How Is the Mass of a Board Determined Using Torque and Tension?

[andy7793]

Hi All,

I'm confused with a particular question and I'm not entirely sure how the answer was derived.

A one-meter board with uniform density hangs in static equilibrium from a rope with tension T--The rope is connected .2m from the left end. A weight of 3kg hangs from the left end of the board. What is the mass of the board?
-----> |T=?
__(.2m)___|____________________Board
|
|
3kg

A 1kg
B 2kg <---Answer
C 3kg
D 4kg

Thank you in advance!
-A

 

[Chestermiller]

I'm unable to understand your diagram. Can't you provide something better than that, like, say, a real figure uploaded using the snipping tool?

Chet

 

[andy7793]

Sorry about that! Here is the diagram and original question posted!

A

 

[Chestermiller]

Please draw a free body diagram of the board, showing all the forces acting on it.

Chet

 

[andy7793]

I'm not very good at drawing fbd for scenarios like this. Bare with me please &Thank you.

 

[billy_joule]

And what of the force you are asked to find? Where is the centre of gravity for a board with uniform density?

 

[andy7793]

Here is how the problem was reasoned through according to the book:

"The axis of rotation is the point where the rope attached to the board. The hanging weight creates a counter-clockwise torque equal to 3kg x 0.2m. The wight of the board creates a clockwise torque at the distance from the rope attachment to the board's center of mass, which is 0.3m. The net torque is zero, so the clockwise torque equals the counterclockwise torque, so 3kg x 0.2m=0.3m times the weight of the boards. Therefore the wight of the board is 2kg. "

Thank you!

 

[Chestermiller]

Here is how the problem was reasoned through according to the book:

"The axis of rotation is the point where the rope attached to the board. The hanging weight creates a counter-clockwise torque equal to 3kg x 0.2m. The wight of the board creates a clockwise torque at the distance from the rope attachment to the board's center of mass, which is 0.3m. The net torque is zero, so the clockwise torque equals the counterclockwise torque, so 3kg x 0.2m=0.3m times the weight of the boards. Therefore the wight of the board is 2kg. "

Thank you!

This is a correct analysis. It's too bad you had to refer to the book to get a solution.

Chet

 

[andy7793]

I know. I struggled with this one, especially with little practice with problems like this--Thank you

A