Statica question, almost impossible solution.

[BramVandenbo]

Hi,

 

I have been studying Statica for the past month. And I can say with proud I am finally starting to get it. I am stuck however on a problem I made up myself. Something is probably wrong in my way of thinking.

 

Excercice:

Let's say there is a framework of 2 equal horizontal iron pieces both are attached to opposing walls at the same height. In the center they are joined together. Next there's a cable hanging in the connection point. The cable is connected to a weight of 20 Newton. The angle of the cable (alpha) as seen in the picture = 89,99° (or almost PI/2)

 

 

My answer:

I solved it in 3 steps...

1) first of all the sum of the external vertical powers should be 0. This makes that in point A and B there is an vertical power-component of 10N.

 

2) 2*(cos(89,99°) * the_internal_pulling_or_pressure_inside_one_of_the_cables) = 20N

or ...

the_internal_pulling_or_pressure_inside_one_of_the_cables = 10N/cos(89,99°)

The result of the last equation is: 57 296 Newton

 

3) Point A and B have both a vertical power of 57296 Newton

(= sin(89,99°)*57296N)

 

My question towards the readers of this:

First of all is my way of thinking correct? As you can see the weight of 20N created an internal force of 57.296 Newton!!!

And what if alpha would have been 0. Does that mean the internal force becomes infinit? This can't be correct can it?

 

Thank you in advance

[5614]

The way I would go about it is by doing what you did in #2 and I get the same answer: 57295.7798 (exactly)

 

Point about 3) You are rounding, they are not identical, actual value (using unrounded figures) is: 57295.77893... (where the ... represents more decimal places)

 

And if you use the exact (unrounded answers) then use Pythagoras and it all works out...

 

57295.7798 squared - 57295.77893 (with more decimal places) squared = 100 (which is 10 squared)... it works!

 

The VERTICAL = 10N because we know the the sum of the vertical component is 20N because the system is in equilibrium. As each iron thing is at the same angle and because I am assuming this model assumes the two iron things are identical the vertical force supplied by each will be 20/2 = 10N

 

This will act as a second proof for our big 57296 Newton "internal pulling".

 

10 / cos(89.99) = 57296.7798

 

So yes, you are correct.

 

=====

 

And what if alpha = 0 ???

 

Well work it out!!!

 

Then it would be:

 

10 / cos(0) = 10

 

(Because cos(0) = 1)

 

Oh, as there is NO vertical component the force must have all gone into the horizontal component. The whole of the 10 going into one thing results in 10! (That's 10 exclamation mark, not 10 factorial!)

 

=====

 

You could also say that:

 

(internal_pulling_A * sin(89.99)) - (internal_pulling_B * sin(89.99)) = 0

 

Because the system is in equilibrium the 2 horizontal components are identical, so one subtract the other is 0.

[BramVandenbo]

And what if alpha = 0 ???

 

Well work it out!!!

 

Then it would be:

 

10 / cos(0) = 10

 

(Because cos(0) = 1

 

Sorry, I meant "what if alpha is 90°". Not 0°, but 90°... my mistake.

 

F = 10N/cos(alpha)

 

Does this mean...

10N/cos(90°) = 10N/0 = Infinit?

[5614]

If alpha was 90 then you wouldn't need cos and clearly it wouldn't work.

 

If alpha = 90 then it's a vertical rod and (assuming there is only 1) tension would be equal to the weight of the object (assuming the system is in equilibrium).

[5614]

Also I just thought of this today... think of the force triangle.

 

There is an angle of 89.99 which is almost 90. If it was 90 then two lines would be parallel and it would not be a traingle. The angle in this question is just 0.001 degrees off being parallel. Even if the lines were just 1 unit apart it would still be a considerable distance until the two lines touched. In the question the two lines are 10 units apart, that's a big distance for two lines which are so nearly parallel to meet.

[BramVandenbo]

According to my prof and my book there is a difference between a vertical rod and 2 vertical joining rods. And they should be regarded as 2 different subjects. The difference is that there is a turning point present. This changes the way pressure is handled and always makes the forces point in the direction of the pieces.

 

If alpha = 90 then it's a vertical rod and (assuming there is only 1) tension would be equal to the weight of the object (assuming the system is in equilibrium).

 

Actually according to what my book and prof have told me, the pressure and pull always go in the direction of the rod. On the other hand, I must agree that 10N would make sense, It's the first reply I would have given as well. But feelings and emotions have of course nothing to do with Statica .

 

Actually what I didn't show last time, are the numbers below. If you look at the huge numbers I get, just before I get 90. (I copied them below). As you can see the first side goes to +infinit, and the other side goes to -infinit. I guess this gives a total different answer. When the pieces are perfectly horizontal, then suddnely both powers (tension and pull) are present: the huge positive one and the huge negative one. The total pressure or tension is 0, or undefined actually. (Then again nothing is really 100% horizontal in reality.)

