forces in space.ppt
TRANSCRIPT
-
8/11/2019 Forces in space.ppt
1/58
FORCES IN SPACE(Noncoplanar System of Forces)
FS - 1
-
8/11/2019 Forces in space.ppt
2/58
Forces in space
A Force in space : A Force is said to be in space if its line ofaction makes an angle , and with respect to rectangularco-ordinate axes X, Y and Z respectively as shown the Fig. 1.
F Fig. 1. A Force in space
FS - 2
-
8/11/2019 Forces in space.ppt
3/58
Forces in space
Noncoplanar system of forces (Forces inSpace) and Their ClassificationsSystem of forces which do not lie in a single plane is called
noncoplanar system of forces(Forces in space ). A typicalnoncoplanar system of forces (forces in space) is shown inthe Fig. 2. below
Fig. 2 Forces in space (noncoplanar system of forces)
FS - 3
-
8/11/2019 Forces in space.ppt
4/58
Forces in space
Noncoplanar system of forces (Forces in space) can be broadlyclassified into three categories. They are
1. Concurrent noncoplanar system of forces2. Nonconcurrent noncoplanar system of forces3. Noncoplanar parallel system of forces
FS - 4
-
8/11/2019 Forces in space.ppt
5/58
Forces in space
1. Concurrent noncoplanar system of forces: Forces whichmeet at a point with their lines of action do not lie in a planeare called Concurrent noncoplanar system of forces. A
typical system of Concurrent noncoplanar system of forces isshown in the Fig.3.
Fig. 3. Concurrent noncoplanar system of forces
FS - 5
-
8/11/2019 Forces in space.ppt
6/58
Forces in space
2. Nonconcurrent noncoplanar system of forces: Forces whichdo not meet at a point and their lines of action do not lie in a plane, such forces are called Nonconcurrent noncoplanarsystem of forces. A typical system of nonconcurrentnoncoplanar system of forces is shown in the Fig.4.
Fig. 4. Nonconcurrent noncoplanar system of forces
FS - 6
-
8/11/2019 Forces in space.ppt
7/58
Forces in space
3. Noncoplanar parallel system of forces : If lines of action of allthe forces in a system are parallel and they do not lie in a planesuch a system is called Non-coplanar parallel system offorces. If all the forces are pointing in one direction then theyare called Like parallel forces otherwise they are called unlike
parallel forces as shown in the Fig.5.
Fig. 5 Noncoplanar parallel system of forces
FS - 7
-
8/11/2019 Forces in space.ppt
8/58
Forces in space
Fig. 6. Resolving a force in space into rectangular components
Rectangular components of a force in space
FS - 8
-
8/11/2019 Forces in space.ppt
9/58
Rectangular components of a force in spaceIn the Fig.6(a) a force F is acting at the origin O of the systemof rectangular coordinate axes X,Y,Z. Consider OBAC planepassing through the force F. This plane makes an angle withrespect to XOY plane. Force F makes an angle
y with
respect to Y-axis.
Forces in space
Fig.6(a)
FS - 9
-
8/11/2019 Forces in space.ppt
10/58
Rectangular components of a force in space
In the Fig.6(b), the force F is resolved in the vertical (Y- axis) andhorizontal direction (X axis) as
Fy = F Cos y and
Fh = F Sin y respectively.
Forces in space
Fig.6(b)
FS - 10
-
8/11/2019 Forces in space.ppt
11/58
Rectangular components of a force in space
In the Fig 6(c) the horizontal component Fh is again resolved in the X andZ axes directions. These components are
Fx = Fh cos = F sin y cos
Fz = F h Sin = F sin y sin
Forces in space
Fig.6(c)
FS - 11
-
8/11/2019 Forces in space.ppt
12/58
Now applying Pythagorean theorem to the triangles OAB
and OCDF2 = (OA) 2 = OB 2 + BA 2 = F y2 +F h2 ----------------(1)
Fh2 = OC 2 = OD 2 + DC 2 = F x2 +F z2 ----------------(2)
Substituting equation (2) into the equation (1), we getF2 = Fx2 +F y2 + F z2
F = F x2 + F y2 + F z2 ----------------(3)
Forces in space FS - 12
-
8/11/2019 Forces in space.ppt
13/58
Forces in spaceThe relationship existing between the force F and its three components Fx,
Fy, Fz is more easily visualized if a box having Fx, Fy, Fz for edges is drawnas shown below. The force F is then represented by the original OA of this
box.
