# Calculation Help Needed

1. Apr 7, 2007

### Air

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?

2. Apr 7, 2007

### Hootenanny

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?

3. Apr 7, 2007

### Air

Yes, You are correct.

4. Apr 7, 2007

### Hootenanny

Staff Emeritus
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. Apr 7, 2007

### Air

Erm...Not sure but guessing zero.

6. Apr 7, 2007

### Hootenanny

Staff Emeritus
Sounds good to me and where will this reation force be applied?

7. Apr 7, 2007

### Air

Either at the angle at the end of Y or near the mass?

8. Apr 7, 2007

### Hootenanny

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?

9. Apr 7, 2007

### Air

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. Apr 7, 2007

### Hootenanny

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?

Last edited: Apr 7, 2007
11. Apr 7, 2007

### Air

Yes, I do.

12. Apr 7, 2007

### Air

Can Tension be calculated without mass?

13. Apr 7, 2007

### arildno

You still have not set up the equalibrium of FORCES, only of torques..

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