 

force angle

------- ------

10,000 0

10,002 1

10,006 2

10,014 3

10,024 4

10,038 5

10,055 6

10,075 7

10,098 8

10,125 9

10,154 10

10,187 11

10,223 12

10,263 13

10,306 14

10,353 15

10,403 16

10,457 17

10,515 18

10,576 19

10,642 20

10,711 21

10,785 22

10,864 23

10,946 24

11,034 25

11,126 26

11,223 27

11,326 28

11,434 29

11,547 30

11,666 31

11,792 32

11,924 33

12,062 34

12,208 35

12,361 36

12,521 37

12,690 38

12,868 39

13,054 40

13,250 41

13,456 42

13,673 43

13,902 44

14,142 45

14,396 46

14,663 47

14,945 48

15,243 49

15,557 50

15,890 51

16,243 52

16,616 53

17,013 54

17,434 55

17,883 56

18,361 57

18,871 58

19,416 59

20,000 60

20,627 61

21,301 62

22,027 63

22,812 64

23,662 65

24,586 66

25,593 67

26,695 68

27,904 69

29,238 70

30,716 71

32,361 72

34,203 73

36,280 74

38,637 75

41,336 76

44,454 77

48,097 78

52,408 79

57,588 80

63,925 81

71,853 82

82,055 83

95,668 84

114,737 85

143,356 86

191,073 87

286,537 88

572,987 89

??? 90 <<== Not Inifinit , but unknown +oo -oo = Undefined

-572,987 91

-286,537 92

-191,073 93

-143,356 94

-114,737 95

-95,668 96

-82,055 97

-71,853 98

-63,925 99

-57,588 100

-52,408 101

-48,097 102

-44,454 103

-41,336 104

-38,637 105

-36,280 106

-34,203 107

-32,361 108

-30,716 109

-29,238 110

-27,904 111

-26,695 112

-25,593 113

-24,586 114

-23,662 115

-22,812 116

-22,027 117

-21,301 118

-20,627 119

-20,000 120

-19,416 121

-18,871 122

-18,361 123

-17,883 124

-17,434 125

-17,013 126

-16,616 127

-16,243 128

-15,890 129

-15,557 130

-15,243 131

-14,945 132

-14,663 133

-14,396 134

-14,142 135

-13,902 136

-13,673 137

-13,456 138

-13,250 139

-13,054 140

-12,868 141

-12,690 142

-12,521 143

-12,361 144

-12,208 145

-12,062 146

-11,924 147

-11,792 148

-11,666 149

-11,547 150

-11,434 151

-11,326 152

-11,223 153

-11,126 154

-11,034 155

-10,946 156

-10,864 157

-10,785 158

-10,711 159

-10,642 160

-10,576 161

-10,515 162

-10,457 163

-10,403 164

-10,353 165

-10,306 166

-10,263 167

-10,223 168

-10,187 169

-10,154 170

-10,125 171

-10,098 172

-10,075 173

-10,055 174

-10,038 175

-10,024 176

-10,014 177

-10,006 178

-10,002 179

-10,000 180

 

How come constructions like this don't break in real life? There's no material that can hold nearly Infinit pressure or tension.

[5614]

According to my prof and my book there is a difference between a vertical rod and 2 vertical joining rods. And they should be regarded as 2 different subjects. The difference is that there is a turning point present. This changes the way pressure is handled and always makes the forces point in the direction of the pieces.

Well the difference between one vertical rod and 2 vertical joining rods would be the joint. Assuming the only forces acting are the mass and this is a vertical force we can say that the net pulling across both vertical rods and the joint together will add to the weight of the object.

 

If you start adding the complication of the joint having different properties to the rod then you need more detail (ie. there may be more internal pulling within the joint relative to the rod), but other than that there isn't really a difference between a vertical rod and a vertical rod with a joint in the middle of it (when the only force is a vertical force).

 

======

 

I made a mistake in what I said before, when alpha is 90 the rod is horizontal (not vertical), I was thinking of the wrong angle, appologies.

 

So when alpha is zero and the rod is vertical then the tension within the rod is equal to the weight of the object at the end of it.

 

When alpha is 90 and the rod is horizontal then the hypotenuse and one of the sides of the triangle have merged... it is no longer a triangle.

 

Although I see what you mean, as alpha tends to 90 so tension tends to infinity, this is illogical as a simple mass cannot supply an infinite force... I'm gonna ask someone else about this tomorrow.

[BramVandenbo]

I just had my exam this morning. Big success! Don't know my score yet, but probably pretty high.

 

Thank you for helping me in my studies.

[5614]

I asked my Maths teacher (Cambridge Maths graduate) and he said that it is absolutely correct that as theta tends to 90 so tension tends to infinity.

 

 

That bridge explains the problem... the road is a straight line, so the tension within it is massive, this is why the vertical support lines are required. To stop the bridge collapsing under it's own weight.

