Why is this so?Why Does the Scale Show a Different Weight Than Expected?

[KingNothing]

Hi everyone...this one is really really confusing me.

My teacher showed a picture up on the overhead of a diagram sort of like this (I attached it).

Basically a little scale thing is being pulled by 100N weights on btoh sides (we assume that this 100 N each is fully transferred so its 100N sideways.).

Anyhoo, the question was: what will the scale show for the weight/force? I answered 200, the correct answer was 100. My teacher swears on it. I just don't understand how it can be that way. The curly things on the right of the scale are springs and the circles are pulleys.

Can someone explain this to me?

 

[Hydr0matic]

What would happen if you removed the left weight and instead held the scale with your hand ? - the scale would show 100N right ? .. And what force do you need to apply with your hand to hold the scale & weight ? - 100N. So in a sense, the left weight is only "Holding" the right weight. ... or the other way around.

 

[HallsofIvy]

Or the opposite of Hydr0matic's suggestion: suppose there were a 100N force pulling on one side and NO force on the other! Now what would the scale read? What would happen?

There ALWAYS has to be the same force pulling on both sides in order for a scale to register a correct value.

(To me the mystery is why all the scales I use register about 20 pounds more than I KNOW I weigh!)

 

[gta-maloy]]

A good way to explain this as I know it would be:

Take a spring loaded scale that only needs weight aplied to register the Newtons, it would read 100.

Now I know what makes you think 200N, 200N is the force applied to the "shaft" of the scale, yet registered Newtons are 100 because to find out the weight or force of something a counter force/weight needs to be applied.

 

[KingNothing]

So, the actual force would be 200 N, yet the scale would show 100N?

 

[Janitor]

No, the actual tension in the shaft is 100 N.

 

[KingNothing]

I see. So one could conclude that the tension on the scale is always equal to the lesser of the two forces applied, because any difference in them simply results in increased acceleration rather than tension?

 

[Doc Al]

So one could conclude that the tension on the scale is always equal to the lesser of the two forces applied, because any difference in them simply results in increased acceleration rather than tension?

I'm not sure what you are saying here. If by "forces applied" you are referring to the hanging weights, then the answer is no. For example, if the 100N weight on the right is replaced by a 200N weight, then the masses will accelerate, and the tension in the cables (and the force measured by the scale) will turn out to be about 133N. (Apply Newton's 2nd law; I assume that the cables and scale have negligible mass.)

Just to be clear, if the mass of the scale is small enough to be ignored then you will find that the tension is the same on both sides of the scale--no matter what you try to do. This tension is the force that the scale will read.