Download - Ch (2) Mech. I- Space Concurrent W 2015
1
Chapter 2
: (Concurrent Forces)Space Statics
Resultant and Equilibrium of a Particle
Prof. Imam Morgan
Head of MCTR Department
(2)
TBH TBG
FA
FD
P, TBG , TBH …. three Nonconcurrent
Forces (known)
P, Q1 ,Q2 …. three concurrent Known
Forces
F
Forces in Space (in 3D)
P, TBG , TBH and FA , FD five Nonconcurrent
[Resultant ? ??]
[Find F and T for Equilibrium]
or: Add F =? To get [Equilibrium] for
the four forces
[Resultant ]
Prof. Imam Morgan
Head of MCTR Department
(3)
Forces in Space ? …The forces are called in space as long as they
are not contained in one plane.
►The tensions in the
cables supporting the
platform are in space.
The platform is
considered as a particle.
►The tensions in the
cables supporting the
container are also
space forces. The
tensions intersect at
one point.
We study:
■ Resultant of
concurrent forces ■ Equilibrium of a particle.
Concurrent Forces
Lines of action of
forces intersect at
a point
Statics in Space [3D]? 1
Prof. Imam Morgan
Head of MCTR Department
(4)
►The plane forces exerted by the
four tugboats on the sub-marine
can be replaced by a single
equivalent force exerted by one
tugboat. (Resultant)
►Three or more space forces direct
the shipping boat along certain
direction. They could be replaced by
two boats for the same action.
(Resultant)
Non-concurrent
Prof. Imam Morgan
Head of MCTR Department
(5)
►Three space forces act
on a bracket. It is
required to replace them
by an equivalent effect
at O.
►The sign plate is kept in
Equilibrium in the shown
position by means of two
cables and a support at A
Non-concurrent We need to study:
■ Equivalent Systems
■ Equilibrium of
Rigid Body.
Prof. Imam Morgan
Head of MCTR Department
(6)
►► Resolve into horizontal and
vertical components:
F
Finally, has been resolved into three
rectangular components Fx , Fy , and Fz F
2 Force ( ) in Vector Form F
► Consider a force acting at the origin O of the orthogonal rectangular frame. The
angle defines the plane containing
while θy defines the position of in that plane.
F
F F
►►►Then, resolve Fh into rectangular components:
Fx = Fh cos = (F sin θy) cos
and Fz = (F sin θy))sin
Prof. Imam Morgan
Head of MCTR Department
(7)
θ θx
θz
kFjFiFF zyx
Therefore: Force in vector form
F
F
Where: are called Rectangular Components zyx F,F,F
θx , θy , θz are called the direction angles of
cos θx , cos θy , cos θz are the direction cosines
k,j,i are the unit vectors along x, y, and z, respectively
of
Prof. Imam Morgan
Head of MCTR Department
(8)
► In general, the space can be divided
into eight regions as shown in the figure.
The regions are generated using three
intersecting planes which are mutually
orthogonal. Their point of intersection is
the origin O of the shown chosen frame.
The location of any force, given in vector
form, can be easily specified in space.
►Three forces are shown in
figure. Although they have the same
magnitude (see later), however they are
completely different in their effects due to
the different signs of the components.
