I learnt about an experiment to show that acceleration is proportional to force. It was done by placing a trolley (like a toy car) on a smooth track. At the end of the smooth horizontal track was a pulley connected to the trolley by a string which hung masses over the edge of the table. So the trolley experienced acceleration due to the weight from the masses.

My problem is in what my textbook says to do next. They say you have to take a mass being hung over the edge, which results in less force applied to the trolley. But then they tell you to put the mass onto the trolley, to ensure constant mass in the system.

The problem is in placing the mass onto the trolley. Surely the trolley's mass should stay constant throughout, since that is what we are measuring the acceleration, and not the system as a whole?

I hope you can explain this to me, or if my book is incorrect. I'll expand on the experiment if you find my description confusing.


1 Answer 1


since that is what we are measuring the acceleration, and not the system as a whole.

No, you are measuring the acceleration of the whole system.

You may think of the string as a flexible (not elastic!) rod that connects 2 pieces: the trolley and the hung masses. When the masses fall and the trolley moves forward, they are both accelerating as a whole. The only difference is that the masses are accelerating down and the trolley is accelerating forward (the pulley does that), but you can see easily that the speed of these two pieces must be the same, so their accelerations must be equal too.

That is, the weight of the hung masses is the applied force, and the mass to which it is applied is the sum of the masses plus the trolley.

One easy way to check that you must add the masses of the trolley and the hung masses is to consider what would happen if you make the mass of the trolley near to 0, that is, almost no trolley at all. If you don't count the mass of the hung masses, the acceleration would go to infinity! But surely the total acceleration would never be over $g$ (gravity).

  • $\begingroup$ Your final paragraph is amazingly helpful. Thank you. $\endgroup$
    – Cataline
    Commented Dec 16, 2015 at 21:52

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