Debate over the experimental procedure for Newton's 2nd Law

[ekrabappel]

Hi,

I'm currently having an ongoing debate with some teachers regarding a practical for Newton's Second Law. The prac involves a cart attached to some weights via a string on a pulley. The protocol as it stands has students add increasing mass to the end of the string (not changing the mass of the cart) while measuring the time it takes for the cart to travel a certain distance. The students run the experiment with several different masses and calculate the cart acceleration using time and distance. They then plot the mass at the end of the string against acceleration. The slope is the mass of the cart thus supporting Newton's second Law. I have been arguing that the cart and weight at the end of the string should be used as a total mass and weight should be added to the string by removing it from the cart thus keeping the total mass constant. The calculated mass from the slope is the mass of the entire setup. This has been counter argued that if you had 2 people pulling a car along the road with a rope and you added a third person then you wouldn't take the mass of the people pulling the rope into consideration. Who is right here?

 

[mfb]

You are right.

I have been arguing that the cart and weight at the end of the string should be used as a total mass and weight should be added to the string by removing it from the cart thus keeping the total mass constant.

That would be better, yes.

If the mass of the weights is small compared to the mass of the car it is only a small correction, but you can consider extreme cases: Use an elephant as weight. Will the acceleration depend on the mass of the cart? Not measurably, the elephant will just fall down freely. If you use two elephants as weight you get the same acceleration.

People pulling on a rope is different because the people don't move - there is no force involved in accelerating these people.

 

[Dale]

Summary: Correct method for Newton's 2nd Law experiment

Who is right here?

If you work out the math for this problem you get $$a=g\frac{m_{\text{string}}}{m_{\text{string}}+m_{\text{car}}}$$

So if you keep the total mass fixed then the denominator is a constant and the acceleration is proportional to the mass on the string. If you keep the car mass fixed you shouldn’t get a straight line (of course with friction it probably doesn’t matter anyway)

Summary: Correct method for Newton's 2nd Law experiment

This has been counter argued that if you had 2 people pulling a car along the road with a rope and you added a third person then you wouldn't take the mass of the people pulling the rope into consideration.

This scenario is simple enough to analyze directly. No need to make imperfect analogies with other scenarios.

 

[A.T.]

This has been counter argued that if you had 2 people pulling a car along the road with a rope and you added a third person then you wouldn't take the mass of the people pulling the rope into consideration.

If the people themselves are accelerating at the same rate as the car, then you would have to include their masses to get the acceleration from the total external force at their feet. Otherwise the scenario/question is not equivalent.

 

[Delta2]

If you work out the math for this problem you get $$a=g\frac{m_{\text{string}}}{m_{\text{string}}+m_{\text{car}}}$$

I am not sure I have understood the setup 100% (unfortunately for me no figure provided) but from what I understand I think the acceleration of the system is $$a=g\frac{m_{string}-m_{car}}{m_{string}+m_{car}}$$

EDIT: OK i think i see my mistake now, the car is supported on a horizontal surface...

 

[Dale]

unfortunately for me no figure provided

I agree. A figure is always helpful for this sort of thing.

 

[kuruman]

If you design the experiment following the constant mass protocol (moving mass from the end of the string to the cart) suggested by @Dale in post #3, you can write$$m_{string}g=Ma$$where ##M## is the constant combined mass ##m_{string}+m_{cart}##. This equation shows that even though the Earth exerts an unbalanced force on only part of the two-mass system, it's the entire mass ##M## that has the measured acceleration ##a##. Should you choose to go this way, I would suggest that ##m_{string}## not be used as an independent variable. Consider plotting the two sides of the equation on the same graph versus run number. Then see whether the points overlap to within the error bars for each run. This should persuade your detractors that they are using the wrong analogy. It's the Earth that does the "pulling" and, of course, its mass is not included anywhere.

BTW, if you want only the mass of the cart to be your system, then you will need to measure the tension independently, perhaps with a force gauge, and plot measured tension versus measured acceleration. Under ideal conditions, the slope will be the mass of the cart.