FORCES IN SPACE ( Noncoplanar System of Forces)

FORCES IN SPACE(Noncoplanar System of Forces)

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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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Forces in space

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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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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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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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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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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Forces in space

Fig. 6. Resolving a force in space into rectangular components

Rectangular components of a force in space

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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.

Forces in space

Fig.6(a)

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In the Fig.6(b), the force F is resolved in the vertical (Y- axis) and horizontal direction (X – axis) as

Fy = F Cosy and

Fh = F Siny respectively.

Forces in space

Fig.6(b)

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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 siny cos

Fz = Fh Sin = F siny sin

Forces in space

Fig.6(c)

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Now applying Pythagorean theorem to the triangles OAB and OCDF2 = (OA)2 = OB2 + BA2 = Fy

2 +Fh2 ----------------(1)

Fh2 = OC2 = OD2 + DC2 = Fx

2 +Fz2 ----------------(2)

Substituting equation (2) into the equation (1), we get

F2 = Fx2 +Fy

2 + Fz2

F = Fx2 + Fy

2 + Fz2 ----------------(3)

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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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From the above Figure (Fig. 7) Fx = F Cos x, Fy = F Cosy, Fz = F Cosz ------------(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; Cosy = Fy/F; Cosz = Fz/F Substituting equation (4) into the equation (3), we get

F = Fx2 + Fy

2 + Fz2

F = F2Cos2x + F2Cos2y + F2Cos2z F2 = F2 ( Cos2x + Cos2y + Cos2z ) 1 = Cos2x + Cos2y + Cos2z -------------(5)



From the adjacent Fig. 8.

dx= d Cos x, dy = d Cosy, dz = d Cosz

----(6) d = dx

2 + dy2 + dz

2 ---(7)

Dividing member by member the relations (4)

and (6), we obtain

Fx /dx = Fy/dy = Fz/dz = F/d ----------------------------(8)

Forces in space

Force Defined by its magnitude and two points on its line of action

Fig 8

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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

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 + Ry

2 + Rz2

Angles x, y, z resultant forms with the axes of coordinates are obtained by

R

RCos

R

RCos

R

RCos z

zy

yx

x ;;

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Problems:(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.



Solution: (a) Here dx = 50m, dy = 200m, dz = -100m

Total distance A to B

d = dx2 + dy

2 + dz2

= (50)2 + (200)2 + (-100)2

= 229.13 m

Using the equation (8) Fx /dx = Fy/dy = Fz/dz = F/d

Fx = dx . (F/d) = (50 x 6000)/ 229.13 = 1309.3 N

Fy = dy . (F/d) = (200 x 6000)/ 229.13 = 5237.20 N

Fz = dz . (F/d) = (-100 x 6000)/ 229.13 = -2618.6 N



(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) = 115.88



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.



Fh = 250 Cos60 = 125 N Fy = 250 Sin60 = 216.5 N Fx = 125 Cos25 = 113.29 N Fz = 125Sin25 = -52.83 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) = 102.11o

Components of 250 N in the x, y, z axes directions are Fx = 113.29N Fy = 216.5 N Fz = -52.83N



Problem 3. Find the resultant of the system of forces as shown below



Solution

Components of Force F1 = 3000 N:

Fy1 = 3000 x Cos40o = 2298.13 N

Fh1 = 3000 x Sin40o = 1928.36 N

Fx1 = 1928.36 x Cos30o =1670 N

Fz1 = 1928.36 x Sin30o = 964.18 N



Components of Force F2 = 2000 N:

Fy2 = 2000 x Cos20 = 1879.39 N

Fh2 = 2000 x Sin20 = 684.04 N

Fx2 = 684.04 x Sin35 = 392.35 N

Fz2 = 684.04 x Cos35 = 560.33N



Rx = Fx = Fx1 + Fx2 = 1670 – 392.35 = 1277.65

Ry = Fy = Fy1 + Fy2 = 2298.13 + 1879.39 = 4177.52 N

Rz = Fz = Fz1 + Fz2 = 964.18 + 560.33 = 1524.51 N

Resultant R = Rx2 + Ry

2 +Rz2

= 1277.652 +4177.522 +1524.512

= 4626.9 NIts inclinations with respect to x, y and z axes are calculated 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) = 70o45’36”



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.



