How Does Child Movement Affect Torque and Force Distribution on a Plank?

AI Thread Summary
The discussion focuses on calculating the forces required to maintain static equilibrium on a plank when a child moves. Initially, the forces exerted by the parents are F1=(1/4)mg and F2=(3/4)mg. When the child shifts position, the father's force decreases to 0.60mg, prompting a reevaluation of the child's new position. Using torque and net force equations, the new position of the child is determined to be X=(0.4L), with the movement calculated as (3L/20). The calculations confirm the correct application of static equilibrium principles to find the child's displacement.
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Homework Statement


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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...
 
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Show you calculations. You made a mistake somewhere.
 
SteamKing said:
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
 
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)
 
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SteamKing said:
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 said:
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
 
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