Engineering statics questions

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Homework Statement


Hello everyone,

I have a few questions regarding the problem below.

1. I did the process for solving γ correct however I took the 60° angle instead of the 120° angle. The below solution states that γ>90°. Why must is be greater, than 90°, and when would it have been smaller?

2. I got a big confused when solving for the projected portion of the triangle. What I did initially was solve for the hypotenuse of the projection which I thought would be

F'=450sin(45°), and just use that F' to find the x, and y components, however that was completely wrong it seems.


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The Attempt at a Solution

 

Answers and Replies

  • #2
Simon Bridge
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Homework Statement


Hello everyone,

I have a few questions regarding the problem below.

1. I did the process for solving γ correct however I took the 60° angle instead of the 120° angle. The below solution states that γ>90°. Why must is be greater, than 90°, and when would it have been smaller?
Where is γ measured from?
If F2 pointing above pr below the x-y plane?

2. I got a big confused when solving for the projected portion of the triangle. What I did initially was solve for the hypotenuse of the projection which I thought would be

F'=450sin(45°), and just use that F' to find the x, and y components, however that was completely wrong it seems.
F1 forms the hypotenuse of a 45deg triangle, Is this the one you mean?

The x-y projection would start with what you did[*]: F'=F1/√2
How did you get the x and y components from there?

Note: the angles are very friendly.
i.e. sin(45)=cos(45)=1/√2, sin(30)=cos(30)=1/2, etc.

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

[*] except it's the cosine - doesn't matter to the result though.
Does matter for the next results.
 
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  • #3
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γ is measured from the z axis to F2. It seems that F2 is below the xy plane. For the projection what I mean was for the 30degree the way my mind analyzes it is that we get the x components of both triangles e.g 30 and 45 degree.

Which would mean that 450*cos(45) would be the x component for the 45 degree right triangle, and cos(30) would be for the 30 degree triangle.

Which would make the i comonent. 450*cos(45)cos(30), however the solution above has it as 450*cos(45) sin(30)
 
  • #4
Simon Bridge
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γ is measured from the z axis to F2. It seems that F2 is below the xy plane.
... and? Relate this answer back to your question:
The below solution states that γ>90°. Why must is be greater, than 90°, and when would it have been smaller?

What does your answer mean for the size of γ?
If F2 were exactly in the x-y plane, what would γ equal?
Since it is below the x-y plane, would γ be bigger or smaller than this value?



For the projection what I mean was for the 30degree the way my mind analyzes it is that we get the x components of both triangles e.g 30 and 45 degree.

Which would mean that 450*cos(45) would be the x component for the 45 degree right triangle, and cos(30) would be for the 30 degree triangle.

Which would make the i comonent. 450*cos(45)cos(30), however the solution above has it as 450*cos(45) sin(30)
Looks like you got the sines and cosines mixed up.
You got it right for if the 30deg angle was measured from the +x axis like normal - but it isn't.
It can help if you sketch the triangle out separately and label the sides before working out which is the adjacent and which the opposite sides.
 
  • #5
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Oh because since it is below the xy plane that would mean it would be in either quadrants 3 or 4, but would that mean it should be γ>180 not 90?. I do understand though that the max a triangle can be is 180 I just don't understand that relation.

For the projection of the 30 degree triangle I now see that it is on the yz plane. But still dont see how that would help me. Even if I labeled the sides, for example I know that cos(30) would be on the y axis now. But I cant see how that is in any way related to the X axis.
 
  • #6
Simon Bridge
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Oh because since it is below the xy plane that would mean it would be in either quadrants 3 or 4, but would that mean it should be γ>180 not 90?. I do understand though that the max a triangle can be is 180 I just don't understand that relation.
Careful: ##\gamma## cannot be bigger than 180deg.

What is the angle between the z-axis and anywhere on the x-y plane?

For the projection of the 30 degree triangle I now see that it is on the yz plane.
It isn't. But the angle is to the y axis in the x-y plane. The formula ##x=r\cos\theta## only works for when ##\theta## is measured from the x axis.

But still don't see how that would help me. Even if I labeled the sides, for example I know that cos(30) would be on the y axis now. But I cant see how that is in any way related to the X axis.
If ##\cos(30)## is on the y axis, then what is the trig function that would be on the x axis?
 

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