# Homework Help: Why is there more than 1 value for load W where there is no rotation?

1. Nov 16, 2013

### s3a

1. The problem statement, all variables and given/known data
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.”

2. Relevant equations
Vectorial torque and force summations.

3. 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!

#### Attached Files:

• ###### TheProblemAndSolution.jpg
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2. Nov 16, 2013

### haruspex

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. Nov 17, 2013

### s3a

Sorry, I double-posted.

4. Nov 17, 2013

### s3a

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. Nov 17, 2013

### haruspex

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.
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.
Yes.