Rotational dynamics: TORQUE

Tags:
1. Nov 24, 2014

Nnk

1. The problem statement, all variables and given/known data

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?

2. Relevant 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

3. The attempt at a solution
I have no Idea how to find the distance, as when I tried it, the "L" cancels out...

2. Nov 24, 2014

SteamKing

Staff Emeritus
Show you calculations. You made a mistake somewhere.

3. Nov 24, 2014

Nnk

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

4. Nov 24, 2014

SteamKing

Staff Emeritus
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.

5. Nov 24, 2014

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)

Last edited: Nov 24, 2014
6. Nov 25, 2014

Nnk

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