Experiment to show F is proportional to a

JakeTheMsitake

Experiment from my physics book:


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This 'diagram' is supposed to show a friction-compensating slope with a trolley on it. The trolley is connected by string to a thing with weights in it at the bottom. The more weights placed in the thing, the more the trolley accelerates. Using a ticker tape timer, the acceleration for each force was found.
What's puzzling me though is that it says that the weights are transferred between the trolley and the thing (I can't think of the name), in order to keep the mass of the system constant.
But surely the trolley is the mass that is being accelerated and it's mass should be constant, and weights are just put on/taken off the thing. No?

Civilian_Target

The trolley should be of constant weight. You're trying to prove F prop. a, so the third factor, mass should be constant. In this case, more weight on the pulley system will increase the pulling Force, not any system mass.

You can measure it with different masses on the trolley and compensate, of course, but it seems like a strange way of doing things. Have you mentioned this to your teacher?

BTW, if it's an "acceleration due to gravity" experiment, then what's being done could potentially be correct.

JoseJones

because F=ma, mass must be kept constant in order to prove F proportional to a. This mass refers to the mass of the whole system, ie: the trolley, string and scale pan. Remember, the whole system is being accelerated, so transferring mass from one part of the system to the other doesn't change the overall mass, if you have .25kg on the trolley, and .25kg on the scale pan, and then move .1kg from the trolley to the pan, you now have .15kg on the trolley and .35kg on the pan. This means the overall mass is still .5kg, but the system's acceleration will increase.

SOL

it really doesn't matter how much each weighs as long as you know how much it weighs because when you come back to your F=ma question you will just be putting in your known values for f and m and get the same a each time, f against m on a graph would have a slope a...

JoseJones

but your not graphing F against M, your investigating if F prop. a. that's why m must be constant. F goes on the X axis, a on the Y axis, and the slope=1/m

JakeTheMsitake

Can it be classed as an exernal force then?