Engineering Mechanics (II)

Dynamics
second Semester 2023-2024

LECTURE 5
Engineering Mechanics: II 2023/2024
Example 2.3
The baggage truck A shown in the photo has a weight
of 900 lb and tows a 550-lb cart B and a 325-lb cart
C. For a short time the driving frictional force
developed at the wheels of the truck is FA = (40t) lb,
where t is in seconds. If the truck starts from rest,
determine its speed in 2 seconds. Also, what is the
horizontal force acting on the coupling between the
truck and cart B
at this instant?
Neglect the size
of the truck and
carts.
Free-Body Diagram.
Equation of Motion

  Fx  ma x

900  550  325

40t  ( )a
32 .2
a  0.7256t
 a  0.7256t

by the relation a  dv / dt associated with the initial

condition that v0  0 at t  0

v t
0 dv  0 0.7256 t dt
2t
v  0  0.3628 t
0

v  0.3628 t 2

At t=2 sec
v  1.45 ft/s
free-body diagram of the truck

Equation of Motion.
When t = 2 s, then

  Fx  ma x
900 a  0.7256t
(40 )( 2)  T  ( ) (0.7256)(2)
32 .2
FA  40t
T  39.4 Ib
Example 2.5

The 100-kg block A

shown in Fig. is
released from rest.
If the masses of the
pulleys and the cord
are neglected,
determine the speed
of the 20-kg block
B in 2 s.
Free-Body Diagrams. Kinematics.
2s A  s B  l
2a A   a B (1)
Equations of Motion.
  Fy  ma y
For block A: 981  2T  100 a A (2)
For block B: 196 .2  T  20 aB (3)
2a A   a B (1)
Then solving equations (1), (2) and (3)
simultaneously, we get
T  327 N
a A  3.27 m/s 2

aB  6.54 m/s 2
Hence when block A accelerates downward, block B
accelerates upward.
a A  3.27 m/s 2 aB  6.54 m/s 2

Since aB is constant, the velocity of block B in 2 s is

v  v0  aBt
 0  (6.54 )( 2)

 13.1 m/s
The negative sign indicates that block B is moving
upward.
Example 2.4
A smooth 2-kg collar
C, shown in Fig., is
attached to a spring
having a stiffness
k = 3 N/m and an
unstretched length of
0.75 m. If the collar is
released from rest at A,
determine its acceleration
and the normal force of
the rod on the collar at
the instant y =1 m.
W = (2)(9.81) =19.62 N.

  Fx  ma x
 NC  Fs cos  0 (1)

  Fy  ma y
19.62  Fs sin  2a (2)
Here the unstretched length is
lu  AB  0.75 m
where the final length is
y
l  CB  y  (0.75)
2 2 tan  
0.75
 lu  AB  0.75 m l  CB  y 2  (0.75) 2

s  l  lu  y  (0.75)  0.75
2 2

y
Fs  k s tan   (4)
0.75
As k = 3 N/m

Fs  ks  3( y  (0.75)  0.75)
2 2
(3)
Fs  ks  3( y  (0.75)  0.75)
2 2
(3)
y
tan   (4)
0.75
At y = 1 m equations (3) and (4) gives
Fs  1.5 N   53.1 o

using these results in equations (1) and (2), we get

 NC  Fs cos  0 (1) N C  0 .9 N

19.62  Fs sin  2a (2) a  9.21 m/s 2

Example 2.2
A 10-kg projectile is fired vertically upward from the
ground, with an initial velocity of 50 m/s.
Determine the maximum height to which it will travel
if (a) atmospheric resistance is
neglected; and (b) atmospheric
resistance is measured as
FD  (0.01v 2 ) N,
where v is the speed at
any instant, measured in m/s.
Part (a): Free-Body Diagram is drawn as shown
W = mg = 10(9.81) = 98.1 N.
   Fy  ma y  98.1  10 a
a  9.81 m/s 2
Initially, y0 = 0 and v0 = 50 m/s, and at the
maximum height y = h, v =0 . Since the
acceleration is constant, then
v 2
 v0
2
 2a( y  y0 )

