Explanation for Newton II giving negative mass in my Physics lab results please

[SammyS]

The tension in the string cannot be more than the weight of m2, which is round about 0.2 N. Something is wrong with your measurements.

I agree with this !

(0.02 kg)(10 N/kg) = 0.2 N

 

[haruspex]

The tension in the string cannot be more than the weight of m2, which is round about 0.2 N. Something is wrong with your measurements.

Good point. According to the measurements, $m_1$ must be about 2kg and $m_2$ about 0.05kg.

 

[member 731016]

But in your table you have written positive values for acceleration, so did m₂ accelerate upwards? If so, I guess the mass must be negative.

Yes, but you plotted against a. What is x according to what you wrote above?

As I wrote in post #22, in a single drop all the values should be the same: all the forces should be the same and all the accelerations should be the same. Your data have the acceleration increasing at first. To explain that, there must be some systematic error in the early readings. I would ignore all the data before it settles down to a fairly constant acceleration. The slope you found merely tracks how the error evolves during the drop.

1. Discard the dodgy data,
2. Take the average T and average a of what remains,
3. Correct the sign error in T=(m2)(a+g), and
4. … use that to find the mass, not T=(m2)a.

Thank you for your reply @haruspex!

True, $m_2$ did accelerate downwards, so I guess I should change my convention to make the downward acceleration of $m_2$ positive.

The new equation I will use is:

$m_2g - T = m_2a$
$m_2g - m_2a = T$
$m_2(g - a) = T$
Where $y = T$, $(g - a) = x$, and $m = m_2$

I will complete the steps and post on how I get on. However, do you please know why take the average T and average a for 2.? Why not perform linear regression?

Many thanks!

 

[member 731016]

Yeah, I can see that! What I’ve been trying to get you to realize with this escapade is that you better think again…

So which way is $m_2$ accelerating in this experiment w.r.t your convention? is $a$ positive or negative? I'm asking because you are plotting it as a positive value...which is fine so long as your equation describing $T$ makes sense for $a$ being a positive value. Does your equation describing $T$ make sense for $a$ being a positive value?

Thank you for your replies @erboz!

Yeah I realized now that I was following different conventions for the equation and data. The correct equation for tension is in post #38.

Many thanks!

 

[member 731016]

Hi @ChiralSuperfields. It might help if you take a few steps back. Can you work-through the questions below?

Take the positive direction for each mass to be the direction in which it accelerates.

Since $m_1$ and $m_2$ have the same magnitude acceleration, use the same symbol, $a$, for the acceleration of each,

$T$ is the tension in the string and $g$ is acceleration due to gravity.

Q1. What is the net force on $m_1$ alone?

Q2. Using your answer to Q1, apply Newton’s 2nd law to $m_1$ to write an equation for $m_1$.

Q3. Give an expression for the net force on $m_2$ alone.

Q4. Using your answer to Q3, apply Newton’s 2nd law to $m_2$ to write an equation for $m_2$.

Q5. Combine your two equations (from Q2 and Q4) by eliminating $T$. Reaarange the resulting equation to make $a$ the subject.

Post your answers.

To verify Newton’s second law you will need to check that (within the limits of experimental error) your values of $a, m_1$ and $m_2$ [edit: and $g$] agree with the (correct) equation from Q5.

Thank you for your reply @Steve4Physics! It is nice to see a different way to verify Newton II than I thought of!

Answers:
Q1
Tension
Q2
$T = m_1a$
Q3
$F_{net} = m_2g - T$
Q4
$m_2g - T = m_2a$
Q5
$m_2g - m_1a = m_2a$

I assume that the string is inextensible and massless so that the acceleration of the cart on small mass are the same.

$m_2g = m_2a + m_1a$
$\frac{m_2g}{m_2 + m_1} = a$

Many thanks!

 

[member 731016]

If $m_1$ is the mass of the cart along with everything on the cart, then as I said, the tension is related to $m_1$ by $T=m_1\, a$.

For the hanging mass, we have $m_2\, g -T = m_2\, a$.

By the way:
You can eliminate $T$ from these two equations, to get $\displaystyle \quad a=\dfrac{m_2}{m_1+m_2}\,g$.

For the values that you have, this gives an acceleration of approximately $0.22\,\rm{m/s^2 }$ . This agrees fairly well with your data.

However, this gives a value for the tension of approximately $0.19\,\rm N$.
This is problematic.

