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[Air]

Homework Statement

Find the mass:

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?

 

[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?

 

[Air]

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.

 

[Hootenanny]

So, we know that since the system is in equilibrium, the sum of the moments about any point should be zero. However, let's 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?

 

[Air]

So, we know that since the system is in equilibrium, the sum of the moments about any point should be zero. However, let's 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.

 

[Hootenanny]

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

 

[Air]

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?

 

[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?

 

[Air]

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).

 

[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?

 

[Air]

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.

 

[Air]

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?

 

[arildno]

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

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