Proof of Tan (theta) = 1 + (2M / m)

[daniel123]

In my Physics 231 class (Calculus Based) I was given a homework problem asking me to prove the following equation. It had a picture attached but the equation was:

Tan (theta) = 1 + (2M / m)

The two (m) represent different masses balanced in equilibrium while hanging with 3 tension forces none of which are connected to the other mass. If you need anymore information just post what you need below. Thank you.


|
/ |
/ |
| / (theta)
| /__________
| \ M
\ /
\ /
45 Degrees \ / 45 Degrees
______|______
|
m

 

[arildno]

Do you find your own image comprehensible? I don't.

First, describe the situation properly!

The two (m) represent different masses balanced in equilibrium while hanging with 3 tension forces none of which are connected to the other mass.

1. What is M? One of the two m's?
2. "balanced in equilibrium": By means of what??
Some sort of weight?
3. "while hanging with 3 tension forces none of which are connected to the other mass." Eeh?
First, are there any ropes in the image providing the tension?
Is each mass connected to 3 ropes?

And what, precisely, is the angle?

 

[Poncho]

Is this the setup?

$$Tan (\theta) = 1 + (\frac{2M}{m})$$

-Dave

 

[arildno]

Probably!
Or, at least, congrats that you managed to get more sense out of OP's description than I did.

 

[daniel123]

Is this the setup?

$$Tan (\theta) = 1 + (\frac{2M}{m})$$

-Dave

That is the setup, yes. The angles on either side of the smaller hanging mass are both 45 degrees. Sorry my drawing got messed up during the post. On my problem the angle identified as theta is the angle opposite of theta in yours.

 

[lightgrav]

looking at the knot that holds up m,
you find from x: T_left = T_middle ;
vertical T components total = mg.

The vertical component of T_right
has to hold up all of M and ½ of m.
(do you see the m+2M hiding?) :tongue2:

Now write the horizontal Sum(F) =0.