EK physics #61 (torque)

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yoyohomieg5432

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The question is attached. The answer is 16N i have no idea why. There explanation makes no sense. I'm getting 12N. (3m to hte right of the nail *4m force).

I attached the solution explanation as well

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Not sure how to make it clearer than the solution is. But basically for board Y not to move it MUST be in static equilibrium. That means sum of all forces and torques must be zero!

Choose for your pivot point the point about which the board would move if the nail weren't holding it in place.... it would rotate around the end of the board that is touching board X. So you make that the origin of a coordinate system and do sum of all torque = 0. I use the convention that a counterclockwise rotation is positive and a clockwise rotation is negative.

So from the end of the board you get:

sum of torques = (4 m)*(4 N) + (1m)*(F_nail) = 0

Solving yields F=16N

Note that for these problems, just like the solution "NOTE", you can actually choose WHATEVER point along the board you want and set up a torque equation. You usually pick a point that will give you ONLY two torques.. 1) the one that you know (a given force and distance), and 2) one that you don't. That way you end up with one equation and one unknown.

For example, if you choose the pivot point to be the nail and set up another static equilibrium equation:

Sum of all torque on board Y = 0.
sum of torques = (3 m)*(4 N) - (1 m)*(x) = 0

Solving for x gives you 12N, which must be the force of board X on the edge of the board Y to stop it from rotating. That's where they got the 12N at the edge of the board in the solution figure.. You don't need that to get the 16N answer above, but their point is that if the system is in static equilibrium you can choose any point along perpendicular to the acting forces and the sum of torques must be zero. Once you know that, you can choose whatever point you want on the board that will give you the unknown force or distance and a bunch of torques you do know. From there you can solve.
 
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Not sure how to make it clearer than the solution is. But basically for board Y not to move it MUST be in static equilibrium. That means sum of all forces and torques must be zero!

Choose for your pivot point the point about which the board would move if the nail weren't holding it in place.... it would rotate around the end of the board that is touching board X. So you make that the origin of a coordinate system and do sum of all torque = 0. I use the convention that a counterclockwise rotation is positive and a clockwise rotation is negative.

So from the end of the board you get:

sum of torques = (4 m)*(4 N) + (1m)*(F_nail) = 0

Solving yields F=16N

Note that for these problems, just like the solution "NOTE", you can actually choose WHATEVER point along the board you want and set up a torque equation. You usually pick a point that will give you ONLY two torques.. 1) the one that you know (a given force and distance), and 2) one that you don't. That way you end up with one equation and one unknown.

For example, if you choose the pivot point to be the nail and set up another static equilibrium equation:

Sum of all torque on board Y = 0.
sum of torques = (3 m)*(4 N) - (1 m)*(x) = 0

Solving for x gives you 12N, which must be the force of board X on the edge of the board Y to stop it from rotating. That's where they got the 12N at the edge of the board in the solution figure.. You don't need that to get the 16N answer above, but their point is that if the system is in static equilibrium you can choose any point along perpendicular to the acting forces and the sum of torques must be zero. Once you know that, you can choose whatever point you want on the board that will give you the unknown force or distance and a bunch of torques you do know. From there you can solve.
I understand your solution for how you got 16N (thanks), but at the end you mentioned that you get 12N by doing it the other way. That is what I did. I don't understand why we get 2 different answers? Or is there another torque I'm supposed to be factoring in if I choose the point of rotation to be where the nail is?
 
Yes, they are two different torques! There are 3 torques in this problem in total, and if you put the origin of your coordinate system on top of one of them you will eliminate it from the sum of all torques.

Both the torques in question here are acting at the same time to opposite the applied 4N force (which creates the third torque), but depending on how you choose your axis you can get the one you don't care about (isn't asked about) to disappear. When you choose your origin, any torques through that point do not show up in your sum of all torques because the distance from the "axis of rotation" you are choosing is zero, and the torque due to any force is Force times Distance from Axis. In that way, you end up making whatever torques go through that point disappear. In other words, if you have a door attached to a hinge, and you press on the door, it will swing due to torque producing angular acceleration. If you press really hard in the same direction right on the hinge, the door won't move because while F > 0, the distance from the axis of rotation equals zero, therefore the torque ends up being zero and the door won't rotate around the hinge.

You're doing the mathematical equivalent of that in this problem when you choose to analyze the system from the end of board Y. That 12N force IS at that point the whole time, but it produces no torque because it goes through the axis of rotation of board Y (i.e. the center of our coordinate system).

However, if we move our coordinate system and "assume" our board is rotating around the nail, we can get the nail torque to go to zero and calculate the torque at the edge of the board instead (12N).

The lesson here is this: If you're trying to calculate a torque, DON'T choose your coordinate system origin to be at the point you care about (which is opposite of advice that works for most other physics problems) otherwise you'll be getting an equation for every torque EXCEPT the one you care about.

Also, consider this: If someone/something is applying at 4N force to the free end of board Y, and the friction force is 16N, why isn't board Y accelerating? After all, if a system is in static equilibrium, the sum of all torques AND forces must be zero. So you would expect that if the only two forces acting on the board were the 4N push and the 16N of friction, you'd expect the board to experience an accelerating force of 12N. Since the board isn't moving, we know there MUST be another 12N force opposing the motion of the board. So it's not coincidence that we calculated 12N to be the force at the edge of board Y. That's the force of the rigid board X pushing back against board Y when someone tries to lift the edge of board Y (4N) while they're nailed together.
 
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