1. FORCES IN SPACE (Noncoplanar System of Forces)
FS - 1
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2. FS - 2
Forces in space
A Force in space: A Force is said to be in space if its line of action makes an angle α, β and γ with respect to rectangular co-ordinate axes X, Y and Z respectively as shown the Fig. 1.
F Fig. 1. A Force in space
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3. Fig. 2 Forces in space (noncoplanar system of forces)
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Noncoplanar system of forces (Forces in Space) and Their Classifications
System of forces which do not lie in a single plane is called noncoplanar system of forces(Forces in space ). A typical noncoplanar system of forces (forces in space) is shown in the Fig. 2. below
Fig. 2 Forces in space (noncoplanar system of forces)
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4. FS - 4
Forces in space
Noncoplanar system of forces(Forces in space) can be broadly classified into three categories. They are
1. Concurrent noncoplanar system of forces
2. Nonconcurrent noncoplanar system of forces
3. Noncoplanar parallel system of forces
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5. Fig. 3. Concurrent noncoplanar system of forces
FS - 5
Forces in space
1. Concurrent noncoplanar system of forces: Forces which meet at a point with their lines of action do not lie in a plane are called “Concurrent noncoplanar system of forces”. A typical system of Concurrent noncoplanar system of forces is shown in the Fig.3.
Fig. 3. Concurrent noncoplanar system of forces
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6. Fig. 4. Nonconcurrent noncoplanar system of forces
FS - 6
Forces in space
2. Nonconcurrent noncoplanar system of forces: Forces which do not meet at a point and their lines of action do not lie in a plane, such forces are called “Nonconcurrent noncoplanar system of forces”. A typical system of nonconcurrent noncoplanar system of forces is shown in the Fig.4.
Fig. 4. Nonconcurrent noncoplanar system of forces
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7. FS - 7
Forces in space
3. Noncoplanar parallel system of forces: If lines of action of all the forces in a system are parallel and they do not lie in a plane such a system is called Non-coplanar parallel system of forces. If all the forces are pointing in one direction then they are called Like parallel forces otherwise they are called unlike parallel forces as shown in the Fig.5.
Fig. 5 Noncoplanar parallel system of forces
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8. Fig. 6. Resolving a force in space into rectangular components
Forces in space
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Rectangular components of a force in space
Fig. 6. Resolving a force in space into rectangular components
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9. Rectangular components of a force in space
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Forces in space
   Rectangular components of a force in space
In the Fig.6(a) a force F is acting at the origin O of the system of rectangular coordinate axes X,Y,Z. Consider OBAC plane passing through the force F. This plane makes an angle  with respect to XOY plane. Force F makes an angle θy with respect to Y-axis.
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Fig.6(a)

10. Rectangular components of a force in space
FS - 10
Forces in space
   Rectangular components of a force in space
In the Fig.6(b), the force F is resolved in the vertical (Y- axis) and horizontal direction (X – axis) as
Fy = F Cosy and
Fh = F Siny respectively. 
Fig.6(b)
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11. Rectangular components of a force in space
Forces in space
FS - 11
   Rectangular components of a force in space
In the Fig 6(c) the horizontal component Fh is again resolved in the X and Z axes directions. These components are
Fx = Fh cos = F siny cos
Fz = Fh Sin = F siny sin
Fig.6(c)
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12. Now applying Pythagorean theorem to the triangles OAB and OCD
Forces in space
FS - 12
Now applying Pythagorean theorem to the triangles OAB and OCD
F2 = (OA)2 = OB2 + BA2 = Fy2 +Fh (1)
Fh2 = OC2 = OD2 + DC2 = Fx2 +Fz (2)
Substituting equation (2) into the equation (1), we get
F2 = Fx2 +Fy2 + Fz2
F =  Fx2 + Fy2 + Fz (3)
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13. FS - 13
Forces in space
The 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 drawn as 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
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14. Forces in space FS - 14 From the above Figure (Fig. 7)
Fx = F Cos x, Fy = F Cosy, Fz = 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 = Fx/F; Cosy = Fy/F; Cosz = Fz/F
Substituting equation (4) into the equation (3), we get
F =  Fx2 + Fy2 + Fz2
F =  F2Cos2x + F2Cos2y + F2Cos2z
F2 = F2 ( Cos2x + Cos2y + Cos2z )
1 = Cos2x + Cos2y + Cos2z (5)
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15. Forces in space Fig 8 From the adjacent Fig. 8.
