See also: An Atwood's Machine (involves tension, torque)

You are given a system that is at rest; you know the **mass** of the object, and the **two angles** of the strings. In this example problem, there are two strings, one with an angle of 25 degrees, and the other with an angle of 65 degrees, and a mass: 5 kilograms. **Label the tension from the strings as T1 and T2, respectively. **

The first thing to notice is that because the system is at rest, the forces in the x and y directions balance each other out. We can now create two equations.

We first create an equal for the horizontal forces.
$$T_1\cos(25) = T_2\cos(65)$$
This is because the X-vector of each string equals Force $\cdot\cos(\theta)$.
We now have to balance the vertical forces.
$$T_1\sin(25) + T_2\sin(65) = F_g$$
$F_g$ is the only force we know. It equals mass $\cdot$ gravity, which is $5 \cdot 9.8 = 49$. So, we can make these equations:
$$T_1 \sin(25) + T_2\sin(65) = 49$$
$$T_1 \cos(25) = T_2\cos(65)$$
So, by expressing $T_2$ in terms of $T_1$ and plugging it back into the first equatino, we can find a value for $T_1$, then figure out $T_2$.

## Doing the Math

This section is just solving the problem. We have no more physics left to do, just algebra. We start by isolating $T_2$ in the section equation.
$$T_2 = T_1 \dfrac{\cos(25)}{\cos(65)}$$
$$T_2 = 2.145 T_1$$
Now, we plug this back into the first equation.
$$T_1\sin(25) + (2.145 T_1) \sin(65) = 49$$
$$\left(\sin(25) + 2.145\sin(65)\right)T_1 = 49$$
$$2.366T_1 = 49$$
$$\boxed{T_1 = 20.708 N}$$

We now plug $T_1$ into the first equation. $$T_2 = 2.145(20.708) = \boxed{44.4193 N}$$

Now, we have our final answer. The tension in string 1 is $20.708 N$, and the tension in string 2 is $44.419 N$.

We now plug $T_1$ into the first equation. $$T_2 = 2.145(20.708) = \boxed{44.4193 N}$$

Now, we have our final answer. The tension in string 1 is $20.708 N$, and the tension in string 2 is $44.419 N$.

David Witten