Center of Gravity

Bob and Jimmy are sitting in a jungle gym. Jane is sliding downward on a diagonal part of the jungle gym inclined at \(30^\circ\). The normal force acting on her by the incline is \(500 \,\mathrm{N}\). Bob has a mass of \(60\,\mathrm{kg}\), and Jimmy has a mass of \(70\,\mathrm{kg}\). From the leftmost point of the jungle gym, Bob is \(1.5\,\mathrm{m}\) away, Jimmy is \(2.75\,\mathrm{m}\) away, and Jane is \(5\,\mathrm{m}\) away. Where is the center of gravity of Bob, Jimmy, and Jane, as a system, with respect to the leftmost point of the jungle gym?

Remember our formula for the center of gravity:

$$x_{cm} =\frac{\sum w_i x_i}{\sum w_i}\\\mathrm{.}$$

First, let's find all of the weights. Remember that we have to multiply \(60\,\mathrm{kg}\) by \(10\,\mathrm{\frac{N}{kg}\\}\) to get Bob's weight, which is the force on Bob due to gravity. Therefore,

$$60\,\mathrm{kg}\times 10\,\mathrm{\frac{N}{kg}\\}=600\,\mathrm{N},$$

is Bob's weight.

Now, we'll move on to Jimmy:

$$70\,\mathrm{kg}\times 10\,\mathrm{\frac{N}{kg}\\}=700\,\mathrm{N.}$$

To find Jane's weight, we need to calculate the upward normal force exerted on her. To do this, we treat her position like a right triangle with the forces acting on her as side lengths.

The purple vector in the image above labeled \(\vec{F}_\mathrm{N}\) is the normal force acting on Jane, whose magnitude is given to us in the problem as \(500 \,\mathrm{N}\). To find the magnitude of Jane's weight, recall that the normal force is equal to the perpendicular component of the weight, as depicted in figure 4. From this relationship, we can solve for the weight.

$$\begin{align*} |\vec F_{\mathrm{N}}| &= F_\mathrm{N} \\ \vec |w| &= w \\[20pt] w &= mg \\ F_\mathrm{N} &= mg\cos{30.0^\circ } \\ F_\mathrm{N} &= w\cos{30.0^\circ } \\ w&= \frac{F_\mathrm{N}}{\cos{30.0^\circ .}} \\ \end{align*}$$

Let's now calculate her weight using the above result:

$$w=\frac{5.00\times 10^2 \,\mathrm{N}}{\cos{30.0^\circ }}=577\,\mathrm{N.}$$

Now that we know all the weights, we are ready to use the formula for the center of gravity.

$$\begin{aligned} x_{\text{cw}} &= \sum\frac{w_ix_i}{w_i}\\[8pt] x_{\text{cw}} &= \frac{(6.00\times10^2\,\mathrm{N})1.50\;\mathrm{m}+ (7.00\times10^2\,\mathrm{N}) 2.75\;\mathrm{m} + (577\,\mathrm{N}) 5\;\mathrm{m}}{600\,\mathrm{N} + 700\,\mathrm{N} + 577\,\mathrm{N}}\\[8pt]x_{\text{cw}} &= 3.04\,\mathrm{m}\end{aligned}$$

Therefore, the system's center of gravity is \(3.04\,\mathrm{m}\) from the leftmost part of the gym. Now that was a physics workout!

Imagine that we wanted to design a megastructure and require to determine the center of gravity of a \(3\,000\,\mathrm{km}\) long rod. We can model the rod as one-dimensional, so we can describe its mass distribution with a linear mass density function \( \lambda (y) \). The rod needs to be partially buried so only \(2\,000\,\mathrm{km}\) will be visible. For your calculations, use Earth's center as the origin and consider the radius of Earth as \(6\,371\,\mathrm{km}\)). At what height would the center of gravity for this rod be?

Solution

The linear mass density of a straight rod is the derivative of the rod's mass with respect to the position(in this case, it is constant):

$$\lambda = \frac{\mathrm{d}m}{\mathrm{d}y}.$$

This means that

$$\mathrm{d}m = \lambda\mathrm{d}y$$

Recall that the gravitational force exerted by the Earth on a particle of mass \( m \) is the particle's weight. And we can calculate it with the formula

$$ w = F_g =G\frac{M_{\text{Earth}}m}{r^2} $$

In this case, we have to use this formula, and we can't consider the gravitational strength constant because the length of our system (\( 3\,000\;\mathrm{km} \)) is not negligible when compared to the radius of the Earth (\(6\,371\,\mathrm{km}\)). Now, we consider the rod as comprised of infinitely many pieces of mass, \( dm, \) located at a distance \( y \) from the center of the Earth. Therefore the equation for the weight of that piece of mass becomes

$$ \mathrm{d}w =G\frac{M_{\text{Earth}}\mathrm{d}m}{y^2}.$$

We are almost ready to use our formula for the center of mass:

$$\vec y_{\mathrm{cw}} =\frac{\int y \mathrm{d}w}{\int \mathrm{d}w}\\\mathrm{.}$$

