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

[member 731016]

Homework Statement

Please see below

Relevant Equations

Newton Second Law

I am trying to verify Newton II. The setup I am using is,

Where $m_1 = 0.887 kg$ is a cart and $m_2 = 0.02016 kg$ is a small hanging mass. There is a force sensor on $m_1$ to measure the force acting on it from the string and the acceleration of the cart.

To verify Newton's Second Law, we tried to plot we plot acceleration of $m_2$ vs force $F$ which is the same as the tension $T$.

Apply Newton II to the $m_2$: $T = m_2a + m_2g$ which is a linear equation ($y = mx + c$) of the form,
which is of the form $y = T, m = m_2, x = a$ and $c = m_2g$

Therefore we graphed T vs a to get:

However, do you please know why I am getting $m_2 = -0.2152 kg$ from the graph when we measure it to be 0.02kg? I am not sure why the negative sign and why it is x10 larger than the measured $m_2$. Also when I solve $c = 0.5069 = m_2g$, for $m_2$ I also get a very different value for $m_2$

Any help greatly appreciated!

Many thanks!

 

[berkeman]

You mention a "cart" but I see no wheels at m1. Is it a cart with frictionless wheel bearings, or is it on a frictionless surface in your experimental setup, or should you have included a friction term in your equations?

 

[member 731016]

You mention a "cart" but I see no wheels at m1. Is it a cart with frictionless wheel bearings, or is it on a frictionless surface in your experimental setup, or should you have included a friction term in your equations?

Thank you for your reply @berkeman ! Yes we were told to ignore friction (in the wheel bearings).

Many thanks!

 

[erobz]

Homework Statement: Please see below
Relevant Equations: Newton Second Law

I am trying to verify Newton II. The setup I am using is,
View attachment 326971Apply Newton II to the $m_2$: $T = m_2a + m_2g$ which is a linear equation ($y = mx + c$) of the form,

If $a$ = $g$ what does $T$ equal?

 

[member 731016]

If $a$ = $g$ what does $T$ equal?

Thank you for your reply @erobz!

$T = 2m_2g$

 

[erobz]

Thank you for your reply @erobz!

$T = 2m_2g$

Does that make sense to you?

Spoiler: Hint

It shouldn't.

 

[SammyS]

@ChiralSuperfields ,

Is the mass of the force sensor included as contributing to $m_1$ ?

 

[scottdave]

As @erobz hinted at, if T is zero, what situation is that similar to?

 

[member 731016]

Does that make sense to you?

Spoiler: Hint

It shouldn't.

Thank you for your reply @erobz!

Sorry how not?

Many thanks!

 

[member 731016]

@ChiralSuperfields ,

Is the mass of the force sensor included as contributing to $m_1$ ?

Thank you for your reply @SammyS!

Yes

Many thanks!

 

[member 731016]

As @erobz hinted at, if T is zero, what situation is that similar to?

Thank you for your reply @scottdave !

Free fall.

Many thanks!

 

[erobz]

Thank you for your reply @erobz!

Sorry how not?

Many thanks!

$a=g$ implies $m_2$ is in free fall. Is there anything tugging on $m_2$ (holding it back) in that case?

 

[SammyS]

Thank you for your reply @SammyS!

Yes

Many thanks!

I'm surprised at that.

The data clearly show that $m_1$ must be approximately 2 kg.

Afterall, you have the net force, $T$, exerted on $m_1$ and you have its acceleration ,

If this were an ideal situation, no friction and no stretching of the string, the acceleration and the tension would not vary throughout your table.

 

[member 731016]

$a=g$ implies $m_2$ is in free fall. Is there anything tugging on $m_2$ (holding it back) in that case?

Thank you for your reply @erboz!

Yes the inertia of $m_1$ is decreaseing the acceleration of $m_2$

Many thanks!

 

[member 731016]

I'm surprised at that.

The data clearly show that $m_1$ must be approximately 2 kg.

Afterall, you have the net force, $T$, exerted on $m_1$ and its acceleration ,

If this were an ideal situation, no friction and no stretching of the string, the acceleration and the tension would not vary throughout your table.

Thank you for your reply @SammyS!

Do you please know why the data would give a wrong mass for $m_1$?

Many thanks !

 

[erobz]

Thank you for your reply @erboz!

Yes the inertia of $m_1$ is decreaseing the acceleration of $m_2$

Many thanks!

No. If $a=g$, there is only a single force acting on $m_2$. What is that force? You need to be thinking about what happens in the limit as $m_1 \to 0$

 

[member 731016]

No. If $a=g$, there is only a single force acting on $m_2$. What is that force? You need to be thinking about what happens in the limit as $m_1 \to 0$

Thank you for your reply @erobz!

If $a = g$ then the only force acting on the system is force of gravity on $m_2$

Many thanks!

