How Is Acceleration Determined When a Weight Is Suspended on a String?

[redshift]

A weight weighing 0.2 kg is attached to a string that is raised/lowered vertically by hand. When the pulling force is 2.5 N, I'm supposed to find the acceleration.

Since the normal force in this case is N=mg=0.2(9.8)= 1.96 N, would the acceleration simply be the net force (2.5 N - 1.96 N = 0.54 N) divided by the mass?

Many thanks in advance

 

[Chen]

Why do you have a normal force? Isn't the weight hanging in the air? I guess you are talking about the string tension... so yes, your calcuation is correct.

 

[HallsofIvy]

Yes. When you are simply "holding" the weight, with no acceleration, you are applying a force equal to its weight, 0.2(9.8)= 1.96 N. It is the force beyond that, 2.5- 1.96 that causes the acceleration. Alternatively, you could divide the 2.5 N force by the mass to determine the acceleration you would get from that force alone, then subtract the acceleration due to gravity. It is, of course, exactly the same thing.

 

[redshift]

Thanks. After thinking about this some more, I'm a little confused about how this accerlation could be the same when the string is raised and lowered. That is, does wouldn't the pulling force be less when the weight is being lowered?

 

[arildno]

Thanks. After thinking about this some more, I'm a little confused about how this accerlation could be the same when the string is raised and lowered. That is, does wouldn't the pulling force be less when the weight is being lowered?

You are absolutely correct in that you will have different accelerations in lowering/raising, when applying a force with stated value 2.5N

If we by "pulling downwards" means that your hand is always below the weight and the string is taut, then
the acceleration experienced by the weight is found as:
(2.5+1.96)/0.2.

 

[redshift]

OK, I get it. As long as the hand is above the weight, regardless of direction, than the accerlation is the difference of a and g.

 

[arildno]

Eh, no:
If your hand is above the weight, holding the one end of the string, and you apply a downwards force of 2.5N ON THE STRING (You must assume the string has a mass dm here..), this will happen:
a) The string will loosen, and not transmit a force on the weight
(This happens, because any tensile force in a string only counteracts stretch in the string, never compression!)
b) Only gravity affects the weight now, it will experience free fall.
c) Since we may assume that the mass of the string is much less than the mass of the weight, the acceleration of the center of mass of the string will be much greater than 2.5/0.2+9.8 in the downwards direction.
(It's a bit difficult to consistently apply a force of 2.5N to the string, but that's a condition in the exercise)
The string will crumple up, your hand will either slam into the falling weight, or whizz past it dragging the string with it as well.

 

[redshift]

OK, NOW i get it. That's why the question said "pulling". And as you mentioned before, while it's possible to also "pull" downward, that's probably not what's meant.

 

[arildno]

I agree, however this MIGHT be a trap from the one that gave the exercise:
If you assume that that the hand is always above the weight, you get two distinct accelerations:
1. Proper pulling (the difference expression)
2. Free fall (g only!) when you're "pushing" the string downwards.
I suggest you make a careful reading of the text to see if this is what he had in mind)