Fig. 7 Relationship between Force F and its components Fx, Fy and Fz
FS - 13
-
8/11/2019 Forces in space.ppt
14/58
From the above Figure (Fig. 7)
Fx = F Cos x, F y = F Cos y, F z = F Cos z ------------(4)
Where x, y, z are the angles formed by the force F with X, Y, Z axes respectively Fx,Fy,Fz are the rectangular components of the force F in the directions of X, Y, Z
axes respectively.
Cos x = F x/F; Cos y = F y/F; Cos z = F z/F
Substituting equation (4) into the equation (3), we get
F = F x2 + F y2 + F z2
F = F 2Cos 2 x + F 2Cos 2 y + F 2Cos 2 z F2 = F 2 ( Cos 2 x + Cos 2 y + Cos 2 z )
1 = Cos 2 x + Cos 2 y + Cos 2 z -------------(5)
Forces in space FS - 14
-
8/11/2019 Forces in space.ppt
15/58
From the adjacent Fig. 8.
dx= d Cos x, dy = d Cos y, dz = d Cos z
----(6)d = d x2 + d y2 + d z2
---(7)Dividing member by
member the relations (4) and(6), we obtain
Fx /dx = F y/dy = F z/dz = F/d----------------------------(8)
Forces in space
Force Defined by its magnitude and two points on its line of action
Fig 8
FS - 15
-
8/11/2019 Forces in space.ppt
16/58
Forces in space
Resultant of concurrent forces in Space :- Resolve all the forces into their rectangular components in X, Y and Z axes directions.Adding algebraically all the horizontal components in the x direction gives
R x = F x,Similarly adding algebraically all the components in y and z directions yield the
following relationsR y = F y, R z = F z
Thus magnitude of resultant
R = R x2 + R y2 + R z2
Angles x, y, z resultant forms with the axes of coordinates are obtained by
R
RCos
R
RCos
R
RCos z
z
y
y
x
x
;;
FS - 16
-
8/11/2019 Forces in space.ppt
17/58
P r o b l e m s :
(1) A tower guy wire is anchored by means of a bolt at A is shown in the followingFigure. The tension in the wire is 6000N. Determine(a) The components F x, Fy, F z of the forces acting on the bolt.(b) The angles x, y, z defining the direction of the force.
Forces in space FS - 17
-
8/11/2019 Forces in space.ppt
18/58
Solution: (a) Here d x = 50m, d y = 200m, d z = -100m
Total distance A to B
d = d x2 + d y2 + d z2
= (50) 2 + (200) 2 + (-100) 2 = 229.13 m
Using the equation (8) F x /dx = F y/dy = F z/dz = F/d
Fx = d x . (F/d) = (50 x 6000)/ 229.13 = 1309.3 N
Fy = d y . (F/d) = (200 x 6000)/ 229.13 = 5237.20 N
Fz = d z . (F/d) = (-100 x 6000)/ 229.13 = -2618.6 N
Forces in space FS - 18
-
8/11/2019 Forces in space.ppt
19/58
(b) Directions of the force:
Cos x = d x /d , x = Cos 1 (50/229.13) = 77.4
y = Cos 1 (d y /d) = Cos 1 (200/229.13) = 29.2
z = Cos 1 (d z /d) = Cos 1 (-100/229.13) = 115.88
Forces in space FS - 19
-
8/11/2019 Forces in space.ppt
20/58
Problem(2) Determine (a) the x , y and z components of the 250 Nforce acting as shown below
(b) the angles x, y, z that the force forms with the coordinateaxes.
