Intro to static equilibrium

In summary, the equation that results from choosing the pivot point to be the right end of the plank (where T_L acts) is W(x)+T_R(L) = T_L. This equation is expressed in terms of T_L, T_R, W, and the dimensions L and x. However, it is important to note that not all of these variables may show up in the solution.
  • #1
jaded18
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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.
____
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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  • #2
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?
 
  • #3
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.
 
  • #4
is it -(T_L)(x/L)+(T_R)((L-x)/L)??
 
  • #5
jaded18 said:
is it -(T_L)(x/L)+(T_R)((L-x)/L)??

Why are you dividing by L?
 
  • #6
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?
 
  • #7
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.
 
  • #8
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..
 
  • #9
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?
 

1. What is static equilibrium?

Static equilibrium is a state in which all forces acting on an object are balanced, resulting in no movement or acceleration of the object.

2. What is the difference between static and dynamic equilibrium?

Static equilibrium occurs when an object is at rest, while dynamic equilibrium occurs when an object is moving at a constant velocity.

3. How do you determine if an object is in static equilibrium?

To determine if an object is in static equilibrium, you must first draw a free body diagram and analyze all the forces acting on the object. If the sum of all forces is equal to zero, then the object is in static equilibrium.

4. What is the significance of static equilibrium in real-world applications?

Static equilibrium is important in many real-world applications, such as building structures and bridges, as it ensures that the forces acting on these structures are balanced and can withstand external forces without collapsing or breaking.

5. Can an object be in static equilibrium if it is moving?

No, an object cannot be in static equilibrium if it is moving. Static equilibrium only occurs when the net force on an object is zero, meaning there is no acceleration or movement. If the object is moving, then there must be unbalanced forces acting on it.

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