forces in space.ppt

8/11/2019 Forces in space.ppt

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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 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

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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)

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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

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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

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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

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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

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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 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)

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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)

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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)

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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)

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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

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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)



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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

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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

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

;;

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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.



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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



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(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



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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.



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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



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



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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



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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



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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.



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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

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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



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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



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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



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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



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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)



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Forces in space

Solution :

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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



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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



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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

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Forces in


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

loading as shown below

Forces inspace

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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


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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



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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)



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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)



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2) Three cables are connected at D and support, the 400kN load as shown in the Fig given below. Determine thetensions in each cable



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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



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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


i


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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


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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)



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Solving equations (1), (2) and (3), we getFDB = 304.1kN

FDA = 411.8 kN

FDC = 102 kN


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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 )

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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 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 )

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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 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)

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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 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)

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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 = 7.28kN

(Ans: T AB= 2.88 kN , T AC = 5.76 kN, T AD= 3.6kN)

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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) 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

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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 Question

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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 )

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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: T AB = 500 NT AC = 459 NT AD = 516 N

Practice Question


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forces in space.ppt

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