Chapter 5 – Force and Motion I
I.
Newton’s first law.
II. Newton’s second law.
III. Particular forces:
- Gravitational
- Weight
- Normal
- Tension
IV. Newton’s third law.
In the figure below, mblock=8.5kg and θ=30º. Find (a) Tension in the
cord.
(b) Normal force acting on the block. (c) If the cord is cut, find the magnitude
of the block’s acceleration.
N
T
Fg
(a) a  0  Fgx  mg sin 30  (8.5kg)(9.8m / s 2 )0.5  41.65N
(b) N  Fgy  mg cos sin 30  72.14 N
(c) T  0  Fgx  ma  41.65 N  8.5a  a  4.9m / s 2
Chapter 6 – Force and Motion II
I. Frictional forces.
II. Drag forces and terminal speed.
III. Uniform circular motion.
I. Frictional force
Counter force that appears when an external force tends to slide a body
along a surface. It is directed parallel to the surface and opposite to the
sliding motion.
-Static: (fs) compensates the applied force, the body does


not move.
f s   F//
-Kinetic: (fk) appears after a large enough external force is
applied and the body loses its intimate contact with
the surface, sliding along it.
No motion
Acceleration
F
(applied
force)
Constant velocity
f k  f s ,max
f s,max  s N
If F//  f s ,max  body slides Friction coefficients
f k  k N
After the body starts sliding, fk decreases.
II. Drag force and terminal speed
-Fluid: anything that can flow. Example: gas, liquid.
-Drag force: D
- Appears when there is a relative velocity between a fluid and
a body.
- Opposes the relative motion of a body in a fluid.
- Points in the direction in which the fluid flows.
Assumptions:
* Fluid = air.
* Body is baseball.
* Fast relative motion  turbulent air.
1
D  CAv 2
2
C = drag coefficient (0.4-1).
ρ = air density (mass/volume).
A= effective body’s cross sectional area  area perpendicular to v
-Terminal speed: vt
- Reached when the acceleration of an object that experiences a vertical
movement through the air becomes zero  Fg=D
1
D  Fg  ma  if a  0  CAv 2  Fg  0
2
vt 
2 Fg
C A
III. Uniform circular motion
-Centripetal acceleration:
v2
a
r
v, a are constant, but direction changes during motion.
A centripetal force accelerates a body by changing the direction of the
body’s velocity without changing its speed.
-Centripetal force:
v2
F m
R
a, F are directed toward the center of curvature
of the particle’s path.
A puck of mass m slides on a frictionless table while
attached to a hanging cylinder of mass M by a cord
through a hole in the table. What speed
keeps the cylinder at rest?
N
T
mg
For M  T  Mg
T
Mg
2
2
v
v
Mgr
For m  T  m  Mg  m  v 
r
r
m
Calculate the drag force on a missile 53 cm in diameter
cruising with a speed of 250 m/s at low altitude, where the
density of air is 1.2 kg/m3. Assume C=0.75
1
2
2
3
2
D  CAv  0.5  0.75  (1.2kg / m )    (0.53m / 2) 250m / s   6.2kN
2
The terminal speed of a ski diver is 160 km/h in the
spread eagle position and 310 km/h in the nosedive position. Assuming that the diver’s drag
coefficient C does not change from one point to
another, find the ratio of the effective cross
sectional area A in the slower position to that of the
faster position.
2 Fg
2 Fg
160km / h
vt 


CA
310km / h
CAE
2 Fg
CAD

AD
AE

 3.7
AD
AE
Two blocks of weights 3.6N and 7.2N, are connected by a massless
string and slide down a 30º inclined plane. The coefficient of kinetic
friction between the lighter block and the plane is 0.10; that between
the heavier block and the plane is 0.20. Assuming that the lighter
block leads, find (a) the magnitude of the acceleration of the blocks
and (b) the tension in the string.
NB
Block A
NA
NB
T
FgxA
Block B
fkA
FgyA
NA
B
T
T
FgxB
fkB
T
FgyB
Light block A leads
fk,A
A
FgA
fk,B
FgB
Light block A leads
a  3.49m / s 2
T  0 .2 N
Block B weighs 711N. The coefficient of static friction between the
block and the table is 0.25; assume that
the cord between B and the knot is
N
horizontal. Find the maximum weight
T2
of block A for which the system will
f
be stationary.
T1
T1 T3
T3
System stationary  f s ,max   s N
FgB
FgA
Block B  N  mB g
T1  f s ,max  0  T1  0.25  711N  177.75 N
Knot  T1  T2 x  T2 cos 30  T2 
177.75 N
 205.25 N

cos 30
T2 y  T2 sin 30  T3
Block A  T3  mA g  T2 sin 30  0.5  205.25 N  102.62 N
Blocks A and B have weights of 44N and 22N, respectively. (a)
Determine the minimum weight of block C to keep A from sliding
if μPs between A and the table is 0.2. (b) Block C suddenly is lifted
of A. What is the acceleration of block A if μk between A and the
table is 0.15?
(a)
f  f s ,max   s N
Block A  a  0  T  f s ,max  0  T   s N
(1)
N
f
T
Block B  T  mg  0  T  22 N
Wc
(2)
WA=44N
T
T
WB=22N
22 N
(1)  (2) N 

 110 N
s
0.2
Blocks A, B  N  WA  WC  WC  110 N  44 N  66 N
Blocks A and B have weights of 44N and 22N, respectively. (a)
Determine the minimum weight of block C to keep A from sliding
if μPs between A and the table is 0.2. (b) Block C suddenly is lifted
of A. What is the acceleration of block A if μk between A and the
table is 0.15?
C disappears  N  m A g  44 N
(b)
T  k N  mAa
T  mB g   mB a
N
f
T
Wc
WA=44N
T
WB=22N
T  6.6  4.5a
T  22  2.2a
 a  2.3m / s 2
 T  17 N
The two blocks (with m=16kg and m=88kg) shown in the figure below are
not attached. The coefficient of static friction between the blocks is:
μs=0.38 but the surface beneath the larger block is frictionless.
What is the minimum value of the
N
f
horizontal force F required to keep the
F’
F’
smaller block from slipping
down the larger block?
mg
Fmin required to keep m from sliding down?
Mg
Treat both blocks as a single system sliding
across a frictionless floor
Movement
F
F  mtotala  a 
mM
 F 
Small block  F  F '  ma  m

m

M


f s  mg  0   s F ' mg  0
(1)  (2)
(1)
( 2)
mg  m  M 
 F 
  mg  F 

  488 N
s  M 
mM 
 s m
An amusement park ride consists of a car moving in a vertical circle
on the end of a rigid boom of negligible mass. The combined
weight of the car and riders is 5 kN, and the radius of the circle is
10 m.
What are the magnitude and the direction of the force of the boom
on the car at the top of the circle if the car’s speed is (a) 5m/s (b)
mg m/s?
12
The force of the boom on the car is capable of pointing any direction
 v2
FB  W  m 
 R
(a) v  5m / s  FB  3.7 N up
y
FB
W


v2 
  FB  W 1 



Rg 


(b) v  12m / s  FB  2.3 down