Rotational dynamics: TORQUE

[Nnk]

Homework Statement

A child of mass m is supported on a light plank by his parents, who exert the forces F1 and F2 as indicated.
Find the forces required to keep the plank in static equilibrium. Use the right end of the planks as the axis of rotation. ( Answer: F1=(1/4)mg, F2=(3/4)mg )
Suppose the child moves to a new position, with the result that the force exerted by the father is reduced to 0.60mg. How far did the child move?

Homework Equations

Torque = F*r
Net force acting on plant equal to zero: F1+F2-mg=0
Net torque acting on plant equal to zero: -F1(L) + mg(L/4) = 0

The Attempt at a Solution

I have no Idea how to find the distance, as when I tried it, the "L" cancels out...

 

[SteamKing]

Show you calculations. You made a mistake somewhere.

 

[Nnk]

Show you calculations. You made a mistake somewhere.

I did 0.60mg*L -mg(?L) = 0 because that's all I could think of, but obviously the L's cancel.. so I'm not sure how to do it

 

[SteamKing]

You apply the same equations of static equilibrium that applied to the original problem.

Torque = F*r
Net force acting on plank equal to zero: F1+F2-mg=0
Net torque acting on plank equal to zero: -F1(L) + mg(X) = 0

You know that F2 = 0.6 mg. X is the location of the child from his father.

 

[Malabeh]

Based on what SteamKing posted, I got .4L. Can anyone confirm?

Edit: X=(.4L) This is the new position, NOT the amount the baby moved.
So you have:

F1+F2-mg=0...(1)
-F1L+mg(X)=0...(2)

Solve for F1 in eq(1):
F1+(.6)mg-mg=0
F1=(.4mg)

Now plug into eq(2) and solve for X:
(-.4mg)L+mg(X)=0
mg(X)=(.4mg)L
X=(.4L)

Initial position was L/4, and new position is .4L=2L/5; so X(i)-X(f)=(2L/5)-L/4=(3L/20)

 

[Nnk]

You apply the same equations of static equilibrium that applied to the original problem.

Torque = F*r
Net force acting on plank equal to zero: F1+F2-mg=0
Net torque acting on plank equal to zero: -F1(L) + mg(X) = 0

You know that F2 = 0.6 mg. X is the location of the child from his father.

Based on what SteamKing posted, I got .4L. Can anyone confirm?

Edit: X=(.4L) This is the new position, NOT the amount the baby moved.
So you have:

F1+F2-mg=0...(1)
-F1L+mg(X)=0...(2)

Solve for F1 in eq(1):
F1+(.6)mg-mg=0
F1=(.4mg)

Now plug into eq(2) and solve for X:
(-.4mg)L+mg(X)=0
mg(X)=(.4mg)L
X=(.4L)

Initial position was L/4, and new position is .4L=2L/5; so X(i)-X(f)=(2L/5)-L/4=(3L/20)

Thank you! It is indeed correct. My mistake was using (X)L instead of X, which is why it went wrong