Acceleration of a ball on a plane

Question

I have a plane ($ax+by+cz+d=0$) in a 3D world, and a gravity vector $\vec{g}$ (say it's $[0, 0, -9.81]$.) How would I find the acceleration vector of an object on this plane, ignoring friction?

Answer 1

What you want to do is project $\vec{g}$ on the plane. First we will consider a new plane: $ax+by+cz=0$. This plane is parallel to our previous one, so our result won't change, but it goes through the origin. Then we need a normal vector of this plane. In this case, we're lucky because we have this particular form of the equation of the plane. A normal vector is in this case $\vec{n}=[a,b,c]$.

Now we will create a line going through $\vec{g}$, and parallel to $\vec{n}$. Then we get the following parameter-equation (I don't know the proper term): $$\vec{x}=\lambda\vec{n}+\vec{g}$$ Where $\lambda$ is the parameter. The intersection of this line and our plane will be $\vec{g}$ projected on our plane. Let's call this intersection-point $\vec{p}$. For $\vec{p}$ it must be true that: $$\vec{p}\text{ is on our line: }ap_x+bp_y+cp_z=\vec{p}\cdot\vec{n}=0$$ $$\vec{p}\text{ is on our plane: }\lambda\vec{n}+\vec{g}=\vec{p}$$ Plugging the second equation into the first we get: $$(\lambda\vec{n}+\vec{g})\cdot\vec{n}=\lambda\vec{n}\cdot\vec{n}+\vec{g}\cdot\vec{n}=0$$ $$\Rightarrow\lambda=-\frac{\vec{g}\cdot\vec{n}}{\vec{n}\cdot\vec{n}}$$ Plugging this into the first equation gives us: $$\vec{p}=\vec{g}-\frac{\vec{g}\cdot\vec{n}}{\vec{n}\cdot\vec{n}}\vec{n}$$ In your example ($\vec{g}=[0, 0, -9.81]$) it looks like: $$\vec{p}=\begin{bmatrix} \frac{9.81c}{a^2+b^2+c^2}a\\ \frac{9.81c}{a^2+b^2+c^2}b\\ \frac{9.81c}{a^2+b^2+c^2}c-9.81\\ \end{bmatrix}$$

Answer 2

you are looking to find the component of the gravity vector in the direction of the line of greatest slope of the plane.

Taking the gravity vector as vertically downwards, the vector parallel to the plane which is horizontal is $$\underline{n}\times\underline{\hat{g}}=\left(\begin{matrix}-b\\a\\0\end{matrix}\right)$$

Then the line of greatest slope is $$\left(\begin{matrix}-b\\a\\0\end{matrix}\right)\times\underline{n}=\left(\begin{matrix}ac\\bc\\-a^2-b^2\end{matrix}\right)$$

Now calculate the component of $$\left(\begin{matrix}0\\0\\-g\end{matrix}\right)$$ in this direction, which is $$\frac{g(a^2+b^2)}{\sqrt{a^2c^2+b^2c^2+(a^2+b^2)^2}}$$

Answer 3

I like to think in terms of projections. The projection of a vector $v$ onto a plane with normal $n$ is $$p= v - n \;(v\cdot n) /(n \cdot n)$$ where $(x_1, y_1, z_1) \cdot (x_2,y_2,z_2) = x_1 x_2 + y_1 y_2 + z_1 z_2$.

So, the projection of the gravity vector onto your plane is just $$ p = g - n\; (g\cdot n) /(n \cdot n).$$

Setting $g=(0,0,-9)$ and $n=(a,b,c)$, we get $$ \begin{aligned} p &= (0,0,-9) - (a,b,c) \frac{(0,0,-9)\cdot (a,b,c)}{ a^2+b^2+c^2}\\ &= (0,0,-9) + (a,b,c)\frac{9c}{ a^2+b^2+c^2}. \end{aligned} $$ (This matches the answers by WE-Tutotial-School, David Quinn, and CodingDragon04.)