# Find theta 1 and theta 2. Tension exercice.

#### topsquark

Forum Staff
Let's use the point where the three strings meet as an object for a FBD.

There are three tensions, T1 acting up and to the right, T2 acting up and to the left, and T3 acting downward.

Okay, Newtons's 2nd.
Since there only two tensions acting in the horizontal direction, and that the point is stationary then we know

$$\displaystyle T_1 ~ cos( \theta _1 ) - T_2 ~ cos( \theta _2 ) = 0$$

In the vertical direction we get
$$\displaystyle T_1 ~ sin( \theta _1 ) + T_2 ~ sin( \theta _2 ) - T_3 = 0$$

Now, we know that $$\displaystyle T_1 = (1.50 ~ \text{kg} )g$$ (why?) and likewise $$\displaystyle T_2 = (2.00 ~ \text{kg} )g$$ and $$\displaystyle T_3 = (2.5 ~ \text{kg} )g$$.
(Make sure you know why this is.)

So, you've got two equations (the Newton's Law equations) in two unknowns. Can you take it from here?

-Dan

Cervesa

#### Henistein

I am sorry but, the answers are theta 1 = 36.9º and for theta 2 = 53,1º. In fact, your explanation makes sense but could you help me to get a final answer, beacuse I am struggling in this exercise for some hours. I am sorry again for the incovenience.

#### topsquark

Forum Staff
I am sorry but, the answers are theta 1 = 36.9º and for theta 2 = 53,1º. In fact, your explanation makes sense but could you help me to get a final answer, beacuse I am struggling in this exercise for some hours. I am sorry again for the incovenience.
Go ahead and post what you were able to do. It would be better to find where you went wrong rather than give you the answer.

-Dan

#### studiot

Or better have another go.

But instead of playing with trigonometry, try to make four equations (it's not that bad) so that

$$\displaystyle a = \cos {\theta _1}$$

$$\displaystyle b = \cos {\theta _2}$$

$$\displaystyle c = \sin {\theta _1}$$

$$\displaystyle d = \sin {\theta _2}$$

$$\displaystyle {a^2} + {c^2} = 1$$
$$\displaystyle {b^2} + {d^2} = 1$$