Two people place a motor on a $2$m uniform board that weighs $200$ newtons. They lift it, On one side one person lifts with a force of $600$ newtons and on another end the other person lifts it with a force of $400$ newtons. Where is the center of gravity of the motor.
Here is my work: I know that together it took them $1000$ newtons to lift, and I know the board weighs $200$. That leaves $800$ newtons as the weight of the motor. Let $x$ be the distance from the $400$ Newton's end to the center of gravity of the motor. Let the $400$ newton end be the pivot. The sum of the torques must add to $0$ so we have,
$$600(2)-800x-200(1)=0$$
Which suggests that $x=1.25$ meters from the $400$ newton force or $0.75$ meters from the $600$ newton force.
However this is not the answer, here is the answer.
It is problem $8$. I don't really understand their work, but It seems to me there are calculating the center of mass of the whole system instead. Am I right?
Thanks in advance.