Sinking a toy boat with lead pellets

[brotherbobby]

Homework Statement

A plastic "boat" with a $25\; \text{cm}^2$ square cross section floats in a liquid. One by one, you place lead pellets of $m = 50\; \text{g}$ each inside the boat and measure how far the boat extends ($x$ cm) below the surface. Your data are as follows: (1) m = 50 g : x = 2.9 cm, (2) m = 100 g : x = 5.0 cm, (3) m = 150 g : x = 6.6 cm and (4) m = 200 g : x = 8.6 cm.

Draw an appropriate graph of the data and, from the slope and intercept of the best-fit line, determine (1) the mass of the boat and the (2) density of the liquid.

Relevant Equations

Law of flotation : Mass of the body = Mass of liquid displaced ($m_B = \Delta m_L$)

Attempt : I begin by plotting the graph of depth x against mass of pellets m as shown alongside.

Using $m_B = \Delta m_L\Rightarrow m_B = \rho_L A_B\, x(B)$ where $\rho_L$ is the density of the liquid, $A_B$ is the cross-sectional area of the boat and $x(B)$ is the depth to which the boat submerge into the liquid due to its own weight without the pellets($m = 0$). As we can see from the graph, $x(B) = 0.8\; \text{cm}$, the y-intercept in the graph. See $x(B)$ as "x due to B only".

From above, $\frac{m_B}{x(B)} = \rho_L A_B$. The graph has an incline of $26^{\circ}$. Hence, from mathematics, $\frac{\Delta x}{\Delta m} = \tan 26^{\circ}\Rightarrow \frac{\Delta m}{\Delta x} = \cot 26^{\circ} = 2.05$. Thus, the density of the liquid $\rho_L = \frac{2.05}{A_B} = \frac{2.05}{25\times 10^-4} = \boxed{820\; \text{kg/m}^3}$.

The mass of the boat $m_B = \rho_L A_B\, x(B) = 820\times (25\times 10^{-4})\times (8\times 10^{-3}) = \boxed{16.4\; \text{g}}$.

My answers are way different from those in the text : (1) Mass of the boat = $\boxed{m_B = 29\; \text{g}}$, (2) Density of the liquid = $\boxed{\rho_L = 1100\; \text{kg/m}^3}$.

Any hint as to where am going wrong?

(A possible error could be in the incline and hence the slope of the graph. I have obtained conflicting answers using the "ruler" (slope = 20.5)in the software and finding the slope manually (slope = 2.44). Still I am mistaken, the slope from the book is 2.67. Is my logic mistaken? Any help would be welcome).

 

[jbriggs444]

What are the units for that slope? Why are you trying to use the slope to find the angle if you are only going to use the angle to find the slope?

 

[Gordianus]

I made a rough estimation from the graph.
The boat sinks approximately 7 cm when the pellets' mass is 200 grams; equating to the weight of the displaced liquid I found:
Displaced volume= $$7 \rm cm x 25 cm^2= 175 cm^3$$

Density of liquid: $$ \rho_{l}=\frac{200 \rm gr}{175 \rm cm^3}=1.14\frac{\rm gr}{\rm cm^3}=1140 \frac{\rm kg}{\rm m^3}$$

 

[brotherbobby]

What are the units for that slope? Why are you trying to use the slope to find the angle if you are only going to use the angle to find the slope?

I am trying to find the slope in two ways : $s = \tan \phi$ and $s = \frac{x_2-x_1}{m_2-m_1}$. With the former, where I have to measure $\phi$, I am having less success, in that I am further from the answers in the text.

The slope I found $s = \tan 26^{\circ} = 0.49 \; \text{cm/g}$.

 

[brotherbobby]

I made a rough estimation from the graph.
The boat sinks approximately 7 cm when the pellets' mass is 200 grams; equating to the weight of the displaced liquid I found:
Displaced volume= $$7 \rm cm x 25 cm^2= 175 cm^3$$

Density of liquid: $$ \rho_{l}=\frac{200 \rm gr}{175 \rm cm^3}=1.14\frac{\rm gr}{\rm cm^3}=1140 \frac{\rm kg}{\rm m^3}$$

I don't think we can do that - given the instruction in the problem. There are several values of depth versus mass to take to find out the answer.

I plot the location in the graph nonetheless. 7 cm of depth corresponds to about 152 gm of pellet mass. From the data, a mass of 200 gm of pelllets displaces 8.6 cm.