The tension in the string cannot be more than the weight of m2, which is round about 0.2 N. Something is wrong with your measurements.

Good point. According to the measurements, $m_1$ must be about 2kg and $m_2$ about 0.05kg.

Thank you for your replies @SammyS, @Grelbr42 , and @haruspex!

I am now thinking about just ignoring the force (tension) data completely because as you all helpfully pointed out, the tension is much larger than it theoretical value so the experimental value is inaccurate. I think maybe we didn't calibrate the force sensor leading to this systematic calibration error since the values for the force are relatively precise.

So I am thinking about verifying Newton II using @Steve4Physics method where I will find the average value of the acceleration for where the data with relatively constant acceleration and compare it with the calculated value since the acceleration values are more accurate.

Many thanks!

 

[haruspex]

Why not perform linear regression?

Idealised, the following equations should apply to your experiment:
1. $m_1a=T$
2. $m_2a=m_2g-T$
The solution to those is
$a=\frac{m_2g}{m_1+m_2}$
$T=\frac{m_1m_2g}{m_1+m_2}$
Had that been accurate, your data 'plot' of $T$ against $a$ would have consisted of the single point $(\frac{m_1m_2g}{m_1+m_2}, \frac{m_2g}{m_1+m_2})$.
No opportunity for a regression line there.

Looking at your data, the acceleration started very low. It is as though there is a rough patch where $m_1$ started.
There is also the possibility that we should include the inertia of the pulley. Suppose it has mass $m_3$, radius $r$ and MoI $\frac 12m_3r^2$.

Modelling that we get
1. $m_1a=T_1-F_f$
2. $m_2a=m_2g-T_2$
3. $\frac 12m_3a=T_2-T_1$
where $F_f$ is variable and $T_1, T_2, a$ vary as a result.

These yield $(m_1+m_2+\frac 12m_3)a=m_2g-F_f$ and $T_1=m_2g-(m_2+\frac 12m_3)a$.

For the slope, $\frac{dT_1}{da}=-(m_2+\frac 12m_3)$.
So it is negative, and its magnitude exceeds $m_2$ by half the mass of the pulley.

 

[member 731016]

Idealised, the following equations should apply to your experiment:
1. $m_1a=T$
2. $m_2a=m_2g-T$
The solution to those is
$a=\frac{m_2g}{m_1+m_2}$
$T=\frac{m_1m_2g}{m_1+m_2}$
Had that been accurate, your data 'plot' of $T$ against $a$ would have consisted of the single point $(\frac{m_1m_2g}{m_1+m_2}, \frac{m_2g}{m_1+m_2})$.
No opportunity for a regression line there.

Looking at your data, the acceleration started very low. It is as though there is a rough patch where $m_1$ started.
There is also the possibility that we should include the inertia of the pulley. Suppose it has mass $m_3$, radius $r$ and MoI $\frac 12m_3r^2$.

Modelling that we get
1. $m_1a=T_1-F_f$
2. $m_2a=m_2g-T_2$
3. $\frac 12m_3a=T_2-T_1$
where $F_f$ is variable and $T, a$ vary as a result.

These yield $(m_1+m_2+\frac 12m_3)a=m_2g-F_f$ and $T_1=m_2g-(m_2+\frac 12m_3)a$.

For the slope, $\frac{dT_1}{da}=-(m_2+\frac 12m_3)$.
So it is negative, and its magnitude exceeds $m_2$ by half the mass of the pulley.

Thank you for help @haruspex! That is very interesting!

Many thanks!

 

[malawi_glenn]

What force sensor did you use?

I do this lab with my students using a Pasco smart cart. First they make sure that the force sensor is calibrated by letting the cart hang in string vertically. The force sensor should read mg.

Then we basically do this but with higher sample rate. Now we get one data point (T, a). For more data points, they place additional weights on the cart and repeat the experiment.

 

[member 731016]

What force sensor did you use?

I do this lab with my students using a Pasco smart cart. First they make sure that the force sensor is calibrated by letting the cart hang in string vertically. The force sensor should read mg.

Then we basically do this but with higher sample rate. Now we get one data point (T, a). For more data points, they place additional weights on the cart and repeat the experiment.

Thank you for your reply @malawi_glenn! Sorry I am not entirely sure. I think we use LoggerPro for the collecting the data in the lab.

Many thanks!