321 ,, FFF
kjiF ˆ3ˆ4ˆ61
kjiF ˆ3ˆ4ˆ62
kjiF ˆ3ˆ4ˆ63
4N
6N
4N
3N
Prof. Imam Morgan
Head of MCTR Department
(9)
θ θx
θz
Given: The force in vector form
kFjFiFF zyx
Required: ● its magnitude,
● its direction
►Magnitude: 222
zyx FFFF
►Direction:
F
F
F
F
F
F
zz
y
y
xx
cos
cos
cos
NF
kjiF
70206030
206030
222
For example: a force is given by:
◄
and the direction angles are:
7020
7060
7030
z
y
x
cos
cos
cos
θx = 64.6o
θy= 31o
θz = 73.4o
A Magnitude and Direction of ( ) F
Basic Relations
x
y
z
A
B
C
D
E
O Fx
Fy
Fz
F
x
y
z
A
B
C
D
E
O Fx
Fy
Fz
F
x
x
y
z
A
B
C
D
E
O Fx
Fy
Fz
F
y
z
The Three direction
angles can be
obtained as:
F
F
F
F
F
F
zz
yy
xx
cos
cos
cos
More Details
Prof. Imam Morgan
Head of MCTR Department
(10)
x
Prof. Imam Morgan
Head of MCTR Department
(11)
θθx
θz
θθx
θz
Unit Vector of a Force ( ) F
Given : The force in vector form
kFjFiFF zyx
Required : ● a unit vector ( ) directed
along its line of action.
F
In general the unit vector is obtained by
dividing the vector by its magnitude.
F
FF
For example, find the unit vector for the
vector given in the previous example
kji
kjiF
7
2
7
6
7
3
70
206030
Note that the components of the
unit vector are the direction cosines of the same vector
unit vector F
B
Prof. Imam Morgan
Head of MCTR Department
(12)
θ θx
θz
Given : the magnitude F, of a force, as
well as its direction (θx , θy , θz)
Required : F in vector form.
We have:
zz
yy
xx
cosFF
cosFF
cosFF
Direction Angles are dependent
The direction angles of the force vector
are dependent. They are related by:
1222 zyx coscoscos
Therefore, only two angles are
sufficient for determination of
the direction of any force in
space (fact).
D
Determination of ( ) on the bases
of its magnitude and direction
F
C
Prof. Imam Morgan
Head of MCTR Department
(13)
Example (1)
The magnitude of the shown force is
300 N. Write down this force in vector form
and hence find its rectangular components.
Solution:
From the figure θx= 60o and θy= 45o
1222 zyx coscoscos
50
2502
.cos
.cos
z
z
θz = 60o or 120o Consider θz= 120o
NcosFF
N.cosFF
NcosFF
zz
yy
xx
150
13212
150
kj.iF 15013212150
z
y
x
F
θz
60o
45o
refused
First Method
F1x F1h
F1y
F1z
Prof. Imam Morgan
Head of MCTR Department
(14)
Example (2)
x
y
z
Express each force in vector form and
hence determine its direction with
respect to the coordinate frame.
k.j.i.F 363588636351
1F
Solution:
► Force
Second Method
The angles 60 and 45 defining the direction of
are not direction angles. Therefore, we must
use the geometry to resolve this force.
As shown: F1 is resolved into F1y and F1h. Then,
F1h is resolved into F1x and F1z:
F1y= 100 sin 60 = +86.6 lb
and F1h= 100 cos 60= 50 lb
F1x = 50 sin45 = 35.36 N
F1z = + 50 cos45= + 35.36 N
1F
Prof. Imam Morgan
Head of MCTR Department
(15)
x
y
z
ozz
oyy
oxxDirection
3.69100
36.35cos
30100
8.86cos
111100
36.35cos
x
y
z
ozz
oyy
oxx
z
x
hy
direction
kjiF
IbF
IbF
IbFandIbF
FForce
3.69300
07.106cos
135300
13.212cos
2.52300
71.183cos:
ˆ07.106ˆ13.212ˆ71.183
07.10630sin13.212
71.18330cos13.212
13.21245cos30013.21245sin300
:
2
2
2
22
2
Prof. Imam Morgan
Head of MCTR Department
(16)
In many applications, the forces are applied along certain defined directions.
● For example the tension in the chain AB is directed through the shown
direction where the two points A and B are well defined.
● Also, the force in the cable AB is directed as shown.
Prof. Imam Morgan
Head of MCTR Department
(17)
Position Vector
►The position vector ( ) is the vector that specifies the position
of point B with respect to point A. In other words, the position of
B as seen from point A. It is drawn from A to B.