 

Well done on the exam... what exam was it?

[BramVandenbo]

Statica and Dynamica:

 

I had 3 question:

 

Question 1:

a boat is floating in the sea.

(v = speed, a = acceleration, t = time, s = distance)

The boat has a speed of v =0,8t. And it is floating in a direction shaped like a circle with a radius of 120m. What is its speed and acceleration after floating a certain distance, lets say it was 20meters?

 

Answer 1:

I solved this in 4 steps.

a) Because the speed was declared in function of time, I started by calculating how long it would take to reach the 20 meters.

 

v = ds/dt

v dt = ds

int(v dt) = 20

int(0,8t dt) = 20

(0,4t²) = 20

t² = 20/0,4 = 50

t = sqrt(50) or (50)^(1/2) = 7,07 seconds according to my calculator

 

b) now it's easy to find the speed namely:

v = 0,8*7,07 = 5,66 m/s

 

c) now the acceleration:

We have 2 components. There's a centropetal acceleration (an) and a straight acceleration (at) touching the curved shape direction of the boat. Both cross in an angle of 90°.

 

For the force in the direction of the boat:

at = v/dt = 0,8 m/s²

 

For the centropetal acceleration (aka middlepoint-seeking acceleration):

an = v²/r = 5,66/20 = 0,28 m/s²

luckely we know the speed (5,66m/s) allready!

 

d) Now all what's left is bringing the 2 forces together

that's easy ofcourse: sqrt(0,8² + 0,28²) = 0,85 m/s²

 

Question 2:

 

Here we go again. Jup a giant framework .

As you can see there are 2 external vertical forces. At the right the framework has a little wheel. And at the left it is fixed to the ground in all directions. Oh yeah, and those 4 rods at the bottum each have a length of 1,5m. And the height of those 2 small vertical rods is 2m each. The question is: what is the pressure or tension inside the 2 red rods. And give all rods that have no tension or pressure at all?

 

Answer 2:

a) I allready marked all rods without forces, this can be done on sight: the pink ones.

 

b) Next I calculated the responses in the left and right point. Because as we know the sum of the external vertical forces has to be 0. The question is how much at the left, and how much at the right?

 

c) There's an easy trick to do this. Namely: the sum of the moments around a point has to be 0. So I calculated the moments around the bottum right point.

7,5*(2*1,5) + 7,5*1,5 + external_force_bot_left*6 = 0

external_force_bot_left = -5,63kN

The right one can be calculated the same way. Or a faster way now:

Sum of all vertical powers = 0.

7,5 + 7,5 - 5,63 + X = 0

X = -9,37kN

 

d) Now that we know the vertical power in bottum right. We can easely calculate the right red rod tension. Because it's the only rod with a vertical force-component. For this we need to calculate it's length first though:

sqrt(1,5² + 2²) = sqrt(6,25) = 2,5m

(not that it matters, but that's a 3-4-5 ratio).

The vertical force component is:

Tension_Or_Pressure = -9,37*(2,5/2)= -11,71kN.

So far so good. That's the first rod.

 

e)Now we are at it it is easy to calculate the horizontal rod in the right-side corner. Because the sum of the horizontal forces has to be 0. This makes:

horizontal_rod = (11,71/2,5)*1,5 = 7,03 kN.

 

f) The 2nd rod is harder... We have to calculate a moment again. For these I am cutting out some part of the diagram. What a nice drawing !

 

So I cut some part off, and there's a moment around the green point in the top. I took the point at the top, because that makes it easy to calculate:

rod_bot_left* 2m + 7,03*2m = 0

rod_bot_left = -7,03N

Yup the same size, but other direction.

 

Question 2 solved!

 

Question 3:

 

This one is actually the easiest. There's a metal construction attached to a wall with a cable. The cable is stretched in the diagonal 3D direction. And the metal construction is also connected to a smooth vertical tube. But as you can see the tube has a square shape, this means it will give horizontal reactions, but no vertical-reactions. The Question is: calculate the tension in the cable and the horizontal responses from the connection to the vertical tube.

This picture isn't correct though. That 8m is 6m not 8! Sorry, don't feel like redrawing it .

 

Well actually the question was a bit more complex, because I first had to calculate those 3 sizes. They were not given directly.

 

Answer 3:

 

Well ofcourse the vertical tension of the cable equals the weight of 45N.

So that's where it all starts...

cableZ = 45N

cableX = 4*45/6 = 30N

cableY = 12*45/6 = 90N

 

Now we need to calculate the size of that force:

Tension = sqrt(45²+ 30² + 90²) = 105N

 

The horizontal responses from the connection are:

-30N and -90N ofcourse. ==> the opposit direction of the cable.

[BramVandenbo]

Lol, actually that last drawing is a mistake. I made some mistakes in the perspective... nevermind That 4m is placed incorrectly. Looks more like an Esscher Drawing.