope connecting Block A and B.
b) What is the weight of Block C?
c) If the rope connecting A and B were cut, what would be the acceleration of C?
FN
FN
Ff
FT
Ff
F T (C )
mB g cos  36.9
36.9
mB g sin 36.9
F T ( A)
FG
FG
Block A
F
x
0
F f ( A)  F T ( A)  0
  k FN  FT  A  0
  k mA g  FT  A  0
Block B
F
x
0
F f  B   F T  A  F Gx  F T C   0
 k FN  B   FT  A  mB g sin  36.9   FT C   0
 k mB cos  36.9  g  FT  A  mB g sin  36.9   FT C   0
Block C
FT C   mC g
Example (Force Vectors)
Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N
each, and the coefficient of kinetic friction between the blocks and the surface is 0.35. Block c descends with constant
velocity.
a) Determine the tension in the rope connecting Block A and B.
b) What is the weight of Block C?
c) If the rope connecting A and B were cut, what would be the acceleration of C?
FN
Ff
FN
FT
Ff
F T (C )
mB g cos  36.9
36.9
mB g sin 36.9
F T ( A)
FG
Block A
k mA g  FT  A  0
FG
Block C
Block B
k mB cos  36.9 g  FT  A  mB g sin  36.9  FT C   0
FT C   mC g
k mB cos  36.9 g  FT  A  mB g sin  36.9  mC g  0
  0.35  25 N  mC g  k mB cos  36.9  g  FT  A  mB g sin  36.9 
 8.75 N
Solving
for25the
  0.35 
N  cos  36.9   8.75
N   25 N
 sin  36.9 
Applying
Block
tension between
 8.8 N
C’s
equation to
 30.757 N
The
weight
block A and B
Block
of Block
C B
 31N
FT  A  k mA g
Example (Force Vectors)
Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N
each, and the coefficient of kinetic friction between the blocks and the surface is 0.35. Block c descends with constant
velocity.
a) Determine the tension in the rope connecting Block A and B.
b) What is the weight of Block C?
c) If the rope connecting A and B were cut, what would be the acceleration of C?
FN
Ff
FN
FT
Ff
F T (C )
mB g cos  36.9
36.9
mB g sin 36.9
F T ( A)
FG
FG
When the rope is cut:
 F  ma
mC g  Ff  B   mB g sin  36.9    mB  mC  a
a) FT  A  8.8 N
b) mC g  31N
c) a  1.5
m
s2
a
mC g  Ff  B   mB g sin  36.9 
mB  mC
30.7 N   0.35  25 N  cos  36.9    25 N  sin  36.9 
2.55kg  3.14kg
m
 1.53 2
s

Example
A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an
angle of 9.20 with the vertical as the truck is accelerating forward. Determine the
magnitude of the acceleration.
9.2
Example
(Free Body Diagram)
A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an
angle of 9.20 with the vertical as the truck is accelerating forward. Determine the
magnitude of the acceleration.
We will look at this from outside the truck (ie the ground) because we would
prefer an inertial frame of reference.
Tension
broken into
components
Tension
FT
FT sin  9.2F T
FT cos 9.2
9.2
Object
9.2
Gravity
FG
FG
Example
(Force Vectors)
A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an
angle of 9.20 with the vertical as the truck is accelerating forward. Determine the
magnitude of the acceleration.
FT sin  9.2F T
FT cos 9.2
9.2
FG
Vertical Forces
F
F
x
y
0
F G  F Ty  0
mg  FT cos  9.2   0
FT 
Horizontal Forces
mg
cos  9.2 
 max
FT sin  9.2   max
FT sin  9.2 
m
sin  9.2 
mg

cos  9.2 
m
ax 

g sin  9.2 
cos  9.2 
ax  g tan  9.2 
m

  9.8 2  tan  9.2 
s 

m
 1.6 2
s
Example
Calculate the acceleration of a box that experiences a magnetic force of repulsion of 12N
from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is
also being pushed against the wall with a force of 25N at 300 to the horizontal.
5.0 kg
25N
30
Example
(Free Body Diagram)
Calculate the acceleration of an box that experiences a magnetic force of repulsion of
12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box
is also being pushed against the wall with a force of 25N at 300 to the horizontal.
Applied
Friction
Object
25sin 30
Magnetic
Since the gravity force
down (5x9.8) is greater
than force up (25sin(30),
the box slides down, so
friction is up.
5.0 kg
25cos 30
Normal
25N
30
Gravity
Example (Vector Forces)
Calculate the acceleration of an box that experiences a magnetic force of repulsion of
12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box
is also being pushed against the wall with a force of 25N at 300 to the horizontal.
25sin 30
+x
5.0 kg
a
25cos 30
+y
Horizontal
F
x
0
F Ax  F M  F N  0
 FAx  FM  FN  0
FN  FAx  FM
FN  25 N cos  30   FM
Vertical
F
y
25N
30
 ma y
F Ay  F f  F G  ma y
 FAy  Ff  FG  ma y
ay 
 FAy  Ff  FG
m
25 N sin  30   k FN  mg

m
25 N sin  30   k  25 N cos  30   FM   mg

m
Example (Insert Numbers)
Calculate the acceleration of an box that experiences a magnetic force of repulsion of
12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box
is also being pushed against the wall with a force of 25N at 300 to the horizontal.
25sin 30
+x
5.0 kg
a
25cos 30
+y
25N
30
ay 
25 N sin  30   k  25 N cos  30   FM   mg
m
m

