Board pivoted at one end with rope pulling down at L/3

[jaded18]

A uniform board of mass m and length L is pivoted on one end and is supported in the horizontal position by a rope attached to the other end. Another rope, attached to the board a distance L/3 from the pivot point, is being pulled straight down with a constant force of magnitude F.

1) How far is the axis of rotation from the center of mass of the board?
Express your answer in terms of quantities given in the problem introduction.

2) How far is the axis of rotation from the point of application of force F?
Express your answer in terms of quantities given in the problem introduction.
____
ok... what? There are no diagrams for this Q. And I'm a visual person. Care to explain what's going on?

 

[Anadyne]

The board is horizontal right now. The length of the board is L so if you pretend there's a coordinate plane: the pivot at one end is at x=0 and the other end is at x=L. Can you fill in the rest?

 

[jaded18]

i think so ... so if the axis of rotation is L/2 from the center of mass of the board ... then the axis of rotation is L/3 from the point of application of force F? I'm not quite sure on the second part

 

[jaded18]

Ah, I think I see it now... So: Suddenly, the rope attached to the end of the board breaks. How many forces are acting on the board after the rope attached to the end of the board breaks?

First there were three forces... one on each side and one from below? and now there is only one??

 

[Anadyne]

The only difference is that the force at L is gone, so now there are two forces. Which one did you forget?

 

[jaded18]

actually there are three... what are they?? the ones on the left and right and gravity?

 

[learningphysics]

so the answers are 1)L/2 and 2)L/3

are there more parts to the problem?

 

[jaded18]

another Q: Find the angular acceleration alpha of the board immediately after the rope breaks.

so i know that Torque = I(alpha) ... mhm... how would i go about using the quantities introduced in #1 box to find the net torque ...

 

[jaded18]

here's the diagram! http://session.masteringphysics.com/problemAsset/1000056/16/104883.jpg

 

[learningphysics]

here's the diagram! http://session.masteringphysics.com/problemAsset/1000056/16/104883.jpg

So the rope at the end (at a distance L) breaks?

Then you'd say the

torque about the pivot = (moment of inertia about the end)*alpha

for the torque, use the F force and the weight mg...

 

[jaded18]

thanks for the response :)