ABr /
kzzjyyixxr
rrr
kzjyixrrandkzjyixrr
ABABABAB
ABAB
BBBOBBAAAOAA
ˆ)(ˆ)(ˆ)(
ˆˆˆˆˆˆ
/
/
//
Note that if a force is applied along AB and directed from A to B,
then, both will have the same unit vector. F
ABrandF /
z
y
x
ixx ABˆ
jyy ABˆ
kzz ABˆ
ABr /
O
z
y
x
ABr /
O
Prof. Imam Morgan
Head of MCTR Department
(18)
F
Very Important Force Defined by its Magnitude and
Two Points on its Line of Action
In this case, the force is known in
magnitude (F) and its line of action
passes through two given points, for
example M and N .
■ Find the unit vector from: F
212
2
12
2
12
12212
zzyyxx
kzzjyyixxF
■ Then, find the force in vector form as:
FFF
Third Method
The two steps
are carried out
in one step.
Prof. Imam Morgan
Head of MCTR Department
(19)
Example (3)
A tower guy wire is anchored by means of a
bolt at A. The tension in the wire is 2500 N.
Determine: (a) the rectangular components
the force acting on the bolt,
(b) the angles defining the
direction of the force.
Solution:
(a) Determination of F
kji
kjiF
79521201060
308040
3000804002500
222
Rectangular components
F
(40, 0, -30)
(0, 80, 0)
A = (40,0,-30)
B = (0,80,0)
Prof. Imam Morgan
Head of MCTR Department
(20)
(b) Direction angles at A:
o
zz
o
yy
o
xx
.cos
cos
.cos
5712500
795
322500
2120
11152500
1060
F
NF
kjiF
2500
ˆ795ˆ2120ˆ1060
F
Prof. Imam Morgan
Head of MCTR Department
(21)
Example (4)
The tension in the shown cable is 700 N. Find the
cartesian components of the tension at A.
Determine its direction.
z
y
x
(2,3,-2)
(0,-3,1)
ozz
oyy
oxxDirection
NTCheck
kji
kjiT
4.115700
300cos
0.31700
600cos
4.73700
200cos:
700300600200
ˆ300ˆ600ˆ200
7
ˆ3ˆ6ˆ2700
222
θz=115.4o
θy=31.0o
θx=73.4o
z’
y’
x’
Prof. Imam Morgan
Head of MCTR Department
(22)
►When the particle is acted upon by several
forces, then, these forces can be replaced by
one force called the equivalent or the “Resultant”.
R
As shown, this resultant must pass through the point of intersection (Cocurrent Forces)
kFjFiFF
kFjFiFF
kFjFiFF
nznynxn
zyx
zyx
2222
1111
kRjRiRR zyx
zz
yy
xx
FR
FR
FR
Resultant of Concurrent Forces 3
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
Prof. Imam Morgan
Head of MCTR Department
(23)
Example (5 )
A wall section of precast concrete is
temporarily held by the cables shown.
Knowing that the tension in cables AB and
AC are 840 lb and 1200 lb, respectively,
determine the magnitude and direction of the
resultant of the forces at A.
Solution
Choose the coordinate frame as shown and
then write each force in vector form. 2T
(16,0,-11)
(0,8,0)
(0,8,-27)
kji
kji
kjiT
ˆ440ˆ320ˆ640
21
ˆ11ˆ8ˆ16840
11816
ˆ11ˆ8ˆ16840
2221
1T
1T
Prof. Imam Morgan
Head of MCTR Department
(24)
kji
kji
kjiT
ˆ800ˆ400ˆ800
24
ˆ11ˆ8ˆ161200
16816
ˆ16ˆ8ˆ161200
2222
2T
So, the resultant which must pass through A is:
kji
TTR
ˆ360ˆ720ˆ1440
21
● R = 1650 lb
● θx = 150.8o ● θy = 64.1o ● θz = 102.6o Draw a sketch to show the direction of
the resultant and try to find its point of
intersection, D, with the wall (Option)
Answer: D ≡ (0, 8, - 15) ft
2T
(16,0,-11)
(0,8,-27)
and we have kjiT ˆ440ˆ320ˆ6401
1T
Prof. Imam Morgan
Head of MCTR Department
(25)
Consider a particle A under the
action of several space forces: F1 , F2 , …… , Fn.