25 N sin  30    0.4   25 N cos  30   12 N    5kg   9.8 2 
s 


5.0kg
m
 6.5 2
s
Example
Three hundred identical objects are connected in series on a ramp. Calculate the force
between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force
that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s 2,
and the coefficient of kinetic friction is 0.15.
Example (Free Body Diagram)
Three hundred identical objects are connected in series on a ramp. Calculate the force
between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force
that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s 2,
and the coefficient of kinetic friction is 0.15.
For Masses 121 to 300
Treat objects 300 thru
121 as a single mass
of 180*0.5kg
Normal
Big Mass
FN
F T ( above)
Ff
10
Tension
mg cos 10from above
Gravity
Friction
FG
mg sin 10
Example (Vector Forces)
Three hundred identical objects are connected in series on a ramp. Calculate the force
between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force
that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s 2,
and the coefficient of kinetic friction is 0.15.
For Combined Masses 121 to 300
FN
F T ( above)
a
+y
+x
Ff
10
FG
y-axis
mg cos 10
mg sin 10
x-axis
F
y
0
F N  F Gy  0
FN  FGy  0
FN  mg cos 10   0
FN  mg cos 10 
F
x
 ma x
F f  F Gx  F T  ma x
 Ff  FGx  FT  ma
  FN  mg sin 10   FT  ma
FT  ma   FN  mg sin 10 
FT  ma    mg cos 10    mg sin 10 
Example (Insert Values)
Three hundred identical objects are connected in series on a ramp. Calculate the force
between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force
that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s 2,
and the coefficient of kinetic friction is 0.15.
For Combined Masses 1 to 120
FN
F T ( above)
a
+y
+x
Ff
10
FG
mg cos 10
mg sin 10
FT  ma    mg cos 10    mg sin 10 
m
m
m



 180  0.5kg   0.88 2    0.15 180  0.5kg   9.8 2  cos 10   180  0.5kg   9.8 2  sin 10 
s 
s 
s 



 362.6477 N
 360 N
Example
A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left
across a level floor when u=0.400,
a) Find the normal force on the crate
b) Find the force of friction on the crate
c) Find the acceleration of the crate.
300
a
Example (Free Body Diagram)
A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left
across a level floor when u=0.400,
a) Find the normal force on the crate
b) Find the force of friction on the crate
c) Find the acceleration of the crate.
FN
a
Ff
300
+y
Fa
Fg
+x
300
Example (Vector Forces)
A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left
across a level floor when u=0.400,
+y
FN
a) Find the normal force on the crate
b) Find the force of friction on the crate
+x
c) Find the acceleration of the crate.
a
Ff
300
y-axis
F
Fa
y
0
F N  F Gy  F Ay  0
FN  FGy  FAy  0
FN  mg  FA sin 30  0
FN  mg  FA sin 30
Insert Values
Fg
m

FN   20.0kg   9.8 2   175 N  sin  30 
s 

 196 N  87.5 N
 283.5 N
284 N
Example (Vector Forces)
A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left
across a level floor when u=0.400,
FN
+y
a) Find the normal force on the crate
b) Find the force of friction on the crate
c) Find the acceleration of the crate.
+x
a
Ff
300
Friction
Fa
F f  k FN
F f   0.400 283.5N 
F f  113.4N
Ff  113N
Fg
Example (Vector Forces)
A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left
across a level floor when u=0.400,
FN
+y
a) Find the normal force on the crate
b) Find the force of friction on the crate
c) Find the acceleration of the crate.
+x
a
Ff
300
Acceleration
F
x
 max
F f  F Ax  max
Ff  FA cos 30  max
113.4N 175N cos 30  max
max  38.1544N
Fa
38.1554 N
20.0kg
m
ax  1.9077 2
s
m
ax  1.91 2 W 
s
ax 
Fg
Example
Calculate the unknowns for each accelerated block.
a)
3.7
m
s2
18 kg
b)
F
  0.2
 F  ma
FA  Ff  ma
FA  ma  Ff
FA  ma   FN

m
N

FA  18kg   3.7 2    0.2 18kg   9.8 
s 
kg 


FA  102 N
c)
a
6 kg
  0.2
40N
41N
5.6
m
s2
m
  0.3
Example (Solution)
Calculate the unknowns for each accelerated block.
a)
3.7
m
s2
18 kg
F
  0.2
 F  ma
FA  Ff  ma
FA  ma  Ff
FA  ma   FN

m
N

FA  18kg   3.7 2    0.2 18kg   9.8 
s 
kg 


FA  102 N
Example (Solution)
Calculate the unknowns for each accelerated block.
b)
a
6 kg
40N
  0.2
 F  ma
FA  Ff  ma
a
FA  Ff
m
F   FN
a A
m

