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Trouble understanding the answer to this torque question

1. Homework Statement

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2. Homework Equations

The relevant equations are obviously going to be the rotational equilibrium expressions regarding balancing clockwise and counterclockwise movement as it relates to T = Fx lever-arm.

3. The Attempt at a Solution

First...here is the answer that the back of the book gives:

  1. "B is correct. The axis of rotation is the point where the rope attached to the board. The hanging weight creates a counter-clockwise torque equal to 3 kg x 0.2 m. The weight of the board creates a clockwise torque at the distance from the rope attachment to the board's center of mass, which is 0.3 m. The net torque is zero, so the clockwise torque equals the counterclockwise torque, so 3 kg x 0.2 m = 0.3 m times the weight of the board. Therefore, the weight of the board is 2 kg."
My question is.....why is it that when calculating the torque due to the weight of the board.....you measure only to center of mass, especially since there is no fulcrum there or anything. Why wouldn't you measure all the way until the end of the board (0.8m instead of 0.3m) and have your equation be:

(0.8 x mass of board) = (0.2 x mass of board) + (0.2 x 3kg) ----> mass = 1kg

If the answer in the book is correct.....and only that extra 0.3 meters between the rope (center of rotation) and center of mass is relevant...then how come the other 0.5 meters to the right of the board doesn't throw everything out of equilibrium...I mean it is extra weight after all, right?

It's clear I have some sort of fundamental conceptual hole in my understanding of torque as it pertains to equilibrium, but I'm having trouble filling that hole in...any help would be very appreciated, thanks!
 

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kuruman

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My question is.....why is it that when calculating the torque due to the weight of the board.....you measure only to center of mass, especially since there is no fulcrum there or anything.
The additional force of gravity (not shown in the picture) acts at the center of mass because it is an external force. Also note that if the sum of the torques is zero because the object is in static equilibrium, it doesn't matter about what point you calculate torques.
 

jack action

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(0.8 x mass of board) = (0.2 x mass of board) + (0.2 x 3kg) ----> mass = 1kg
The correct way to do it this way would be:
$$(0.2 \times 3) + (0.1 \times 0.2m) = (0.4 \times 0.8m)$$
Where you separate the beam into two masses:
  1. One 20% of the total beam mass ##m## where its center of gravity is at 0.1 m of the rope;
  2. One 80% of the total beam mass ##m## where its center of gravity is at 0.4 m of the rope.
You can also divide the beam into 10 smaller masses if you want:
$$(0.2 \times 3) + (0.15 \times 0.1m) + (0.05 \times 0.1m) = (0.05 \times 0.1m) + (0.15 \times 0.1m) + (0.25 \times 0.1m) + (0.35 \times 0.1m) + (0.45 \times 0.1m) + (0.55 \times 0.1m) + (0.65 \times 0.1m) + (0.75 \times 0.1m)$$
The answer is still the same. Note how the 2 masses on the left side are counterbalanced by their two similar masses, at similar distances, on the right side, thus having no effects on the resultant torque.

You could of course repeat the exercise by dividing the beam into 100s or 1000s of smaller masses and the results would still be the same. That is what a center of gravity defines: A equivalent point where the total weight can be applied without creating any resultant torque.
 
Last edited:
Wow, thanks so much! I'm getting closer to having a good intuition of it now. Since the force of gravity acts at the center of mass, with such a magnitude that the system is in equilibrium, it doesn't matter where the torque is calculated.

Thank you again! That was very helpful and kind.
 

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