 

[erobz]

Thank you for your reply @erobz!

If $a = g$ then the only force acting on the system is force of gravity on $m_2$

Many thanks!

Right, so what is the tension $T$ in that case?

 

[member 731016]

Right, so what is the tension $T$ in that case?

Thank you for you reply @erobz!

$T = 0$ since $m_1 = 0$ basically

Many thanks!

 

[erobz]

Thank you for you reply @erobz!

$T = 0$ since $m_1 = 0$ basically

Many thanks!

But you told me above subbing in $a=g$, that $T=2m_2g$. Does $T=0$ in your equation when $a=g$?

 

[member 731016]

But you told me above subbing in $a=g$, that $T=2m_2g$. Does $T=0$ in your equation when $a=g$?

Thank you for your reply @erobz!

True! No, I don't think T = 0 when a = g

Many thanks!

 

[haruspex]

To verify Newton's Second Law, we tried to plot we plot acceleration of m2 vs force F which is the same as the tension T.

Well, no, the F in Newton II is the net force. There are two forces acting on m₂.

Apply Newton II to the $m_2$: $T = m_2a + m_2g$

The first thing to be clear about is your sign conventions. Which way are you taking as positive for a? Does T act on m₂ in that direction? What about m₁?

we graphed T vs a to get

I don’t understand what you are plotting here. The leftmost column is timestamps in increments of 1/20s, Implying these are measurements in a single drop. These should hardly vary. The only reason they do is some inaccuracy in the readings near the start. The slope of the graph you plotted is of no interest here.

You note that $T = m_2a + m_2g$ is a linear relationship, but if you want the slope to represent the mass you must get it in the form $y=m_2x+c$.

 

[member 731016]

Well, no, the F in Newton II is the net force. There are two forces acting on m₂.

The first thing to be clear about is your sign conventions. Which way are you taking as positive for a? Does T act on m₂ in that direction? What about m₁?

I don’t understand what you are plotting here. The leftmost column is timestamps in increments of 1/20s, Implying these are measurements in a single drop. These should hardly vary. The only reason they do is some inaccuracy in the readings near the start. The slope of the graph you plotted is of no interest here.

You note that $T = m_2a + m_2g$ is a linear relationship, but if you want the slope to represent the mass you must get it in the form $y=m_2x+c$.

Thank you for your reply @haruspex ! That is very helpful!

For $m_2$, I am taking upwards as positive and downwards as negative so if $T < m_2g$ then $a < 0$.

Yeah the measurements are in a single drop. Sorry, how is the slope of the graph of no interest?

Getting that linear relationship for $T$ in $y = mx + c$ form we get $T = (m_2)(a + g)$.

I decided to plot the graph again and set the y-intercept equal to zero since $T = (m_2)(a + g)$ is in $y = mx$ form to get the graph:

This is closer to the original value for $m_2$, however, why is $m_2$ so large from the graph?

Many thanks!

 

[malawi_glenn]

Do you really that linear fit describes your data?

 

[SammyS]

Homework Statement: Please see below
Relevant Equations: Newton Second Law

I am trying to verify Newton II. The setup I am using is,
View attachment 326971
Where $m_1 = 0.887 kg$ is a cart and $m_2 = 0.02016 kg$ is a small hanging mass. There is a force sensor on $m_1$ to measure the force acting on it from the string and the acceleration of the cart.

To verify Newton's Second Law, we tried to plot we plot acceleration of $m_2$ vs force $F$ which is the same as the tension $T$.

Apply Newton II to the $m_2$: $T = m_2a + m_2g$ which is a linear equation ($y = mx + c$) of the form,
which is of the form $y = T, m = m_2, x = a$ and $c = m_2g$

These equations are wrong.

The only force acting on the cart & sensor combination (mass, m_1) is the tension, T, measured by the sensor.

There are two forces acting on the hanging mass, m_2 . Do you have any idea as to what these two forces are?

 

[member 731016]

Do you really that linear fit describes your data?

Thank you for your reply @malawi_glenn !

Yes I do. I don't have any reason to not think that the linear fit is for the data.

Many thanks!

 

[malawi_glenn]

What is the R-squared value?

 

[member 731016]

What is the R-squared value?

Thank you for your reply @malawi_glenn ! The value is 0.95 (2 sig figs).

Many thanks!

 

[member 731016]

These equations are wrong.

The only force acting on the cart & sensor combination (mass, m_1) is the tension, T, measured by the sensor.

There are two forces acting on the hanging mass, m_2 . Do you have any idea as to what these two forces are?

Thank you for your reply @SammyS!

Two forces acting on hanging mass is tension and gravity.

Sorry how are the equations wrong?

Many thanks!

 

[haruspex]

I am taking upwards as positive and downwards as negative

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.

since T=(m2)(a+g) is in y=mx form

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

how is the slope of the graph of no interest?

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.