FS - 15
Force Defined by its magnitude and two points on its line of action
From the adjacent Fig. 8.
dx= d Cos x, dy = d Cosy, dz = d Cosz
----(6)
 d =  dx2 + dy2 + dz2
---(7)
Dividing member by member the relations (4) and (6), we obtain
Fx /dx = Fy/dy = Fz/dz = F/d (8)
Fig 8
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16. Forces in space FS - 16 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
Rx =  Fx,
Similarly adding algebraically all the components in y and z directions yield the following relations
Ry =  Fy,
Rz =  Fz
Thus magnitude of resultant
  R =  Rx2 + Ry2 + Rz2
Angles x, y, z resultant forms with the axes of coordinates are obtained by

17. Forces in space Problems: FS - 17
(1) A tower guy wire is anchored by means of a bolt at A is shown in the following Figure. The tension in the wire is 6000N. Determine
(a) The components Fx, Fy, Fz of the forces acting on the bolt.
(b) The angles x, y, z defining the direction of the force.
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18. Solution: (a) Here dx = 50m, dy = 200m, dz = -100m
Forces in space
FS - 18
Solution: (a) Here dx = 50m, dy = 200m, dz = -100m
Total distance A to B
d =  dx2 + dy2 + dz2
=  (50)2 + (200)2 + (-100)2
= m
Using the equation (8) Fx /dx = Fy/dy = Fz/dz = F/d
Fx = dx . (F/d) = (50 x 6000)/ = N
Fy = dy . (F/d) = (200 x 6000)/ = N
Fz = dz . (F/d) = (-100 x 6000)/ = N
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19. (b) Directions of the force:
Forces in space
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(b) Directions of the force:
Cos x = dx /d , x = Cos–1 (50/229.13) = 77.4
y = Cos–1 (dy /d) = Cos–1 (200/229.13) = 29.2
z = Cos–1 (dz /d) = Cos–1 (-100/229.13) = 
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20. Forces in space
FS - 20
Problem(2) Determine (a) the x , y and z components of the 250 N force acting as shown below
(b) the angles x, y, z that the force forms with the coordinate axes.
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21. Forces in space Fh = 250 Cos60 = 125 N Fy = 250 Sin60 = 216.5 N
FS - 21
Fh = 250 Cos60 = 125 N Fy = 250 Sin60 = N
 Fx = 125 Cos25 = N Fz = 125Sin25 = N
x = Cos–1 ( Fx /F) = Cos–1 (113.29/250) = 63
y = Cos–1 ( Fy /F) = Cos–1 (216.5/250) = 30
z = Cos–1 ( Fz /F) = Cos–1 (-52.83/250) = o
Components of 250 N in the x, y, z axes directions are
Fx = N Fy = N Fz = N
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22. Forces in space
FS - 22
Problem 3. Find the resultant of the system of forces as shown below
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23. Components of Force F1 = 3000 N: Fy1 = 3000 x Cos40o = 2298.13 N
Forces in space
FS - 23
  Solution
Components of Force F1 = 3000 N:
Fy1 = 3000 x Cos40o = N
Fh1 = 3000 x Sin40o = N
Fx1 = x Cos30o =1670 N
Fz1 = x Sin30o = N
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24. Components of Force F2 = 2000 N: Fy2 = 2000 x Cos20 = 1879.39 N
Forces in space
FS - 24
Components of Force F2 = 2000 N:
Fy2 = 2000 x Cos20 = N
Fh2 = 2000 x Sin20 = N
Fx2 = x Sin35 = N
Fz2 = x Cos35 = N
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25. Resultant R = Rx2 + Ry2 +Rz2 = 1277.652 +4177.522 +1524.512
Forces in space
FS - 25
Rx =  Fx = Fx1 + Fx2 = 1670 – =
Ry =  Fy = Fy1 + Fy2 = = N
Rz =  Fz = Fz1 + Fz2 = = N
Resultant R = Rx2 + Ry2 +Rz2
= 
= N
Its inclinations with respect to x, y and z axes are calculated as
x = Cos–1 ( /4626.9) = 73 58' 13.1"
y = Cos–1 ( /4626.9) = 25 27'
z = Cos–1 ( /4626.9) = 70o45’36”
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26. Forces in space
FS - 26
Problem 4. In the Fig shown below, the forces in the cables AB and AC are 100 kN and 150 kN respectively. At the joint ‘A’ loading is as shown in the Fig. Find the resultant of system of forces in space and its inclination with rectangular coordinates x,y and z axes.
