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Finding the Center of Mass of a Toy Boat

Consider two people who visit the gym a substantial amount. One is a girl who loves to lift weights and bench press as much as she possibly can. The other is a guy who focuses much more on his legs, trying to break the world record for squat weight. It just so happens that these two are the same height and have the exact same weight, but the center of their weight is not in the same part of their body. This is because the girl has much more weight in the top half of her body and the boy has more weight in the bottom half. This difference in center of mass is a direct result of the different distributions of mass throughout both of their bodies.

Moments and Mass

There are two main components to finding the center of mass of an object. The first, unsurprisingly, is the mass of the whole object. In this case of the boat example, the mass will be uniform throughout the entire object. This is ideal a majority of the time as it drastically reduces the difficulty of the integration that will need to be done later. The mass of object i is: \[M_i = \int_{a}^{b} \rho f(x) \,dx \] Because the mass of our boat is uniform, the density function \rho is noninportant and will cancel when the total moment is divided by the total mass. The other component of finding the center of mass is the what is known as the moment. The moment of an object is its own mass times its distance from a given point, most frequently the origin. These objects can be of any shape, size, or function but the moment will always be defined the same way. For the toy boat, there are 3 moments to consider: the moment of the boat body, the mast, and the sail. These are the moments specific to this problem. The moment of object i is defined as: \[m_i = \int_{a}^{b} \rho x * f(x) \,dx \] A moment is a vector, so it has a direction in addition to magnitude. This makes it important to separate the calculations in each dimension. For simplicity, the center of mass in the z direction will not be calculated.

The Center of Mass

The center of mass in a given direction is the sum of all of the moments in that direction divided by the total mass of all objects. In equation form: \[ CM = \frac{\sum_{}^{} x_im_i }{\sum{}^{} m_i} \] It is important to note that moments can be negative since an object can be on any side of the origin. This would indicate that while mass cannot be negative, the position of the center of mass can be negative. Additionally, moments are additive. This is very useful as it indicates that the full object can be broken down into smaller pieces that are more easily defined by functions over certain domains.

The Boat in Question

As mentioned previously, the boat that we will be finding the center of mass for is composed of three different parts. These parts will all be considered separately and then summed in both the x and y directions. The following is picture of the front face of the boat, drawn using functions in Desmos:
In this boat, there is a trapezoid, a rectangle, and an iscoscles triangle. The point of reference utlized for the calculations will be the origin of the graph. Each moment in each direction and the mass will be calculated separately then added together in the end to find the center of mass. The reason a boat was chosen was to highlight the importance of center of mass in the real world. Ideally, a boat's center of mass is very close to the water as this will maximize the stability of the boat, preventing it from capsizing due to forces like wind. If at any point the center of mass in not over the hull of the boat, the ship will capsize. This boat is highly simplistic, but the general concept that the center of mass is in the hull should ring true even in this basoc shape.

The Body of the Boat - Part 1

Let's remember the equation for finding the moment of a given shape: \[m_i = \int_{a}^{b} \rho x * f(x) \,dx \] This section of the boat is a trapezoid bounded on 4 sides by the lines: \[ y = 2, y = 0, y= 2x - 2, x = 2 \] Notice how the moment of the trapezoid cannot be calculated with a single integral. This is because the height of rectangle dx is not described by one function over the domain [-2,2]. In this case, the moment should be calcualted as two separate integrals whose two domains add up to the domain on which the trapezoid lies. The moment of this shape with respect to y would be the sum of the two following integrals: \[ my_1 = \int_{-2}^{1} \rho x(2-0) \, dx + \int_{1}^{2} \rho x(2 - (2x -2) \, dx = \int_{-2}^{1} \rho 2x \, dx + \int_{1}^{2} \rho (4x-2x^2) \, dx \] These integrals evaluate to: \[ my_1 =\rho( x^2 \vert_{-2}^{1} + (2x^2 - \frac{2}{3}x^3) \vert_{1}^{2}) = \rho ((1^2 - (-2)^2) + (2(2)^2 - \frac{2}{3}(2)^3) - (2(1)^2 - \frac{2}{3}(1)^3)) = \rho (-3 + \frac{8}{3} - \frac{4}{3}) = -\rho \frac{5}{3} \] This is the first of 3 moments needed to find y coordinate of the center of mass. Next, the moment of the boat body with respect to x will be found. The integral for the moment with respect to x is: \[ mx_1 = \int_{0}^{2} \rho y(\frac{1}{2}y + 1 - (-2)) \, dy = \int_{0}^{2} \rho (\frac{1}{2}y^2 + 3y) \, dy \] \[ mx_1 = \rho \frac{1}{6}y^3 + \frac{3}{2}y^2 \vert_{0}^{2} = \rho ((\frac{1}{6}(2)^3 + \frac{3}{2}(2)) - (0)) = \rho(\frac{4}{3} + 3) = \rho \frac{22}{3} \] This is the moment of the first part with respect to x. Lastly, the mass of this part of the boat needs to be found. Recall the equation for the mass of an object: \[M_i = \int_{a}^{b} \rho f(x) \,dx \] Since mass is a scalar, the direction actually does not matter. This means that for f(x), we can utilize either of the function defined earlier to find the moments. To keep this calculation to one integral, the function used to find the moment with respect to x will be used. \[M_1 = \int_{0}^{2} \rho (\frac{1}{2}y + 3) \,dy = \rho( \frac{1}{4}y^2 + 3y) \vert_{0}^{2} = \frac{1}{4}(2)^2 + 3(2) - 0 = 7\] The body of the boat has a mass of 7 \rho.

