Calculation Help Needed

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  • #1
Air
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Homework Statement


Find the mass:
FindMass.jpg


Homework Equations


Not Sure.


The Attempt at a Solution


I don't know where to start as I believe that there is less information.

All that I have got upto is Tan-1(3/8) but I don't know if this is needed to find the mass? :confused:
 

Answers and Replies

  • #2
Hootenanny
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I'm assuming the frame is massless, pivoted bout the right angle of the triangle and the system is in equilibrium, am I correct?
 
  • #3
Air
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Hootenanny said:
I'm assuming the frame is massless, pivoted bout the right angle of the triangle and the system is in equilibrium, am I correct?
Yes, You are correct.
 
  • #4
Hootenanny
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So, we know that since the system is in equilibrium, the sum of the moments about any point should be zero. However, lets consider the forces; if the system is not translating (i.e. moving up or down) what must be the vector sum of the vertical forces?
 
  • #5
Air
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Hootenanny said:
So, we know that since the system is in equilibrium, the sum of the moments about any point should be zero. However, lets consider the forces; if the system is not translating (i.e. moving up or down) what must be the vector sum of the vertical forces?
Erm...Not sure but guessing zero. :redface:
 
  • #6
Hootenanny
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Sounds good to me :smile: and where will this reation force be applied?
 
  • #7
Air
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Hootenanny said:
Sounds good to me :smile: and where will this reation force be applied?
Either at the angle at the end of Y or near the mass?
 
  • #8
Hootenanny
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Normal forces are applied at the point where the system pivots. Just checking, is the side x attached to a wall of some kind?
 
  • #9
Air
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Hootenanny said:
Normal forces are applied at the point where the system pivots. Just checking, is the side x attached to a wall of some kind?
Nope. The pivot is at the corner of x and y.

x and y dimensions are given if to calculate the angle (I think).
 
  • #10
Hootenanny
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Okay, let us consider the torques about the pivot point;

[tex]0.08T\sin\left(\arctan\left(\frac{3}{8}\right)\right) = 0.4mg[/tex]

Where T is the tension in the diagonal member. Do you follow?
 
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  • #11
Air
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Hootenanny said:
Okay, let us consider the torques about the pivot point;

[tex]0.08T\sin\left(\arctan\left(\frac{3}{8}\right)\right) = 0.4mg[/tex]

Where T is the tension in the diagonal member. Do you follow?
Yes, I do. :smile:
 
  • #12
Air
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Hootenanny said:
Okay, let us consider the torques about the pivot point;

[tex]0.08T\sin\left(\arctan\left(\frac{3}{8}\right)\right) = 0.4mg[/tex]

Where T is the tension in the diagonal member. Do you follow?
Can Tension be calculated without mass? :confused:
 
  • #13
arildno
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You still have not set up the equalibrium of FORCES, only of torques..

Also, for your information,
[tex]\sin(\arctan(x))=\frac{x}{\sqrt{1+x^{2}}}[/tex]
 

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