# A Doubt On Constrained Motions

When solving a constrained motion (particularily in wedge constraints) problem we often, in order to solve it quickly, use a trick or formula which goes by $$\sum \vec N \cdot \vec a = 0$$ where $$\vec N$$ is the normal force acting on a body and $$\vec a$$ is its acceleration.

What I wanted to know is how far is this formula correct, and how is it derived?

• It just encapsulates the fact that the body doesn't fall through the wedge. It means that the normal force always has magnitude and direction such that the body has no acceleration perpendicular to the wedge. Aug 30, 2020 at 13:52
• With due respect can you please elaborate sir...... Aug 30, 2020 at 14:53
• Does this answer your question? Understanding constraint for not losing contact Aug 30, 2020 at 20:50

For arbitrary path $$~x(s)\,,y(s)~$$ where s is the line element, you obtain at arbitrary point P,the tangent vector $$\vec{t}(s)$$ and the normal vector $$\vec{n}(s)$$

the acceleration $$\vec{a}=[\ddot x~,\ddot y]^T~$$ is:

$$\vec{a}=\ddot{s}\,\vec{t}$$

and the normal force vector $$\vec{N}$$ is:

$$\vec{N}=N\,\vec{n}$$

take the dot product you obtain:

$$\vec a\cdot \vec N=\ddot{s}\,\vec{t}\cdot N\,\vec{n}=0$$

the dot product is zero because the tangent vector is perpendicular to normal vector $$\vec t\perp\vec n$$ but $$\ddot{s}\,N$$ is not zero!