N
40 N   0.2  6kg   9.8 
kg 

a
6kg
a  4.7
m
s2
Example (Solution)
Calculate the unknowns for each accelerated block.
c)
5.6
41N
m
s2
m
  0.3
 F  ma
FA  mg  ma
m  a   g   FA
m
m
FA
a  g
41N
m
m


5.6

0.3
9.8






s2 
s2 


m  4.8kg
Example
Determine the acceleration of the cart that is required to prevent Sir Isaac Newtant
from falling. The coefficient of static friction between the block and Newtant is μs
Example
Determine the acceleration of the cart that is required to prevent Sir Isaac Newtant
from falling. The coefficient of static friction between the block and Newtant is μs
If I do not fall, then the
friction force, Ff, must
balance my weight mg,
that is Ff = mg
The only horizontal
force is the Normal
Force .Therefore
F=N=ma
Putting this together we obtain:
Ff   s N
  s ma
mg   s ma
g  s a
a
g
s
Attached bodies on two
inclined planes
smooth peg
m1
m2
1
2
all surfaces frictionless
peg is frictionless
How will the bodies move?
From the free body diagrams for each body, and the chosen coordinate
system for each block, we can apply Newton’s Second Law:
Taking “x” components:
x
y
T1
1) T1 - m1g sin 1 = m1 a1X
2) T2 - m2g sin 2 = m2 a2X
N
But T1 = T2 = T
and -a1X = a2X = a
(constraints)
y
x
N
T2
m2
m1
1
2
m2 g
m1 g
Solving the equations
Using the constraints, we get 2 eqn and 2 unks,
solve the equations.
T - m1gsin 1 = -m1 a
(a)
T - m2gsin 2 = m2 a
(b)
Subtracting (a) from (b) gives:
m1gsin 1 - m2gsin 2 = (m1+m2 )a
So:
a
m 1 sin 1  m2 sin 2
g
m1  m 2
Two-body dynamics

In which case does block m experience a larger acceleration? In (1)
there is a 10 kg mass hanging from a rope. In (2) a hand is
providing a constant downward force of 98.1 N. In both cases the
ropes and pulleys are massless.
m
a
m
a
10kg
F = 98.1 N
Case (1)
(a)
Case (2)
Case (1)
(b) Case (2)
(c)
same
Solution

For case (1) draw FBD and write FNET = ma for each block:
T = ma
mWg -T = mWa

(b)
m
Add (a) and (b):
mWg = (m + mW)a
mW g
a
m  mW

(a)
(a)
Note:
m
T  mW g 
m  mW
a
10kg
mW=10kg
(b)
Solution

For case (2)
a
T = 98.1 N = ma
a
98 .1N
m  10 kg
98 .1N
m
a
98 .1N
m
m
a
m
a
10kg
F = 98.1 N
Case (1)

Case (2)
The answer is (b) Case (2) In this case the block experiences a larger
acceleratioin
Problem: Accelerometer

A weight of mass m is hung from the ceiling of a car with a massless
string. The car travels on a horizontal road, and has an acceleration
a in the x direction. The string makes an angle  with respect to the
vertical (y) axis. Solve for  in terms of a and g.
a
i

Accelerometer...

Draw a free body diagram for the mass:
 What are all of the forces acting?
T (string tension)

m
i
mg (gravitational force)
Accelerometer...

Using components (recommended):
i: FX = TX = T sin  = ma
j: FY = TY  mg
= T cos   mg = 0

T

j
m
i
ma
mg
Accelerometer...

Using components :
i: T sin  = ma
T
j: T cos  - mg = 0

j
m
i
ma

mg
Eliminate T :
T sin  = ma
T cos  = mg
tan  
a
g
Accelerometer...

Alternative solution using vectors (elegant but not as systematic):

Find the total vector force FNET:
T
mg

FTOT
T (string tension)

m
mg (gravitational force)
Accelerometer...



Alternative solution using vectors (elegant but not as systematic):
Find the total vector force FNET:
Recall that FNET = ma:
T (string tension)
T
mg


m
ma
mg (gravitational force)

So
ma a
tan  

mg g
tan  
a
g
Accelerometer...

Let’s put in some numbers:

Say the car goes from 0 to 60 mph in 10 seconds:
 60 mph = 60 x 0.45 m/s = 27 m/s.
 Acceleration a = Δv/Δt = 2.7 m/s2.
 So a/g = 2.7 / 9.8 = 0.28 .

tan  
 = arctan (a/g) = 15.6 deg
a

a
g
Understanding
A person standing on a horizontal floor feels two forces: the
downward pull of gravity and the upward supporting force from the
floor. These two forces
(A)
(B)
(C)
(D)
(E)
Have equal magnitudes and form an action/reaction pair
Have equal magnitudes but do not form an action/reaction pair
Have unequal magnitudes and form an action/reaction pair
Have unequal magnitudes and do not form an action/reaction pair
None of the above
Because the person is not accelerating, the net force they feel is zero.
Therefore the magnitudes must be the same (opposite directions. These
are not action/reaction forces because they act of the same object (the
person). Action/Reaction pairs always act on different objects.