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Calculation Help Needed...

 
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Apr7-07, 07:45 AM   #1
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Calculation Help Needed...


1. The problem statement, all variables and given/known data
Find the mass:


2. Relevant equations
Not Sure.


3. 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?
 
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Apr7-07, 07:48 AM   #2
 
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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?
 
Apr7-07, 08:51 AM   #3
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Quote by Hootenanny
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.
 
Apr7-07, 08:57 AM   #4
 
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Calculation Help Needed...


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?
 
Apr7-07, 08:59 AM   #5
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Quote by Hootenanny
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.
 
Apr7-07, 09:06 AM   #6
 
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Sounds good to me and where will this reation force be applied?
 
Apr7-07, 09:12 AM   #7
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Quote by Hootenanny
Sounds good to me and where will this reation force be applied?
Either at the angle at the end of Y or near the mass?
 
Apr7-07, 09:18 AM   #8
 
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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?
 
Apr7-07, 09:21 AM   #9
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Quote by Hootenanny
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).
 
Apr7-07, 09:31 AM   #10
 
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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?
 
Apr7-07, 09:38 AM   #11
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Quote by Hootenanny
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.
 
Apr7-07, 09:56 AM   #12
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Quote by Hootenanny
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?
 
Apr7-07, 02:50 PM   #13
 
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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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