physics equilibrium problem of see-saw?

Σtorque around the support = 0

-A(3) + 200(2) + 500(2 + 5) = 0

A = 1300 N

ΣVertical = 0

F - A - 200 - 500 = 0

F - 1300 - 200 - 500 = 0

F = 2000 N

to balance the plank we must have equal moments from both sides of the pivot where the moment = force x distance x sin ( F,d ) let's place the pivot at a distance d away from the box. On one side of the pivot we have the weight of the box and on the other side we have the weight of the plank; both forces acting vertically downward so the angle that the force makes with the distance is 90degree moment (of box's weight ) = moment (of plank's weight) the reason is we must keep the plank balanced so weight(box) x distance(away from pivot) = weight(plank) x distance (away from pivot) m(box) x g x d = m(plank) x g x (0.80 m - d ) where weight of plank is applied at midpoint of the plank,so 1.6m/2 = 0.80m (12kg)(d) = 24kg(0.80m - d ) where g is the gravity and is canceled from both sides solve the equation,we get : d = (0.80)(24) / 36 = 0.53m away from the box

I assume you have to find the force at point A in order to have the system be in equilibrium.

You simply make the net torque equal zero.

(2m)*(200N) + (7m)*(500N) = (3m)*X

X = 1300 N