Why is there more than 1 value for load W where there is no rotation?

In summary, the problem involves a 20 ft frame supported at two points, L and M, with a 300 lb load attached at point Q. The question is to determine the range of load W at point P that will keep the frame in equilibrium. The solution involves considering the vectorial torque and force summations, and explains that there are three ranges of W where the frame will be stable, with the middle range being the most stable. This is due to the fact that there are two fulcrums, L and M, which apply forces in opposite directions to prevent rotation. As W varies, the forces from the two fulcrums adjust to maintain balance, but only within certain ranges.
  • #1
s3a
818
8

Homework Statement


The problem along with its solution is attached as TheProblemAndSolution.jpg.

Here is the textual part of the attached image:
“In Fig. 1 a 20 ft-frame PQ is supported at two points L and M, 6 ft and 4 ft respectively from the edges. If a 300 lb load is attached to edge Q, determine the range of load W that must be placed at P to keep the frame in equilibrium.”

Homework Equations


Vectorial torque and force summations.

The Attempt at a Solution


I actually understand the mathematical explanation that the solution gives, but I am trying to fully understand this theoretically.

I get, mathematically, when we considering fulcrum L, for example, to be point from which we consider the rotation, that if W is toward the smaller value in its range, fulcrum M will apply a larger upward force to the frame, but how does the theory behind the vectorial sums take this into account?

Basically, can someone give me a non-mathematical (or less-mathematical) explanation to why there is more than one value for W where there is no rotation (i.e., where the net torque is zero)?

Any input would be greatly appreciated!
 

Attachments

  • TheProblemAndSolution.jpg
    TheProblemAndSolution.jpg
    25.1 KB · Views: 401
Physics news on Phys.org
  • #2
The two downward forces can be equated to a single net downward force at some point along the bar. With no force at P, the net force acts at Q. As the force at P becomes very large, it gets arbitrarily close to P. In between, there is a range of forces at P which make the net force act in MQ, another range where it acts in LM, and a third where it acts in PL. It will only be stable in the middle range.
 
  • #3
Sorry, I double-posted.
 
  • #4
Sorry, I didn't understand very well, and I'm still confused. :(

What do you mean by instability when it's not in the middle range (LM)? According to the math, W doesn't need to be in the middle range (LM) for the net torque to be zero.

When you say that “the two downward forces can be equated to a single net downward force at some point along the bar”, that seems like manipulation of theory, but I'm trying to get an intuitive understanding of what's going on, rather than just relying on the math.

Basically, what is preventing the rotation as the weight, W, varies? It must have something to do with the fact that there are two fulcrums (instead of one). One of the fulcrums must be applying a force oppositely directed (if you don't think about the torque vector being in or out of the page but rather as clockwise or counter-clockwise) to the direction of rotation that would have existed if there were only one fulcrum. I just can't see this more deeply; I'm looking for some way to confirm this with an explanation.
 
  • #5
s3a said:
What do you mean by instability when it's not in the middle range (LM)? According to the math, W doesn't need to be in the middle range (LM) for the net torque to be zero.
Apologies, I missed out something. I meant to say 'consider an arbitrarily large force'. You're right of course that for a more modest W the range is different, but the principle is the same: in one range the beam will tip to the right, in another it will be stable, in a third it will tip to the left.
When you say that “the two downward forces can be equated to a single net downward force at some point along the bar”, that seems like manipulation of theory, but I'm trying to get an intuitive understanding of what's going on, rather than just relying on the math.
I gave you what explains it intuitively to me :shy:. Not everyone's intuition works the same way. Consider this set-up: a large weight rests on a long table. If the weight is up one end of the table, the force from legs onto floor will be greatest at that end. As the weight is slid along to the middle, the forces reduce on those legs and increase on the others. We see the same in a see-saw. The see-saw balances if the two downward forces equate to a single downward force which lines up vertically with the upward force from the fulcrum.
What we have here is two parallel downward forces whose sum must line up with the sum of two upward forces. The two upward forces automatically adjust their relative magnitudes to achieve that, but neither can go negative, so they can only succeed in achieving balance when the downward sum lies between them.
Basically, what is preventing the rotation as the weight, W, varies? It must have something to do with the fact that there are two fulcrums (instead of one). One of the fulcrums must be applying a force oppositely directed (if you don't think about the torque vector being in or out of the page but rather as clockwise or counter-clockwise) to the direction of rotation that would have existed if there were only one fulcrum.
Yes.
 

1. Why does load W have more than one value when there is no rotation?

This phenomenon is due to the principle of static equilibrium in physics. When an object is at rest or not rotating, the sum of all forces acting on it must be equal to zero. This means that there can be multiple combinations of forces that can result in a state of static equilibrium, hence multiple values for load W.

2. How do the different values of load W affect the object's stability?

The different values of load W can affect the object's stability by changing the distribution of weight and forces acting on it. A higher value of load W may shift the center of mass and make the object more prone to tipping over, while a lower value may make it more stable.

3. Can the number of values for load W be infinite?

No, the number of values for load W is limited by the physical properties and dimensions of the object. As the size and shape of the object change, the number of possible values for load W also changes.

4. Are there any real-world applications for this concept?

Yes, this concept is important in engineering and design, particularly for structures and machines that need to remain in a state of static equilibrium. By understanding the different values of load W, engineers can design more stable and safe structures.

5. How can we determine the specific value of load W for a given object?

The specific value of load W can be determined through calculations and experiments. By analyzing the forces and moments acting on the object, engineers can solve for the value of load W that will result in a state of static equilibrium. This value can also be verified through physical testing on the object.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
4K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top