# Calculation Help Needed

Find the mass:

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?

Related Precalculus Mathematics Homework Help News on Phys.org
Hootenanny
Staff Emeritus
Gold Member
I'm assuming the frame is massless, pivoted bout the right angle of the triangle and the system is in equilibrium, am I correct?

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.

Hootenanny
Staff Emeritus
Gold Member
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?

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.

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

Hootenanny said:
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
Staff Emeritus
Gold Member
Normal forces are applied at the point where the system pivots. Just checking, is the side x attached to a wall of some kind?

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).

Hootenanny
Staff Emeritus
Gold Member
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:
Hootenanny said:
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.

Hootenanny said:
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
$$\sin(\arctan(x))=\frac{x}{\sqrt{1+x^{2}}}$$