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27. Forces in space Solution: Force in the cable AB = 100 kN
FS - 27
Solution: Force in the cable AB = 100 kN
Force in the cable AC=150 kN
For the cable AB
dx = -20 m
dy = 15m
dz = 5m
dAB =  dx2 +dy2 + dz2
=  (-20)2 +(15)2 + (5)2 =  = 25.5 m
For the cable AC
dy = 25m
dz = -10m
  dAc = (-20)2 +(25)2 + (-10)2 = 1125 = m 
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28. Forces in space For the cable AB (1) Fx1/dx1 = Fy1/dy1 = Fz1/dz1 = F/d
FS - 28
For the cable AB (1)
  Fx1/dx1 = Fy1/dy1 = Fz1/dz1 = F/d
  Fx1/-20 = Fy1/15 = Fz1/5 = 100/25.5
Fx1 = kN
Fy1 = kN
Fz1 = kN
For the Cable AC (2)
Fx2/-20 = Fy2/25 = Fz2/-10 = 150/33.54
Fx2 = kN
Fy2 = kN
Fz2 = kN
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29. Forces in space Fx4 = 50 kN Component of force 60 kN
FS - 29
Component of force 60 kN
Fx3 = 60 x Cos(70) = kN
Fy3 = 60 x Cos(30o) = kN
Fz3 = 60 x Cos(11123)= kN
Component of the force 50KN
Fx4 = 50 kN
Fy4 = 0
Fz4 = 0
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30. Forces in space
FS - 30
Algebraic summation of Rectangular Components in X, Y and Z axes directions yield:
 Rx =  Fx = Fx1 + Fx2 + Fx3 + Fx4 =
= kN
 Ry =  Fy = Fy1 + Fy2 + Fy3 + Fy4 =
= kN
Rz =  Fz = Fz1 + Fz2 + Fz3 + Fz4
= – –
= kN
 Resultant R =  Rx2 + Ry2 + Rz2
=  (-97.36)2 + (222.59)2 + (-46.99)2
= kN
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31. Forces in space Inclinations of the resultant with X,Y and Z axes:
FS - 31
Inclinations of the resultant with X,Y and Z axes:
x = Cos-1(Rx / R) = Cos-1 (-97.36/247.45) = 113 10’
y = Cos-1(Ry / R) = Cos-1 (222.59/247.45) = 25 54’
z = Cos-1(Rz / R) = Cos-1 (-46.99/247.45) = 100 56’ 47”
Check:
Cos2x + Cos2y + Cos2z = 1
  Cos2(11310’) + Cos2(2554’) + Cos2(10056) = 1
1 = 1
Hence OK
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32. Problem 5. a) Forces F1, F2, and F3 pass through the origin and
FS - 32
Forces in space
Problem 5.   a) Forces F1, F2, and F3 pass through the origin and
points whose coordinates are given. Determine the
resultant of the system of forces.
F1 = 20 kN, (3,-2,1)
F2 = 35 kN, (-2,4,0)
F3= 25 kN, (1,2,-3)
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33. Forces in space Solution: FS - 33
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34. Forces in space Force F1 = 20 kN d =  32 + (-2)2 + 12 =  14 = 3.74
FS - 34
Force F1 = 20 kN
d =  (-2) =  14 = 3.74
Cos x1 = dx / d = 3/3.74 = 0.802
Cos y1 = -2/3.74 = ; Cos z1 = 1/3.74 = 0.267
Force F2 = 35 kN
d =  (-2) = 20 = 4.47
Cos x2 = -2/4.47 = -0.45
Cos y2 = 4/4.47 = 0.9; Cos z2 = 0
Force F3 = 25 kN
d =  (1) (-3)2 = 14 = 3.74
Cos x3 = 1/3.74 = 0.267
Cos y3 = 2/3.74 = 0.535; Cos z3 = -3/3.74 =
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35. Resultant R =  Rx2 + Ry2 + Rz2
Forces in space
FS - 35
Summation of the rectangular components in X, Y and Z axes directions yield:
Rx = Fx = 20 x x (-0.45) +25 x = kN
Ry =  Fy = 20 x (-0.535) + 35 x x = kN
Rz =  Fz = 20 x x x (-0.802) = kN
Resultant R =  Rx2 + Ry2 + Rz2
=  (-14.71)2
= kN
Inclination of the resultant R with respect to X, Y and Z axes are calculated as
x = Cos–1(Rx / R) = Cos-1(6.965/37.85 ) = 79.4
y = Cos–1(Ry /R) = Cos-1(34.175/37.85) = 25.46
z = Cos–1(Rz /R) = Cos-1(-14.71/37.85) = 
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36. Forces in space
FS - 36
Equilibrium of Concurrent non-coplanar system of forces:
When a rigid body subjected to concurrent noncoplanar system of forces F1, F2…. ..FN as shown in the Fig. given below, is in equilibrium, then algebraic summation of all the components of the forces in three mutually perpendicular directions must be equal to zero.
