Spring and acceleration

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1. The problem statement, all variables and given/known data
The olympic commissioners decided to buy and sell gold, hoping to set a new profit record. They conducted all their business in an elevator. They bought and sold at the same price per ounce. An archaic unit of force, which measured with a verticle spring scale scale suspended from the ceiling. They always bought their gold when the elevator had a downward acceleration of 2 m/s^2 and always sold when the acceleration was 2.5 m/s^2 upward. Evaluate their percentage profit based on their buying profit. - means a loss.


3. The attempt at a solution
Alright so i know the following things:

a of buying: 2 m/s^2
a of selling : 2.5 m/s^2

and there is something to do with springs... but how would i start off?

haruspex

Don't worry about the springs. All that matters is that the scale doesn't measure mass, it measures the force. So consider the free body diagram of the gold. For a given mass of gold, what weight will be measured when accelerating up? Down?

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They do not tell us the mass,all they tell us is that accelerating numbers in the elevator. However they are like they sold and bought at the same price per ounce.

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Umm is it like this..

so the t represents the price per ounce so
Fg = mg
mgt = price per ounce

so
when accelerating up it would be like this

ma[up]+mg[up] = mg[down]t
so a[up]+g[up] = g[down]t

right?

and when accelerating down
ma[down]+mg[down]t = mg[up]
a[down] + g[down]t = g[up]

but how would i find the profit?

haruspex

No, that would be the total price for m ounces.
That's a very confused way of writing it. On the left of the equals you have forces, on the right amounts of money. What do you mean by g[up] and g[down]?
When accelerating upwards, what price will they sell m ounces of gold for?

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well they said that the selling and buying price per ounce stays the same in the question. what i meant by g[down] is gravity and g[up] i meant normal force.

haruspex

No, I mean if there really are m ounces of gold, no acceleration, what will they actually sell it for?

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they will sell for only the weight of it right?

haruspex

i don't understand what you mean. Suppose they have a nugget of gold which, weighed in the absence of acceleration, tipped the scales at 'm ounces' (this being a force, m ounces of weight). What would that same nugget 'weigh' when nugget plus scales are undergoing an upward acceleration of a? What would they therefore sell the nugget for?

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they would weigh very little since we are going up against gravity.. i don't know specifically.

haruspex

first, to avoid some possible confusion, I'm going to modify what I said before. Suppose the nugget weighs mg when not accelerating. (I wrote 'm' before.) This means that gravity exerts a downward force mg on the nugget, and in order to achieve equilibrium the spring balance has to push up with force mg. That leads the balance to record a weight of mg ounces.
When the nugget, sitting on the spring balance, is accelerating upwards at rate a, what force must the spring balance be exerting on it?

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The spring balance must be exerting the normal force

haruspex

Yes, that's what it's called, but I mean how big is it? Write an expression for it in terms of m, a and g.

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Fnet + Fn = Fg
ma + mg = mg

haruspex

I know what you mean, but you really must avoid writing equations so carelessly. If ma+mg=mg than ma=0, right? You should write this as
force on spring balance = ma+mg
So, if the price is p per ounce, the fair price for this nugget is mgp. What price will they actually sell it for?

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So if they are selling the price for mgp than they are actually just selling it for the acceleration they go up by mass

haruspex

No.
Suppose the lift is not accelerating. We weigh the nugget and it weighs mg. The fair price for this nugget is therefore mgp. Now the lift accelerates upwards at rate a. You have calculated that the balance will now show the weight as mg+ma. If a buyer buys based on the weight the balance shows now, how much will the buyer pay?