Forces in space FS - 20
-
8/11/2019 Forces in space.ppt
21/58
Fh = 250 Cos60 = 125 N F y = 250 Sin60 = 216.5 NFx = 125 Cos25 = 113.29 N F z = 125Sin25 = -52.83 N
x = Cos 1 ( F x /F) = Cos 1 (113.29/250) = 63 y = Cos 1 ( F y /F) = Cos 1 (216.5/250) = 30 z = Cos
1 ( F z /F) = Cos
1 (-52.83/250) = 102.11
o
Components of 250 N in the x, y, z axes directions are
Fx
= 113.29N Fy = 216.5 N F
z = -52.83N
Forces in space FS - 21
-
8/11/2019 Forces in space.ppt
22/58
Problem 3. Find the resultant of the system of forces as shown below
Forces in space FS - 22
-
8/11/2019 Forces in space.ppt
23/58
-
8/11/2019 Forces in space.ppt
24/58
Components of Force F 2 = 2000 N:
Fy2 = 2000 x Cos20 = 1879.39 NFh2 = 2000 x Sin20 = 684.04 NFx2 = 684.04 x Sin35 = 392.35 NFz
2 = 684.04 x Cos35 = 560.33N
Forces in space FS - 24
-
8/11/2019 Forces in space.ppt
25/58
R x = Fx = Fx 1 + Fx 2 = 1670 392.35 = 1277.65R y = Fy = Fy 1 + Fy 2 = 2298.13 + 1879.39 = 4177.52 NR z = Fz = Fz 1 + Fz 2 = 964.18 + 560.33 = 1524.51 N
Resultant R = R x2 + R y2 +R z2 = 1277.65 2 +4177.52 2 +1524.51 2 = 4626.9 N
Its inclinations with respect to x, y and z axes arecalculated as
x = Cos 1 (1277.65 /4626.9) = 73 58' 13.1"y = Cos 1 (4177.52/4626.9) = 25 27'
z = Cos 1
(1526.51/4626.9) = 70o
4536
Forces in space FS - 25
-
8/11/2019 Forces in space.ppt
26/58
Problem 4. In the Fig shown below, the forces in the cables AB and ACare 100 kN and 150 kN respectively. At the joint A loading is asshown in the Fig. Find the resultant of system of forces in space and itsinclination with rectangular coordinates x,y and z axes.
Forces in space FS - 26
-
8/11/2019 Forces in space.ppt
27/58
Solution: Force in the cable AB = 100 kN
Force in the cable AC=150 kN
For the cable ABdx = -20 mdy = 15mdz = 5mdAB = dx2 +dy 2 + dz 2
= (-20) 2 +(15) 2 + (5) 2 = 650 = 25.5 m
For the cable ACdx = -20 mdy = 25mdz = -10mdAc = (-20) 2 +(25) 2 + (-10) 2 = 1125 = 33.54 m
FS - 27Forces in space
-
8/11/2019 Forces in space.ppt
28/58
For the cable AB (1)
Fx 1/dx 1 = Fy 1/dy 1 = Fz 1/dz 1 = F/dFx 1/-20 = Fy 1/15 = Fz 1/5 = 100/25.5
Fx 1 = -78.41 kN
Fy 1 = 58.82 kNFz 1 = 19.61 kN
For the Cable AC (2)
Fx2/-20 = Fy 2/25 = Fz 2/-10 = 150/33.54
Fx 2 = - 89.45 kNFy 2 = 111.81 kNFz2 = - 44.72 kN
Forces in space FS - 28
-
8/11/2019 Forces in space.ppt
29/58
Component of force 60 kN
Fx3 = 60 x Cos(70 ) = 20.52 kN Fy3 = 60 x Cos(30
o) = 51.96 kNFz3 = 60 x Cos(111 23)= - 21.88 kN
Component of the force 50KN
Fx4 = 50 kN
Fy4 = 0Fz4 = 0
Forces in space FS - 29
-
8/11/2019 Forces in space.ppt
30/58
Algebraic summation of Rectangular Components in X, Y and Zaxes directions yield:Rx = Fx = Fx 1 + Fx 2 + Fx 3 + Fx 4
= -78.43 - 89.45 + 20.52 + 50= -97.36 kN
Ry = Fy = Fy 1 + Fy 2 + Fy 3 + Fy 4 = 58.82 + 111.81 + 51.96 + 0= 222.59 kN
Rz = Fz = Fz 1 + Fz 2 + Fz 3 + Fz 4
= 19.61 44.72 21.88 + 0= -46.99 kN
Resultant R = Rx 2 + Ry 2 + Rz 2 = (-97.36) 2 + (222.59) 2 + (-46.99) 2
= 247.45 kN
Forces in space FS - 30
-
8/11/2019 Forces in space.ppt
31/58
Inclinations of the resultant with X,Y and Z axes:
x = Cos -1(R
x / R) = Cos -1 (-97.36/247.45) = 113 10
y = Cos -1(R y / R) = Cos -1 (222.59/247.45) = 25 54 z = Cos -1(R z / R) = Cos -1 (-46.99/247.45) = 100 56 47
Check:
Cos 2 x + Cos 2 y + Cos 2 z = 1Cos 2(113 10 ) + Cos 2(25 54 ) + Cos 2(100 56 ) = 1
1 = 1 Hence OK
Forces in space FS - 31
-
8/11/2019 Forces in space.ppt
32/58
Problem 5. a) Forces F 1, F 2, and F 3 pass through the origin and points whose coordinates are given. Determine theresultant of the system of forces.