i.e.  Fx = 0
 Fy = 0
 Fz = (1)
Above equations represent the static conditions of equilibrium for concurrent noncoplanar system of forces
Fig. A rigid body subjected to concurrent noncoplanar system of forces
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37. Forces in space
FS - 37
Problem (1)    Find the forces in the rods AB , AC and AO subjected to loading as shown below
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38. Forces in space dx1 = 0 - 4 = - 4 dy1= 8 - 0 = 8 dz1 = 15 – 0 =15
FS - 38
Solution:
For the cable AB:
dx1 = = - 4
dy1= = 8
dz1 = 15 – 0 =15
d1 =  (-4)2 +(8)2 +(15)2 = m
Fx1/-4 = Fy1/8 = Fz1/15 = FAB/17.46
Fx1 = -0.23FAB,
Fy1 = 0.46FAB,
Fz1 = 0.86FAB
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39. Forces in space For the Cable AC: dx2 = 0 - 4 = -4m dy2 = 8 - 0 = 8m
FS - 39
For the Cable AC:
dx2 = = -4m
dy2 = = 8m
dz2 = -20 – 0 = -20m
d2 =  (-4)2 +(8)2 +(-20)2 = 480 = 21.91m
  Fx2/-4 = Fy2/8 = Fz2/-20 = FAC/21.91
  Fx2 = -0.18FAC, Fy2 = 0.365FAc, Fz2 = -0.91FAc
For the Force 120 N:
Fx3 = 120 N, Fy3 = 0, Fz3 = 0
For the force Fx4 = 300 N:
Fx4 = 0, Fy4 = -300N, Fz4 = 0
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40. Forces in space  Fx = 0,  Fy = 0,  Fz = 0 Considering  Fx = 0
FS - 40
Conditions of Equilibrium
 Fx = 0,  Fy = 0,  Fz = 0
Considering  Fx = 0
Fx1 + Fx2 + Fx3 + Fx4 + Fx5= 0
-FAD-0.23F – 0.18FAC = 0
FAD FAB FAC = (1)   
 Fy = 0
0.46FAB FAC – 300 = 0
0.46FAB FAC = (2)
 Fz = 0
0.86 FAB – 0.91 FAC + 0 = (3)
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41. Solving Equations (1), (2) and (3), we get,
Forces in space
FS - 41
Solving Equations (1), (2) and (3), we get,
FAB = N; FAC = N ; FAO = N
Force in the rod AB , FAB = N (Tensile)
Force in the rod AC, FAC = N (Tensile)
Force in the rod AO, FAO = N (Compressive)
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42. Forces in space
FS - 42
2)    Three cables are connected at D and support, the 400 kN load as shown in the Fig given below. Determine the tensions in each cable
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43. dAD =  (-3)2 + (2)2 + (6)2 = 9 + 4 + 36 = 7 m
Forces in space
FS - 43
Let FDA, FDB and FDC are the forces in the cables AD, BD, and CD respectively.
