Finding the Tension of Two Strings with Different Angles

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 equation, 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 $44.419 N$.