ExamKrackers Torque Question

[Mountaineurons]

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I have been seriously struggling with Torque using ExamKrackers. It's the first time in using their materials I have hit a wall, where no matter how much time I have spent trying to understand, I have been unable to get answers to Torque problems. Has anyone encountered this problem - EK Physics Book Lecture 3 Number 60? Even if not using EK, does anyone have a strategy that works who could walk be through this problem?

Question 60:
A one meter board with uniform density, hangs in static equilibrium from a rope with tension T. A 3 kg weight hangs from the left end of the board, and the rope with tension lies 0.2 m to the right of the weight. What is the mass of the board?

 

[cloak25]

Since the board is 1m long, the center of mass (Cm) is 0.5m. The weight is a distance 0.2m from where the rope hangs so it is 0.3m from the Cm. The force on the right side is 0.8m away from the rope and is also 0.3m away from the Cm. So...

torque on left side = torque on right side
x = mass of board
0.2(3)(10) = 0.3(x)(10)
6 = 3x
x = 2 kg

is that the answer?

edit: heres a picture I drew http://imgur.com/cqud7

 

[Mountaineurons]

That is the correct answer - thanks. A question regarding your approach:

My understanding is that the lever arm is the distance from the force to the point of rotation, which here we are saying is at the center of mass. You have the force due to the weight on the left side of your equation multiplied by a lever arm of 0.2 m, when the force is actually being applied 0.5 m from the point of rotation. So, why would you not use 0.5 m as your lever arm on the left?

 

[cloak25]

That is the correct answer - thanks. A question regarding your approach:

My understanding is that the lever arm is the distance from the force to the point of rotation, which here we are saying is at the center of mass. You have the force due to the weight on the left side of your equation multiplied by a lever arm of 0.2 m, when the force is actually being applied 0.5 m from the point of rotation. So, why would you not use 0.5 m as your lever arm on the left?

The pivot point is at 0.2m (where the rope is attached to the board) not 0.5m (the center of mass).

edit: found this on yahoo answers. Maybe it will be of better help! http://answers.yahoo.com/question/index?qid=20100816175342AAYBZcM

 

[Mountaineurons]

I think I'm starting to get it! Somehow I thought the Tension in the rope was contributing to Torque - but it looks like that it's not, it just serves as the point of rotation.

Thank you so much! I really appreciate you talking the time to explain.

 

[Class1P]

the tricky part about this problem is that ( I learned this from TBR) if the pivot is not at the exact center you have to factor in the weight of the lever.

The weight of the lever (the board) puts a torque equal to the block on the other side. To do these problems you act as if the entire weight of the board is at one point in the direct middle of the board.

So in this case the weight of the board (X*10m/s^2) is exactly at the 0.5m mark. So you just pretend that there is a weight at 0.5m equal to the weight of the board.

So in this case the weight of the board at 0.5m is 0.3m from the fulcrum because the fulcrum is 0.2m away from the other side.

Since the weights are balanced the torque of one side equals the other.

so : 3kg * 10m/s^2 * 0.2m = Xkg * 10m/s^2 * 0.3m

Solve for X and you get 2kg

 

[Mountaineurons]

Thank you - that's helpful

 

[HawaiiMedMan]

I was going crazy trying to figure this one out. Thanks for the posts guys.