 

[Gordianus]

I said a rough estimation.
From the first graph (not the second), I see the boat without pellets sinks about 1 cm (due to its own weight) and loaded with 200 grams it sinks 8 cm. That's why I said the boat sinks 7 cm when loaded with 200 grams. From this rough estimation I obtained a density close to the expected one.
Now, if we want a more refined answer we could resort to a least squares algorithm and get a better estimate for the density and the empty boat weight.

 

[brotherbobby]

I said a rough estimation.
From the first graph (not the second), I see the boat without pellets sinks about 1 cm (due to its own weight) and loaded with 200 grams it sinks 8 cm. That's why I said the boat sinks 7 cm when loaded with 200 grams. From this rough estimation I obtained a density close to the expected one.
Now, if we want a more refined answer we could resort to a least squares algorithm and get a better estimate for the density and the empty boat weight.

Can we use the slope of the graph and its y-intercept for getting to the same answer? I have tried to do that but have gone wrong. That is the method which the problem asks for.

 

[jbriggs444]

I am trying to find the slope in two ways : s=tan⁡ϕ and s=x2−x1m2−m1. With the former, where I have to measure ϕ, I am having less success, in that I am further from the answers in the text.

The slope I found s=tan⁡26∘=0.49cm/g.

You say that the units are centimeters per gram.

The markings on the vertical axis of the graph are one centimeter per coarse division.
The markings on the horizontal axis of the graph are 12.5 grams per coarse division.

So if you determine slope from an angle on your graph, you will be off by a factor of 12.5.

One can see this if one computes the slope between the data endpoints: $\frac{8.6 \text{cm} - 2.9 \text{cm}} {200 \text{g} - 50 \text{g}} = 0.038 \frac{\text{cm}}{\text{g}}$

Multiply that by 12.5 and you have 0.475 -- which by no coincidence matches closely with the figure you've presented above.

 

[Steve4Physics]

Any hint as to where am going wrong?

I'm going to chip-in if I may. Using angles (and their tangents and cotangents) in this sort of question is wrong. It makes no sense mathematically.

I'll try to explain why, with an example. Please spend a few minutes reading it and thinking about it.

Suppose you are plotting the graph y = 3x. You know (I hope) that this has a gradient of 3.

You mark the scales on your graph paper at 1cm intervals as follows:
x-axis: 0, 1, 2, 3, 4, 5
y-axis: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 , 13,1 4, 15

Your friend draws the graph but marks the scales at 1cm intervals as follows:
x-axis: 0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0
y-axis: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 , 13,1 4, 15
Note their x-values will be twice as spread-out as yours.

The result is that the two lines will have different angles to the x-axis. The tangents of the angles will be different. But the gradient of 'y = 3x' is 3 and shouldn't depend on how the graph is drawn.

So the advice is: alway express gradents as:$$\frac{change \text{ } in \text{ } y}{change \text{ } in \text{ } x}$$Then it doesn't matter what scales you use.

If you use the angle, then tan(angle) will not give the required gradient (except by accident sometimes).

 

[brotherbobby]

Thank you. It's a lesson to remember what you said above. However, a question remains. The tangent of the angle of the line given by $y = mx+c$ makes with the x axis, $\tan\theta = m$, gives its slope. Does the value of this slope have a physical meaning? Surely changing the scale of the x coordinate from $x \rightarrow x' = kx$ will alter the slope by the factor $k$, as you pointed above. What then can we say of the well known statement : $$\text{the slope of the line y = mx+c is given by m}\; = \tan \theta ?$$

 

[Steve4Physics]

The tangent of the angle of the line given by $y = mx+c$ makes with the x axis, $\tan\theta = m$, gives its slope. Does the value of this slope have a physical meaning? Surely changing the scale of the x coordinate from $x \rightarrow x' = kx$ will alter the slope by the factor $k$, as you pointed above. What then can we say of the well known statement : $$\text{the slope of the line y = mx+c is given by m}\; = \tan \theta ?$$

In (pure) maths, x and y are dimensionless numbers (i.e. they have no units). In this case, when a graph of y = mx + c is plotted using axes with identical scales (which is assumed by default), then m = tanθ . No problem.

But if x and y represent different physical quantities, the situation is very different.

Consider a graph of distance (metres) on the y-axis vs. time (seconds) on the x-axis for an object moving at constant speed.

The gradient has units of metres/second . The gradient equals the speed.

The actual angle (θ in degrees say) will depend on the choice of scales. It is wrong to treat θ on this graph in the same way as θ in the pure maths situation.

In this example, the slope has a clear physical meaning and a unit - the speed in metres/second. In other situations, the meaning and unit of the slope will depend on what quantity is being plotted against what other quantity.