Finding the magnitude of $\vec{F}$

Question

The object with the weight of $80N$ is balanced in the force of $\vec{F}$. So, I want to find the magnitude of $\vec{F}$.

Assuming that the object is in equilibrium, then we have

$$F\sin{37} - 80 = 0$$

Which gives

$$F\sin37=80$$

Thereby, we get

$$F = \dfrac{80}{\sin 37} = \dfrac{80}{0.6} = 133.33N$$

Am I right?

Regards!

Answer 1

In the vertical direction we have that

$$F+F\sin 37° =80 \implies F=\frac{80}{1+\sin 37°}=49.94 N$$