How Do You Calculate Torque in a Static Equilibrium Problem?

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To calculate torque in a static equilibrium problem, the sum of forces and torques must equal zero. The tension forces T_L and T_R, along with the weight W, contribute to the net torque about a chosen pivot point. When selecting the pivot at the point where the weight W acts, the equation involves the distances from the pivot to the points where T_L and T_R act. The signs of the torques depend on the direction of rotation they induce; T_L creates clockwise torque (negative) while T_R induces counterclockwise torque (positive). Understanding the correct distances and signs is crucial for deriving the accurate torque equations.
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http://session.masteringphysics.com/problemAsset/1007460/26/55544.jpg

What is the equation that results from choosing the pivot point to be the point from which the mass hangs (where W acts)? Express your answer in terms of the unknown quantities T_L and T_R and the known lengths x and L. Recall that counterclockwise torque is positive.
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can anyone please give me hints as to how to start this? I know that sum of net external forces in this picture is 0 so sum of forces in the y direction is T_L +T_R - W = 0 and .. i also know that sum of external forces and torque = 0 ...
 
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Think about what forces in the problem will contribute to net torque. If you choose the pivot point to be at the the place where the string is attached, you know that it will not count towards net torque.

What about the center of mass of the rod? Does the gravitational force on it not count towards the net torque as well? Is the rod massless?
 
jaded18 said:
http://session.masteringphysics.com/problemAsset/1007460/26/55544.jpg

What is the equation that results from choosing the pivot point to be the point from which the mass hangs (where W acts)? Express your answer in terms of the unknown quantities T_L and T_R and the known lengths x and L. Recall that counterclockwise torque is positive.
____
can anyone please give me hints as to how to start this? I know that sum of net external forces in this picture is 0 so sum of forces in the y direction is T_L +T_R - W = 0 and .. i also know that sum of external forces and torque = 0 ...
The problem is asking you to sum torques about "W" (and then set them equal to 0 as you have noted). So determine the torque of each of the forces about W, add them algebraically to get the equation.. Watch plus and minus signs and distances. Start by writing down the formula for the torque of each force about a point.
 
is it -(T_L)(x/L)+(T_R)((L-x)/L)??
 
jaded18 said:
is it -(T_L)(x/L)+(T_R)((L-x)/L)??

Why are you dividing by L?
 
Sorry . Torque = Fr .. Revised: -(T_L)(x)+(T_R)((L-x))

can you please explain to me how the signs work, and if my signs are correct?
 
Imagine someone pulling up on the left side of the rod; the other side would go down right? It would rotate clockwise. Now consider if someone was pulling up on the right side of the rord. The other side would go down and the rod would be rotating counterclockwise. It is the same concept here. The tension force on the left side tends to create a clockwise (negative) rotation while the tension force on the right side tends to create a counterclockwise (positive) rotation.
 
What is the equation that results from choosing the pivot point to be the right end of the plank (where T_R acts)?
Express your answer in terms of T_L, T_R, W, and the dimensions L and x. Not all of these variables may show up in the solution.

what about this one then? would it be something like T_R(L-x) - W = 0??

and i assume my answer to the first question was correct?

thanks by the way..
 
jaded18 said:
What is the equation that results from choosing the pivot point to be the right end of the plank (where T_R acts)?
Express your answer in terms of T_L, T_R, W, and the dimensions L and x. Not all of these variables may show up in the solution.

what about this one then? would it be something like T_R(L-x) - W = 0??

and i assume my answer to the first question was correct?

thanks by the way..

No, because the tension from the right side does not contribute to the torque if it is applied at the pivot point. And also, what is the distance from the place where the weight is applied to the pivot point?
 
  • #10
soooo... mhm... if T_R doesn't contribute... then we use T_L and the distance from where the weight is applied to the pivot point at the right would then be ... still (L-x)? so ... my final answer is -T_L(L-x) - W??
 
  • #11
The distance from where the weight force is applied to the pivot point is (L-x) but I do not see that in your equation.

Also, the tension in the left side is not (L-x) from the pivot point and the sign on the weight is wrong.

I know it must be mind-bogglingly frustrating that I am not giving you a direct answer but I want you to understand why your equation and reasoning is not correct.
 
  • #12
edit: i think i get it ...
 
  • #13
jaded18 said:
WOW, this is incredibly confusing :( . you just said that T_R doesn't contribute. so now i am using T_L and now i say that the tension in the left side is (x) from the pivot point. after taking everything that you have said into account, i come up with -T_L(x)+W. it's saying that the variable L needs to be in the answer as well! ... and don't worry about it. although it does get frustrating at times, I'm getting used to it in this forum

Re-read your question. The pivot point is no longer at the point where the weight force is applied; it is at the right side where the right tension is applied. If it was still at the point where the weight is applied, you would have the same answer from your first question as before and you wouldn't have weight in your equation at all.

P.S. That means your equation is still wrong.
 
  • #14
What is the equation that results from choosing the pivot point to be the right end of the plank (where T_L acts)?
Express your answer in terms of T_L, T_R, W, and the dimensions L and x. Not all of these variables may show up in the solution.
I got W(x)+T_R(L) = T_L, but keeps telling me that am wrong, why?
 
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