Rotational Dynamics prob...??? help needed?

Question

Please help out with this Rotational Dynamics prob...???
A rotational axis is directed perpendicular to the plane of a square. Two forces F1 and F2, (a) are applied to diagonally opposite corners, (b) and act along the sides of the square. In each case the net torque produced by the forces is zero. The square is one here meter on a side, and the magnitude of F2 is three times that of F1. Find the distances a and b that locate the axis.

There are two squares

in square (a) F1(leftside of box) is going right and F2(right side of box) is going downward.
in square (b) F1(left side of box) is going downward and F2(right side of box) is going left.

Please help! thX

Answer 1

You have is a couple

r1 x F1= r2 x F2
or
r1 F1 sin(A) = r2 F2 sin(A)
A- the angle between the arm r and force F

since F2=3F1
r1 F1 sin(A) = r2 3 F1 sin(A)
r1=3r2
and r1+r2=D= a sqrt(2) a- side of the square a=1m


r2 = D - r1 = 1.41-r1
finally since r1=3r2
r1=3(1.41-r1)
4r1=4.2
r1=1.06m
r2=r1 / 3=0.35 m

The center will lie on the diagonal
Use simple trig to compute a and b

Answer 2

permit the sector fall ( roll) via a height h P.E. on the authentic is comparable to the sum of the ok.E. by using translation and that by using rotation on the backside. mgh = 0.5mv^2 + 0.5 I?^2 ........(a million) I = (2/5)mr^2 v = r? consequently the above equation (a million) will become mgh = 0.5 mv^2 + (2/10)mv^2 or mgh = (7/10)mv^2 v^2 = (10/7)gh .....(2) Now use, v^2 = u^2 + 2ah the place v is the linear velocity on the backside , u is the preliminary velocity and "a" is the linear acceleration u = 0 and employing (2) (10/7)gh = 2ah a = (5/7)g