Recognitions:
Gold Member

## 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?

 PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
 Blog Entries: 1 Recognitions: Gold Member Science Advisor Staff Emeritus I'm assuming the frame is massless, pivoted bout the right angle of the triangle and the system is in equilibrium, am I correct?

Recognitions:
Gold Member
 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.

Blog Entries: 1
Recognitions:
Gold Member
Staff Emeritus

## 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?

Recognitions:
Gold Member
 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.

 Blog Entries: 1 Recognitions: Gold Member Science Advisor Staff Emeritus Sounds good to me and where will this reation force be applied?

Recognitions:
Gold Member
 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?

 Blog Entries: 1 Recognitions: Gold Member Science Advisor Staff Emeritus Normal forces are applied at the point where the system pivots. Just checking, is the side x attached to a wall of some kind?

Recognitions:
Gold Member
 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).

 Blog Entries: 1 Recognitions: Gold Member Science Advisor Staff Emeritus Okay, let us consider the torques about the pivot point; $$0.08T\sin\left(\arctan\left(\frac{3}{8}\right)\right) = 0.4mg$$ Where T is the tension in the diagonal member. Do you follow?

Recognitions:
Gold Member
 Quote by Hootenanny Okay, let us consider the torques about the pivot point; $$0.08T\sin\left(\arctan\left(\frac{3}{8}\right)\right) = 0.4mg$$ Where T is the tension in the diagonal member. Do you follow?
Yes, I do.

Recognitions:
Gold Member
 Quote by Hootenanny Okay, let us consider the torques about the pivot point; $$0.08T\sin\left(\arctan\left(\frac{3}{8}\right)\right) = 0.4mg$$ Where T is the tension in the diagonal member. Do you follow?
Can Tension be calculated without mass?

 Recognitions: Gold Member Homework Help Science Advisor You still have not set up the equalibrium of FORCES, only of torques.. Also, for your information, $$\sin(\arctan(x))=\frac{x}{\sqrt{1+x^{2}}}$$