(  ) 0  (50)  2(9.81)(h  0)
2

h  127m
Part (b): Free-Body Diagram is drawn as shown in

   Fy  ma y

 98.1  0.01v  10a

2

a  0.001v 2  9.81
In this case the acceleration is not constant.
dv dv
av (0.001v  9.81)  v
2
dy dy
 dv
(0.001v  9.81)  v
2
dy
Separating the variables and integrating, and using the
initial conditions, at t  0 y 0  0 and v  50 m/s
0
(positive upward), we have
y v vdv
0 dy  50 0.001v 2  9.81
v
y  0  500 ln(v  9810)
2
50
v  9810
2
y  500 ln( )
12310
at y = h, v = 0 h  114 m
 Equation of Motion:
Normal and Tangential Coordinates
F  m a
 Ft  mat
 Fn  man
 Fb  0
at  v

v2 [1  (dy / dx) ]
2 3/ 2
an  
 2
d y / dx 2
Example 2.6
Determine the banking angle  for the race track so
that the wheels of the racing cars shown in Fig. will
not have to depend upon friction to prevent any car
from sliding up or down the track. Assume the cars
have negligible size,
a mass m, and travel
around the curve of
radius  with a speed v.
Free-Body Diagram.

an  v / 
2

Equations of Motion. 2
v
   Fb  0 NC cos  mg  0 tan  
g
 v2 2
  Fn  man N C sin   m 1
  tan (
v
)
 g
Example 2.7
The 3-kg disk D is attached to the end of a cord as shown
in Fig. . The other end of the cord is attached to a ball-and
-socket joint located at the center of a platform. If the
platform is rotating rapidly, and the disk is placed on it and
released from rest as shown, determine the time it takes for
the disk to reach a speed great enough to break the cord.
The cord can sustain a maximum tension of 100 N, and the
coefficient of kinetic friction between the disk and the
platform is  k  0.1
Free-Body Diagram.
F  k N D  0.1N D

W = (3)(9.81) = 29.43 N.

an  v / 
2
 Equations of Motion.
 Fb  0 ND  29.43  0 (1)
 Ft  mat 0.1N D  3at (2)
2
 Fn  man v
T  3( ) (3)
1
Setting T = 100 N,

N D  29 .43 N at  0.981 m/s 2 v  5.77 m/s

Kinematics.
vcr  v0  at t vcr  5.77 m/s
5.77  0  (0.981)t t  5.89 s
Example 2.8
Design of the ski jump shown in Fig. requires
knowing the type of forces that will be exerted on the
skier and his approximate trajectory. If in this case the
jump can be approximated by the parabola shown in
Fig., determine the normal
force on the 150-Ib skier at
the instant he arrives at the
end of the jump, point A,
where his velocity
is 65 ft/s. Also,
what is his
acceleration at this point ?
 Free-Body Diagram.

Equations of Motion.
 

2
 Fn  man N A  150  
150

( 65 )  (1)
 32.2   
 150 
 Ft  mat 0  at a  0
 32 . 2  t
The radius of curvature  for the path can be
determined at point A, x  0 and y  200 ft
as follows:

y
1 2
x  200
dy
 0.01 x d2y
 0.01
200 dx dx 2
[1  (dy / dx) 2 ]3 / 2 [1  0]3 / 2
   100 ft
d 2 y / dx 2 0.01
x0

 150  (65) 

2
N A  150    
 32.2   

N A  347 Ib
Kinematics. The acceleration at point A is
v2 (65) 2
an    42 .2 ft/s 2
 100

a A  an  42.2 ft/s 
2
Example 2.9
Packages, each having a mass of 2 kg, are delivered
from a conveyor to a smooth circular ramp with a
velocity of v0  1 m/s
as shown in Fig. .
If the radius of the
ramp is 0.5 m,
determine the angle
   max
at which each
package begins to
leave the surface.
W  (2)(9.81) N  19 .62 N
Equations of Motion.
v2
 Fn  man  N B  19 .62 cos  (2)( )
0.5
(1)

 Ft  mat 19.62sin  2at (2)

At the instant    max

the package leaves the surface of the ramp so that
NB  0

Free-Body Diagram.
 Kinematics.
Using r  0.5 and ds  rd
dv v dv dv
at  v   2v (3)
ds r d d
By substituting Eq. (3) into Eq. (2) and separate the
variables we get
19.62sin  2at (2)
vdv  4.905 sin d
Integrate both sides, and using the initial condition;
v0  1 m/s at   0
 vdv  4.905 sin d v0  1 m/s at   0
v 
1 vdv  4.9050 sin d
2 v 
0.5v   4.905 cos 0
1

v  9.81(1  cos )  1
2 (4)
Substituting Eq. (4) into Eq. (1) with N B  0
and solving for cos max , yields
v2
 N B  19 .62 cos  (2)( ) (1)
0.5
19.62 cos max  4(9.81(1  cos max )  1)
19.62 cos max  4(9.81(1  cos max )  1)

43 .24
cos max 
58 .86

 max  42.7 o