Conditions for Equilibrium in 3 dimension and 2 dimension

Question

The book describes that there are two conditions for equilibrium of rigid body, $$\sum \vec{F} = 0, \quad \sum \vec{M}_{o} = 0 $$ where $\sum \vec{F} $ is the vector sum of all the external forces acting on the body and $\sum \vec{M}_{o} $ is the sum of couple moments and moments of all the forces about any point O.

If we want to express these external forces and couple moments in Cartesian coordinate system , we get 6 equilibrium equations, $$ \sum F_{x} = \sum F_{y} = \sum F_{z} = 0$$ $$ \sum M_{x} = \sum M_{y} = \sum M_{z} = 0$$ I understand this, but in 2 dimensions, the equilibrium equations are $$ \sum F_{x} = \sum F_{y} = \sum M_{o} = 0.$$ Why are there only three equations? Why aren't there four, i.e. $$ \sum F_{x} = \sum F_{y} = \sum M_{x} = \sum M_y = 0.$$

Answer 1

What the book meant by "two dimensions" is actually a system whose forces acting on are lying in a plane, i.e., they are coplanar forces. Since the torque, $$\vec M=\vec r\times\vec F,$$ is a vector product involving forces, it points in a direction perpendicular to that plane. If the forces are in the $xy$ plane, then the torque is in the $z$ direction.

Answer 2

If a force $\vec{F} = \pmatrix{F_x \\ F_y \\ 0}$ lies on a plane at a location (also on the same plane) $\vec{r} = \pmatrix{x \\ y \\ 0}$ then the equipollent moment has only an out-of-plane component

$$ \vec{M}_0 = \vec{r} \times \vec{F} = \pmatrix{x \\ y\\ 0} \times \pmatrix{F_x \\ F_y \\ 0} = \pmatrix{0\\0\\ x F_y - y F_x} $$

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There is something similar going on, but in reverse, with motion. The velocity of a point $\vec{r} = \pmatrix{x \\ y \\ 0}$ due to a rotation about the origin of $\vec{\omega} = \pmatrix{0 \\ 0 \\ \dot{\theta}}$ lies entirely on the plane

$$ \vec{v} = \vec{\omega} \times \vec{r} = \pmatrix{0 \\ 0 \\ \dot{\theta}} \times \pmatrix{x \\ y \\ 0} = \pmatrix{-y \dot{\theta} \\ x \dot{\theta} \\ 0} $$