For Cable DA:
dx1 = (0 – 3) = -3m
dy1 = (6 - 4) =2 m
dz1 = (6 – 0 = 6 m
dAD =  dx2 + dy2 + dz2
dAD =  (-3)2 + (2)2 + (6)2 =  = 7 m
Fx1/dx1 = Fy1/dy1 = Fz1/dz1 = FDA/dDA
Fx1/(-3) = Fy1/2 = Fz1/6 = FDA/7
 Fx1 = -0.43FDA, Fy1 =0.286FDA, Fz1 = 0.857FDA
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44. dBD = d2 = (-3)2 + (2)2 + (-6)2 =  49 = 7 m
Forces in space
FS - 44
For Cable DB:
dx2= (0 – 3) = -3m
dy2 = (6 - 4) =2 m
dz2 = (-6 – 0 = -6 m
dBD = d2 = (-3)2 + (2)2 + (-6)2 =  49 = 7 m
Fx2/dx2 = Fy2/dy2 = Fz2/dz2 = FDB/dDB
Fx2/(-3) = Fy2/2 = Fz2/-6 = FDB/7
Fx2 = -0.43FDB,
Fy2 =0.286FDB,
Fz2 = FDB
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45. Fx3/dx3 = Fy3/dy3 = Fz3/dz3 = FDC/d DC
Forces in space
FS - 45
For Cable DC:
dx3 (0 – 3) = -3m
dy3= (0- 4) = -4m
dz3= (0 0 = 0m
dDC= d3=  (-3)2 + (-4) + (0)2
= 
= 5m
Fx3/dx3 = Fy3/dy3 = Fz3/dz3 = FDC/d DC
Fx3/(-3) = Fy3/-4 = Fz3/0 = FDC/5
Fx3 = -0.6FDC,
Fy3 =-0.8FDC,
Fz3 = 0
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46. Fx4/dx4 = Fy4/dy4 = Fz4/dz4 = F400/d4
Forces in space
FS - 46
For the force 400KN
dx4 = 12m
dy4= -4m
dz4= 3m
d4= d400 =  (12)2 + (4) + (3)2
= 
= 69 = 13m
Fx4/dx4 = Fy4/dy4 = Fz4/dz4 = F400/d4
Fx4/(12) = Fy4/-4 = Fz4/3 = 400/13
Fx4 = kN,
Fy4 = kN,
Fz4 = kN
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47. i.e.  Fx = 0 -------------(1)  Fy = 0 ------------- (2)
Forces in space
FS - 47
For Equilibrium, the algebraic summation of resolved component in a particular direction is equal to zero.   
i.e.  Fx = (1)
 Fy = (2)
 Fz = (3)
(1)  Fx = Fx1 + Fx2 + Fx3 + Fx4 = 0
+ 0.43FDA FDB + 0.6FDC = (1)
FDA FDB – 0.8FDC = (2)
FDA FDB + 0 = (3)
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48. Solving equations (1), (2) and (3), we get FDB = 304.1kN
Forces in space
FS - 48
Solving equations (1), (2) and (3), we get
FDB = 304.1kN
FDA = kN
FDC = 102 kN
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49. Forces in space Practice Questions FS - 49
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 Fx, Fy and Fz of the force acting on the bolt, (b) the angles θx, θy, θz defining the direction of the force
(Ans: Fx= N , Fy = 2120 N, Fz= N
Θx = o ; Θy= 32.0 o ; Θz= 71.5 o )
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50. Forces in space Practice Questions
FS - 50
2. Determine (a) x, y and z the components of the force 500 N in the below Figure. (b) the angles θx, θy, θz that the force forms with the coordinate axes
(Ans: Fx= N , Fy = N, Fz= N
Θx = 56.2 o ; Θy= 40.0 o ; Θz= 71.3 o )
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51. Forces in space Practice Questions
FS - 51
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 the cable AB is 10 kN, determine the components of the force exerted by the cable AB on the truck
(Ans: Fx= kN , Fy = 6.06 kN, Fz= kN)
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52. Forces in space Practice Questions
FS - 52
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 perpendicular to the wall holds the cylinder in the position shown. Determine the magnitude of the force P and the tension in each cable
(Ans: P = 235 N , TAB = 1401 N, TAC= N )
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53. Forces in space Practice Questions
FS - 53
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 = kN
(Ans: TAB= 2.88 kN , TAC = 5.76 kN, TAD= 3.6 kN)
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54. Forces in space Practice Questions FS - 54
6. 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) the magnitude of the force P which must be applied to the ring to maintain the container in the position shown in figure below, (b) the corresponding the values of the tension in cables AB and AC
(Ans: P = 138 N ,
TAB = 270N,
TAC = 196N )
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55. Forces in space Practice Questions FS - 55
7. 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
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56. Forces in space Practice Questions FS - 56
8. Determine the resultant of the two forces shown below.
(Ans: R = 498 N ,
Θx = 68.9 o ; Θy= 26.3 o ; Θz= 75.1 o )
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57. Forces in space Practice Questions FS - 57
9. A container of weight W = 1165 N is supported by three cables as shown below. Determine the tension in each cable.
Ans: TAB = 500 N
TAC = 459 N
TAD = 516 N
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58. Thank you
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