■ Find the resultant R, then:
►the particle will move along
a space curve.
►the particle will move along a
curvilinear pass contained in xy plane
►the particle will perform a
rectilinear motion along x direction. ■ However, if
0R
The particle
is in case of
equilibrium 0
0
0
zz
yy
xx
FR
FR
FR
Conditions of
Equilibrium
Fn F3
F2 F1
A
kRjRiRRisRIf zyxˆˆˆ
0ˆˆ jRiRRIf yx
00ˆ iRRIf x
Equilibrium of Concurrent Forces 4
Prof. Imam Morgan
Head of MCTR Department
(26)
Example (6)
A 200 kg cylinder is hung by means of
two cables AB and AC. A horizontal force
P holds the cylinder in the shown
position. Determine the magnitude of P
and the tension in each cable.
2
1
(1.2,2,0)
(0,12,8)
(0,12,-10)
Solution:
Write down each force in vector form:
kTjTiT
kjiTT
kTjTiT
kjiTT
jmgW
iPP
ˆ)705.0(ˆ)705.0(ˆ)085.0(
13.14
ˆ10ˆ10ˆ2.1
ˆ)622.0(ˆ)778.0(ˆ)093.0(
86.12
ˆ8ˆ10ˆ2.1
0ˆ0
00ˆ
222
22
111
11
Prof. Imam Morgan
Head of MCTR Department
(27)
Then, apply the conditions of equilibrium:
0085.0093.00 21 TTPFx
0705.0778.019620 21 TTFy
0705.0622.00 21 TTFz
kTjTiT
kjiTT
kTjTiT
kjiTT
jmgW
iPP
ˆ)705.0(ˆ)705.0(ˆ)085.0(
13.14
ˆ10ˆ10ˆ2.1
ˆ)622.0(ˆ)778.0(ˆ)093.0(
86.12
ˆ8ˆ10ˆ2.1
0ˆ0
00ˆ
222
22
111
11
2
1
Prof. Imam Morgan
Head of MCTR Department
(28)
0085.0093.00 21 TTPFx
0705.0778.019620 21 TTFy
0705.0622.00 21 TTFz
T1 = 1402 N
T2 = 1238 N
P = 235 N
Think over!!!
● Determine the location of point C such that the
two tensions will have the same value and find
the tension in this case (distance 1.2 is kept constant).
● If the distance 1.2 is required to be doubled,
find P, T1, and T2 in this case (same cables lengths).
Comment on theobtained results.
● If the two points B and C coincide at one point,
determine the location of this point such that
the particle keeps its equilibrium in the shown
position (consider variable lengths of cables).
Prof. Imam Morgan
Head of MCTR Department
(29)
Example (7)
The 100 kg cylinder is suspended from
the ceiling by cables attached at points
B, C, and D. What are the tensions in
the cables AB, AC, and AD?
Prof. Imam Morgan
Head of MCTR Department
(30)
Solution:
TAD
TAC TAB
W
NT
NT
NT
getweequationstheseSolving
TTT
TTT
TTT
mEquilibriuofConditionsFrom
jjW
kjiTkji
TT
kjiTkji
TT
kjoiTkji
TT
AB
AB
AB
ADACAB
ADACAB
ADACAB
ADABAD
ACACAC
ABABAB
7.168
1.636
1.519
:,
0514.0408.0333.0
981868.0816.0667.0
0514.0408.0667.0
0ˆ9810ˆ81.9100
ˆ514.0ˆ686.0ˆ514.034
34ˆ3
ˆ408.0ˆ816.0ˆ408.024
2ˆ4ˆ2
ˆ333.0ˆ667.ˆ667.036
2ˆ4ˆ4