F1 = 20 kN, (3,-2,1)
F2 = 35 kN, (-2,4,0)F3= 25 kN, (1,2,-3)
Forces in space FS - 32
-
8/11/2019 Forces in space.ppt
33/58
Forces in space
Solution :
FS - 33
-
8/11/2019 Forces in space.ppt
34/58
Force F 1 = 20 kNd = 3 2 + (-2) 2 + 12 = 14 = 3.74Cos x1 = d x / d = 3/3.74 = 0.802Cos y1 = -2/3.74 = -0.535; Cos z1 = 1/3.74 = 0.267
Force F2 = 35 kN
d = (-2) 2 + 4 2 +0 = 20 = 4.47Cos x2 = -2/4.47 = -0.45 Cos y2 = 4/4.47 = 0.9; Cos z2 = 0
Force F 3 = 25 kNd = (1) 2 + 2 2 +(-3) 2 = 14 = 3.74Cos x3 = 1/3.74 = 0.267
Cos y3 = 2/3.74 = 0.535; Cos z3 = -3/3.74 = -0.802
Forces in space FS - 34
-
8/11/2019 Forces in space.ppt
35/58
Summation of the rectangular components in X, Y and Z axes directions
yield: R x = Fx = 20 x 0.802 + 35 x (-0.45) +25 x 0.267 = 6.965 kNR y = Fy = 20 x (-0.535) + 35 x 0.9 + 25 x 0.535 = 34.175 kNR z = Fz = 20 x 0.267 + 35 x 0 + 25 x (-0.802) = -14.71 kN
Resultant R = Rx 2 + Ry 2 + Rz 2 = 6.965 2 + 34.175 2 + (-14.71) 2 = 37.85 kN
Inclination of the resultant R with respect to X, Y and Z axes arecalculated as
x = Cos 1(R x / R) = Cos -1(6.965/37.85 ) = 79.4 y = Cos 1(R y /R) = Cos -1(34.175/37.85) = 25.46
z = Cos 1(R z /R) = Cos -1(-14.71/37.85) = 112.87
Forces in space FS - 35
-
8/11/2019 Forces in space.ppt
36/58
Forces in space
Equilibrium of Concurrent non-coplanar system of forces:
When a rigid body subjected to concurrent noncoplanar system of forcesF1, F 2. ..F N as shown in the Fig. given below, is in equilibrium, thenalgebraic summation of all the components of the forces in three mutually
perpendicular directions must be equal to zero.
