Answer:
The force at right side is 1058N
Step-by-step explanation:
The concept that we need to use to give a solution to this problem are the Static Equation where the force and torque do not experience an acceleration.
Our values are:

We have two tension from the wire, at left and right, then making sum we have




We can do now a sum of moments at right side, then




Replacing in the first equation,



Therefore the force at right side is 1058N