Solution: Force in the cable AB = 100 kN Force in the cable AC=150 kN For the cable AB

dx = -20 mdy = 15mdz = 5mdAB = dx2 +dy2 + dz2

= (-20)2 +(15)2 + (5)2 = 650 = 25.5 m

For the cable ACdx = -20 mdy = 25mdz = -10m

dAc = (-20)2 +(25)2 + (-10)2 = 1125 = 33.54 m

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For the cable AB (1)

Fx1/dx1 = Fy1/dy1 = Fz1/dz1 = F/d

Fx1/-20 = Fy1/15 = Fz1/5 = 100/25.5

Fx1 = -78.41 kN

Fy1 = 58.82 kN

Fz1 = 19.61 kN

For the Cable AC (2) Fx2/-20 = Fy2/25 = Fz2/-10 = 150/33.54

Fx2 = - 89.45 kN

Fy2 = 111.81 kN

Fz2 = - 44.72 kN



Component of force 60 kN

Fx3 = 60 x Cos(70) = 20.52 kNFy3 = 60 x Cos(30o) = 51.96 kNFz3 = 60 x Cos(11123)= - 21.88 kN

Component of the force 50KN

Fx4 = 50 kNFy4 = 0Fz4 = 0



Algebraic summation of Rectangular Components in X, Y and Z axes directions yield: Rx = Fx = Fx1 + Fx2 + Fx3 + Fx4

= -78.43 - 89.45 + 20.52 + 50 = -97.36 kN Ry = Fy = Fy1 + Fy2 + Fy3 + Fy4

= 58.82 + 111.81 + 51.96 + 0 = 222.59 kN Rz = Fz = Fz1 + Fz2 + Fz3 + Fz4

= 19.61 – 44.72 – 21.88 + 0 = -46.99 kN

Resultant R = Rx2 + Ry2 + Rz2 = (-97.36)2 + (222.59)2 + (-46.99)2

= 247.45 kN



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:

Cos2x + Cos2y + Cos2z = 1 Cos2(11310’) + Cos2(2554’) + Cos2(10056) = 1 1 = 1

Hence OK



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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Forces in space

Solution:

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Force F1 = 20 kNd = 32 + (-2)2 + 12 = 14 = 3.74Cos x1 = dx / d = 3/3.74 = 0.802Cos y1 = -2/3.74 = -0.535; Cos z1 = 1/3.74 = 0.267

Force F2 = 35 kNd = (-2)2 + 42 +0 = 20 = 4.47Cos x2 = -2/4.47 = -0.45Cos y2 = 4/4.47 = 0.9; Cos z2 = 0

Force F3 = 25 kNd = (1)2 + 22 +(-3)2 = 14 = 3.74Cos x3 = 1/3.74 = 0.267Cos y3 = 2/3.74 = 0.535; Cos z3 = -3/3.74 = -0.802



Summation of the rectangular components in X, Y and Z axes directions yield:

Rx = Fx = 20 x 0.802 + 35 x (-0.45) +25 x 0.267 = 6.965 kNRy = Fy = 20 x (-0.535) + 35 x 0.9 + 25 x 0.535 = 34.175 kNRz = Fz = 20 x 0.267 + 35 x 0 + 25 x (-0.802) = -14.71 kN

Resultant R = Rx2 + Ry2 + Rz2 = 6.9652 + 34.1752 + (-14.71)2 = 37.85 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 ) = 79.4y = Cos–1(Ry /R) = Cos-1(34.175/37.85) = 25.46z = Cos–1(Rz /R) = Cos-1(-14.71/37.85) = 112.87



Forces in space

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.

