Answer:
F=265.39N
Step-by-step explanation:
In order to solve this problem, we must first draw a diagram of what the problem looks like (see attached picture).
So, in order to find the torque we must use the torque formula:
![\tau=rxF](https://img.qammunity.org/2021/formulas/engineering/college/pz0oskizuscsweeaxj73tk8mb0nli83kog.png)
or
![\tau=rFsin\theta](https://img.qammunity.org/2021/formulas/physics/high-school/xm2wng521js0e63px2j2mj0elyptjdc8we.png)
In this case we can either use 125° or 55°, the answer should be the same. So when solving the formula for F we get:
![F=(\tau)/(r sin\theta)](https://img.qammunity.org/2021/formulas/engineering/college/2zr9kz5se3t8baq722nh4ltdnckyf4ah3r.png)
so we can now substitute values:
![F=(50Nm)/((23x10^(-2)m)sin 125^(o))](https://img.qammunity.org/2021/formulas/engineering/college/x3wgk77qmj1xqr6y4q6259hb2pfag8h6ne.png)
so
F=265.39N