Fig. A rigid body subjected to concurrent
nonco lanar s stem of forces
i.e. Fx = 0 Fy = 0
Fz = 0 (1)
Above equations represent the staticconditions of equilibrium for concurrentnoncoplanar system of forces
FS - 36
Forces in
-
8/11/2019 Forces in space.ppt
37/58
Problem (1) Find the forces in the rods AB , AC and AO subjected to
loading as shown below
Forces inspace
FS - 37
F i
-
8/11/2019 Forces in space.ppt
38/58
Solution:
For the cable AB:
dx1 = 0 - 4 = - 4dy1= 8 - 0 = 8dz1 = 15 0 =15
d1 = (-4) 2 +(8) 2 +(15) 2 = 17.46 m
Fx1/-4 = F y1/8 = F z1/15 = F AB /17.46
Fx1 = -0.23F AB ,Fy1 = 0.46F AB ,F
z1 = 0.86F
AB
Forces in space FS - 38
F i
-
8/11/2019 Forces in space.ppt
39/58
For the Cable AC: dx
2 = 0 - 4 = -4m
dy2 = 8 - 0 = 8mdz2 = -20 0 = -20m
d2 = (-4) 2 +(8) 2 +(-20) 2 = 480 = 21.91mFx2/-4 = F y2/8 = F z2/-20 = F AC/21.91Fx2 = -0.18F AC , F y2 = 0.365F Ac, F z2 = -0.91F Ac
For the Force 120 N:
Fx3 = 120 N, F y3 = 0, F z3 = 0
For the force Fx 4 = 300 N: Fx 4 = 0, Fy 4 = -300N, Fz 4 = 0
Forces in space FS - 39
-
8/11/2019 Forces in space.ppt
40/58
Conditions of Equilibrium Fx = 0, Fy = 0, Fz = 0
Considering Fx = 0Fx 1 + Fx 2 + Fx 3 + Fx 4 + Fx 5= 0-FAD-0.23F 0.18F AC + 120 + 0 = 0
FAD + 0.23F AB + 0.183F AC = 120 ----------------------(1)
Fy = 00.46F AB + 0.365F AC + 0 +0 300 = 00.46F AB + 0.365F AC = 300 ------------------------------(2)
Fz = 00.86 F AB 0.91 F AC + 0 = 0 -----------------------(3)
Forces in space FS - 40
-
8/11/2019 Forces in space.ppt
41/58
Solving Equations (1), (2) and (3), we get,F AB = 372.675 N; F AC = 352.25 N ; F AO = - 30.18 N
Force in the rod AB , F AB = 372.675 N (Tensile)Force in the rod AC, F AC = 352.25 N (Tensile)
Force in the rod AO, F AO = 30.18 N (Compressive)
Forces in space FS - 41
-
8/11/2019 Forces in space.ppt
42/58
2) Three cables are connected at D and support, the 400kN load as shown in the Fig given below. Determine thetensions in each cable
Forces in space FS - 42
-
8/11/2019 Forces in space.ppt
43/58
-
8/11/2019 Forces in space.ppt
44/58
For Cable DB:
dx2= (0 3) = -3mdy2 = (6 - 4) =2 mdz2 = (-6 0 = -6 mdBD = d2 = (-3) 2 + (2) 2 + (-6) 2 = 49 = 7 m
Fx 2/dx 2 = Fy 2/dy 2 = Fz 2/dz 2 = F DB /dDB
Fx 2/(-3) = Fy 2/2 = Fz 2/-6 = F DB /7
Fx 2 = -0.43F DB ,Fy 2 =0.286F DB ,Fz2 = -0.857F DB
Forces in space FS - 44
-
8/11/2019 Forces in space.ppt
45/58
For Cable DC: dx
3 (0 3) = -3m
dy3= (0- 4) = -4mdz3= (0 0 = 0m
dDC= d 3= (-3) 2 + (-4) + (0) 2
= 9 + 16 = 5m
Fx 3/dx 3 = Fy 3/dy 3 = Fz 3/dz 3 = F DC /d DC
Fx 3/(-3) = Fy 3/-4 = Fz 3/0 = F DC /5
Fx 3 = -0.6F DC ,Fy 3 =-0.8F DC ,Fz3 = 0
Forces in space FS - 45
i
-
8/11/2019 Forces in space.ppt
46/58
For the force 400KN
dx4 = 12mdy4= -4mdz4= 3md4= d 400 = (12) 2 + (4) + (3) 2
= 144 +16 + 9 = 69 = 13m
Fx 4/dx 4 = Fy 4/dy 4 = Fz 4/dz 4 = F 400 /d4
Fx4/(12) = Fy
4/-4 = Fz
4/3 = 400/13
Fx 4 = 369.23 kN,Fy 4 = -123.08 kN,
Fz4 = +92.31 kN
Forces in space FS - 46
F i
-
8/11/2019 Forces in space.ppt
47/58
For Equilibrium, the algebraic summation of resolved
component in a particular direction is equal to zero.i.e. Fx = 0 -------------(1)
Fy = 0 ------------- (2)
Fz = 0 --------------(3)
(1) Fx = Fx1 + Fx2 + Fx3 + Fx4 = 0
+ 0.43F DA + 0.43F DB + 0.6F DC = 369.23 ---------(1)+ 0.286F DA + 0.286F DB 0.8F DC = 123.08 ------(2)
+ 0.857F DA - 0.857F DB + 0 = 92.31 ------------(3)
Forces in space FS - 47
-
8/11/2019 Forces in space.ppt
48/58
Solving equations (1), (2) and (3), we getFDB = 304.1kN
FDA = 411.8 kN
FDC = 102 kN
Forces in space FS - 48
FS 49
-
8/11/2019 Forces in space.ppt
49/58
Forces in space
Practice Questions 1. A tower guy wire is anchored by means of a bolt at A as shown below.