The Mast - Part 2

The mast of the boat is the easiest to describe as it is simply a rectangle. The rectangle is bound by the lines: \[x = -.25, x= .25, y = 2, y = 6 \] An interesting property of this object is its symmetry around the reference point (0,0). The shape is symmetrical around the y axis, and therefore will not have an effect on the moment with respect to y. The math will show this in the integral for the moment: \[ my_2 = \int_{-.25}^{.25} \rho x(6-2) \, dx = \rho \int_{-.25}^{.25} 4x \, dx = \rho 2x^2 \vert_{-.25}{.25} = \rho 2(.25^2 - (-.25)^2 = 0 \] The math corroborates the assumption that the moment is unaffected by the symmetrical object. This symmetry does not apply to the momemt with respect to the x-axis however. \[ mx_2 = \int_{2}^{6} \rho x(.25 - (-.25)) \, dx = \rho \int_{2}^{6} \frac{1}{2}x \, dx = \rho \frac{1}{4}x^2 \vert_{2}^{6} = \rho \frac{1}{4}(6^2 - 2^2) = 8 \rho \] This object has a moment of 8 \rho with respect to the x axis. Once again, it does not matter which function is used to find the mass. As long as the function and the bounds match, it will give the same answer. For this case, the function with respect to the y will be used. \[ M_2 = \int_{-.25}^{.25} \rho (6-2) \, dx = \rho 4x \vert_{-.25}^{.25} = 4(.25 -(-.25) = 2 \rho \] The mass of the mast isn't all that massive at 2 \rho.

The Mast - Part 3

Last but not least, the moments and the mast of the sail need to be found. The hardest aspect here is making sure the bounds for each integral are correct. Start with the moment with respect to y: \[ my_3= \int_{.25}^{2} \rho x(-.5x+6.125 -(.5x + 4.125)) \, dx = \rho \int_{.25}^{2} x(-x + 2) \, dx = \rho \int_{.25}^{2} (-x^2 + 2x) \, dx \] \[ my_3 = \rho (-\frac{1}{3}x^3 + x^2) \vert_{.25}^{2} = \rho ((-\frac{1}{3}(2)^3 + 2^2) - (-\frac{1}{3}(.25)^3 + (.25)^2)) = \rho \frac{245}{192} \] The moment with respect to the y-axis for the sail is 245/192 \rho. \[ mx_3= \int_{4.25}^{6} \rho y(-2y + 12.5 -(2y - 8.25)) \, dy = \rho \int_{4.25}^{6} y(-4y + 20.75) \, dy = \rho \int_{4.25}^{6} (-4y^2 + \frac{83}{4}y) \, dy \] \[ mx_3 = \rho (-\frac{4}{3}y^3 + \frac{83}{8}y^2) \vert_{4.25}^{6} = (-\frac{4}{3}(6)^3 + \frac{83}{8}(6)^2) - (-\frac{4}{3}(4.25)^3 + \frac{83}{8}(4.25)^2) = \rho \frac{175}{384} \] This being the final moment that needs to be found. Finding the final mass: \[ M_3 = \int_{.25}^{2} \rho (-x + 2) \, dx = \rho (-\frac{1}{2}x^2 + 2x) \vert_{.25}^{2} = \rho ((-\frac{1}{2}(2)^2 +2(2)) - (-\frac{1}{2}(.25)^2 + 2(.25))) = \rho \frac{49}{32} \] And at long last, the integration is over.

Putting it All Together

Will all of these calculations, the center of mass can be calculated. The y coordinate of the center of mass is the sum of all the moments with respect to x added together over the total mass: \[ y_{CM} = \frac{mx_1 + mx_2 + mx_3}{M_1 + M_2 + M_3} = \frac{mx_1 + mx_2 + mx_3}{M_1 + M_2 + M_3} = 1.499\] It should now be clear that the rhos cancel out from the equation. The same calculation is done but for the y coordinate: \[ x_{CM} = \frac{my_1 + my_2 + my_3}{M_1 + M_2 + M_3} = \frac{mx_1 + mx_2 + mx_3}{M_1 + M_2 + M_3} = -.037\] This indicates that the center of mass is located at coordinates (-.037,1.499) with respect to the origin. For reference, this within the body of the boat on the left side of the top midpoint of the trapezoid.
The red dot is used to mark the center of mass.
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