Note that here, we just consider one dimension so we are using \( y \) instead of \( \vec{r} .\) Since we are measuring the distance with respect to the Earth's center, the distance to the beginning of the rod is \[6\,371\,\mathrm{km}-1000\;\;\mathrm{km}=5371\;\mathrm{km},\]

since the rod's first \( 1\,000\;\mathrm{km} \) are buried. Similarly, the distance of the top of the rod from the center of Earth is

\[6\,371\,\mathrm{km}+2\,000\;\mathrm{km}=8371\;\mathrm{km}.\]

This is because the remaining \( 2\,000\;\mathrm{km} \) protrude from Earth's surface. These values define our limits of integration!

Fig. 4 - This image helps us to visualize the scenario.

Now we have all we need, and we can substitute all our data in the formula for the center of gravity.

$$\begin{aligned}y_{\text{cw}} &= \frac{\int y \textcolor{#00b695}{\mathrm{d}w}}{\int \textcolor{#00b695}{\mathrm{d}w}}\\[8pt]y_{\text{cw}} &= \frac{\displaystyle\int y \textcolor{#00b695}{ \frac{GM_{\text{Earth}}\mathrm{d}m}{y^2}}}{\displaystyle\int \textcolor{#00b695}{\frac{ GM_{\text{Earth}}\mathrm{d}m}{y^2} }}\\[8pt]y_{\text{cw}} &= \frac{\textcolor{#00b695}{\bcancel{\textcolor{black}{G}}} \textcolor{#1478c8}{\bcancel{\textcolor{black}{M_{\text{Earth}}}}} \displaystyle\int y \frac{\mathrm{d}m}{y^2}}{\textcolor{#00b695}{\bcancel{\textcolor{black}{G}}} \textcolor{#1478c8}{\bcancel{\textcolor{black}{M_{\text{Earth}}}}} \displaystyle\int \frac{\mathrm{d}m}{y^2}}\\[8pt]y_{\text{cw}} &= \frac{\displaystyle\int \frac{\mathrm{d}m}{y}}{\displaystyle\int \frac{\mathrm{d}m}{y^2}}\end{aligned}$$

Now let's rewrite the differential of mass using the linear mass density. Recall that we found that \( \mathrm{d}m = \lambda dy \). Using this result allows us to integrate with respect to \( y \) and use the limits of integration that we determined earlier.

$$\begin{aligned} y_{\text{cm}} &= \frac{\displaystyle \int \frac{\textcolor{#00b695}{\mathrm{d}m}}{y}}{\displaystyle \int \frac{\textcolor{#00b695}{\mathrm{d}m}}{y^2}}\\[10pt]y_{\text{cm}} &= \frac{\displaystyle\int \frac{\textcolor{#00b695}{\lambda \mathrm{d}y}}{y}}{\displaystyle \int \frac{\textcolor{#00b695}{\lambda \mathrm{d}y}}{y^2}}\\[10pt]y_{\text{cm}} &= \frac{\bcancel{\lambda} \displaystyle\int \frac{ \mathrm{d}y}{y}}{\bcancel{\lambda} \displaystyle \int \frac{ \mathrm{d}y}{y^2}}\end{aligned}$$

We can take the linear density out of the integral because it is constant. Finally, let's integrate using the limits we found before and simplify.

$$\begin{aligned} y_{\text{cm}} &= \frac{\displaystyle\int_{5\,371}^{8\,371} \frac{ \mathrm{d}y}{y}}{\displaystyle\int_{5\,371}^{8\,371} \frac{ \mathrm{d}y}{y^2}}\\[10pt]y_{\text{cm}} &= \frac{\ln(y)\Big |_{5\,371}^{8\,371}}{\left. -\frac{1}{y}\right|_{5\,371}^{8\,371}}\\[10pt]y_{\text{cm}} &= \frac{\ln(8\,371)-\ln(5\,371)}{\frac{1}{5\,371}-\frac{1}{8\,371}}\\[10pt]y_{\text{cm}} &= 6\, 651\end{aligned}$$

Therefore, the center of gravity of the rod is \(6\, 651\,\mathrm{km}\) from the center of the Earth, which means that the center is

$$6\,651\;\mathrm{km}- 6\,371\,\mathrm{km} = 280\,\mathrm{km}$$

above the Earth's surface. Remember that one half of the rod's length is \( 1\,500\;\mathrm{km} \), but since \( 1\,000\;\mathrm{km} \) are buried down, the geometric center would \( 500\,\mathrm{km} \) above the surface. Therefore, the center of gravity of the rod is not in its geometric center!

Fig. 5 - Notice that the center of mass is not in the same place as the center of gravity.

This is because the gravitational field of Earth changes with distance, and it is stronger near the bottom of the rod, closer to Earth's center of mass.