The tension in the wire is 2500 N. Determine (a) the components F x, Fy and F z of the force acting on the bolt, (b) the angles x, y, z definingthe direction of the force
(Ans: F x= -1060 N , F y = 2120 N, F z= + 795 N x = 115.1 o ; y= 32.0 o ; z= 71.5 o )
FS - 49
FS - 50
-
8/11/2019 Forces in space.ppt
50/58
Forces in space Practice Questions 2. Determine (a) x, y and z the components of the force 500 N in the below
Figure. (b) the angles x, y, z that the force forms with the coordinateaxes
(Ans: F x= + 278 N , F y = + 383 N, F z= + 160.7N
x = 56.2 o ; y= 40.0 o ; z= 71.3 o )
FS - 51
-
8/11/2019 Forces in space.ppt
51/58
Forces in space Practice Questions 3. In order to move a wrecked truck, two cables are attached at A and
pulled by winches B and C as shown. Knowing that the tension in thecable AB is 10 kN, determine the components of the force exerted by thecable AB on the truck
(Ans: F x= -6.30 kN , F y = 6.06 kN, F z= + 4.85kN)
FS - 52
-
8/11/2019 Forces in space.ppt
52/58
Forces in space Practice Questions 4. A 200 kg cylinder is hung by means of two cables AB and AC, which are
attached to the top of a vertical wall. A horizontal force P perpendicularto the wall holds the cylinder in the position shown. Determine themagnitude of the force P and the tension in each cable
(Ans: P = 235 N , T AB = 1401 N, T AC = + 1236 N)
FS - 53
-
8/11/2019 Forces in space.ppt
53/58
Forces in space Practice Questions 5. Three cables are connected at A, where the forces P and Q are applied as
shown. Determine the tension in each cable when P = 0 and Q = 7.28kN
(Ans: T AB= 2.88 kN , T AC = 5.76 kN, T AD= 3.6kN)
FS - 53
FS - 54
-
8/11/2019 Forces in space.ppt
54/58
Forces in space6. A container of weight w = 400 N is supported by cables AB and AC
which are tied to ring A. Knowing that Q = 0, determine (a) themagnitude of the force P which must be applied to the ring to maintainthe container in the position shown in figure below, (b) thecorresponding the values of the tension in cables AB and AC
(Ans: P = 138 N ,T AB = 270N,T AC = 196N )
Practice Question
FS - 55
-
8/11/2019 Forces in space.ppt
55/58
Forces in space7. A container supported by three cables as shown below. Determine the
weight of the container, knowing that the tension in the cable AB is 4 kN
Ans: 9.32 kN
Practice Question
FS - 56
-
8/11/2019 Forces in space.ppt
56/58
Forces in space8. Determine the resultant of the two forces shown below.
Practice Question
(Ans: R = 498 N , x = 68.9 o ; y= 26.3 o ; z= 75.1 o )
FS - 57
-
8/11/2019 Forces in space.ppt
57/58
Forces in space9. A container of weight W = 1165 N is supported by three cables as shown
below. Determine the tension in each cable.
Ans: T AB = 500 NT AC = 459 NT AD = 516 N
Practice Question
-
8/11/2019 Forces in space.ppt
58/58