Fig. A rigid body subjected to concurrent

noncoplanar system of forces

i.e. Fx = 0

Fy = 0

Fz = 0 (1)

Above equations represent the static conditions of equilibrium for concurrent noncoplanar system of forces

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Problem (1) Find the forces in the rods AB , AC and AO subjected to loading as shown below

Forces in space

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Solution:

For the cable AB:

dx1 = 0 - 4 = - 4 dy1= 8 - 0 = 8 dz1 = 15 – 0 =15

d1 = (-4)2 +(8)2 +(15)2 = 17.46 m

Fx1/-4 = Fy1/8 = Fz1/15 = FAB/17.46 Fx1 = -0.23FAB,

Fy1 = 0.46FAB,

Fz1 = 0.86FAB



For the Cable AC:dx2 = 0 - 4 = -4m

dy2 = 8 - 0 = 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



Conditions of Equilibrium

Fx = 0, Fy = 0, Fz = 0 Considering Fx = 0Fx1 + Fx2 + Fx3 + Fx4 + Fx5= 0

-FAD-0.23F – 0.18FAC + 120 + 0 = 0

FAD + 0.23FAB + 0.183FAC = 120 ----------------------(1)

Fy = 00.46FAB + 0.365FAC + 0 +0 – 300 = 0

0.46FAB + 0.365FAC = 300 ------------------------------(2)

Fz = 00.86 FAB – 0.91 FAC + 0 = 0 -----------------------(3)



Solving Equations (1), (2) and (3), we get,FAB = 372.675 N; FAC = 352.25 N ; FAO = - 30.18 N

Force in the rod AB , FAB = 372.675 N (Tensile)Force in the rod AC, FAC = 352.25 N (Tensile)Force in the rod AO, FAO = 30.18 N (Compressive)



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



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 = 9 + 4 + 36 = 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



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 = -0.857FDB



For Cable DC:dx3 (0 – 3) = -3m

dy3= (0- 4) = -4m

dz3= (0 0 = 0m

dDC= d3= (-3)2 + (-4) + (0)2

= 9 + 16 = 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



For the force 400KNdx4 = 12m

dy4= -4m

dz4= 3m

d4= d400 = (12)2 + (4) + (3)2

= 144 +16 + 9 = 69 = 13mFx4/dx4 = Fy4/dy4 = Fz4/dz4 = F400/d4

Fx4/(12) = Fy4/-4 = Fz4/3 = 400/13

Fx4 = 369.23 kN,

Fy4 = -123.08 kN,

Fz4 = +92.31 kN



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

+ 0.43FDA + 0.43FDB + 0.6FDC = 369.23 ---------(1)

+ 0.286FDA + 0.286FDB – 0.8FDC = 123.08 ------(2)

+ 0.857FDA - 0.857FDB + 0 = 92.31 ------------(3)



Solving equations (1), (2) and (3), we get

FDB = 304.1kN

FDA = 411.8 kN

FDC = 102 kN



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 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= -1060 N , Fy = 2120 N, Fz= + 795 N Θx = 115.1 o ; Θy= 32.0 o ; Θz= 71.5 o )

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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 θx, θy, θz that the force forms with the coordinate axes

(Ans: Fx= + 278 N , Fy = + 383 N, Fz= + 160.7 N Θx = 56.2 o ; Θy= 40.0 o ; Θz= 71.3 o )

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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 the cable AB is 10 kN, determine the components of the force exerted by the cable AB on the truck

(Ans: Fx= -6.30 kN , Fy = 6.06 kN, Fz= + 4.85 kN)

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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 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= + 1236 N )

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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 = 7.28 kN

(Ans: TAB= 2.88 kN , TAC = 5.76 kN, TAD= 3.6 kN)

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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) 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 )

Practice Questions

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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 Questions

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Forces in space8. Determine the resultant of the two forces shown below.

Practice Questions

(Ans: R = 498 N , Θx = 68.9 o ; Θy= 26.3 o ; Θz= 75.1 o )

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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: TAB = 500 N TAC = 459 N TAD = 516 N

Practice Questions

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FORCES IN SPACE ( Noncoplanar System of Forces)

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