Constraint Equations for Contacting Surfaces

1. The Fundamental Principle of Contact Constraints

Imagine a block sliding down a wedge. The block and the wedge are separate bodies, but their motions are not independent. The block must remain on the surface of the wedge. This means they cannot move into each other, nor can the block (usually) fly off the surface.

This simple physical idea can be stated with mathematical rigor. Let's consider two bodies, A and B, in contact along a surface. Let $\hat{n}$ be the unit vector that is normal (perpendicular) to the contact surface at a point.

The core constraint is that the component of the relative velocity perpendicular to the surface must be zero. If it weren't, the bodies would either be separating or interpenetrating.
$({\vec{v}}_A-{\vec{v}}_B)\cdot \hat{n}=0$
This is equivalent to stating that their velocity components along the normal direction must be equal:
${\vec{v}}_A\cdot \hat{n}={\vec{v}}_B\cdot \hat{n}$

If the bodies remain in contact, the same logic applies to their accelerations. The components of their acceleration normal to the surface must be equal:
${\vec{a}}_A\cdot \hat{n}={\vec{a}}_B\cdot \hat{n}$

This equation is the geometric constraint equation for surfaces in contact. Our main task in these problems is to define the accelerations ${\vec{a}}_A$ and ${\vec{a}}_B$ and the normal vector $\hat{n}$, and then use this equation to find the relationship between their motions.

2. Method 1: Decomposition in an Inertial Frame

The most direct way to apply the contact constraint is to work in a fixed "ground" reference frame (an inertial frame). We will analyze the classic system of a block sliding on a frictionless movable wedge.

The following video provides a careful, step-by-step derivation of the constraint relationship for just such a system. It uses a tilted coordinate system, but the underlying logic is precisely what we discussed.

Let's break down the logic from the video with a more formal vector approach.

Setup:

• A wedge of mass $M$ and angle $\theta$ rests on a frictionless horizontal surface.
• A block of mass $m$ rests on the frictionless inclined face of the wedge.
• The system is released from rest. The wedge will accelerate horizontally, and the block will slide down.

Derivation:

1. Define Accelerations (in the ground frame):

   • The wedge can only move horizontally. Its acceleration is ${\vec{a}}_W=A\hat{i}$, where $\hat{i}$ is the horizontal unit vector.
   • The block moves both horizontally and vertically. Its acceleration is ${\vec{a}}_B=a_x\hat{i}+a_y\hat{j}$.

2. Define the Constraint Direction:

   • The constraint is applied in the direction normal to the contact surface. Let's call this the $y^{\prime }$ direction, and the direction parallel to the incline the $x^{\prime }$ direction.
   • The normal direction $y^{\prime }$ is tilted at an angle $\theta$ from the vertical.

3. State the Constraint: The components of acceleration of the block and the wedge in the normal direction ($y^{\prime }$) must be equal.
   $a_{B,y^{\prime }}=a_{W,y^{\prime }}$

4. Decompose the Accelerations:

   • Wedge: The wedge's acceleration is purely horizontal, ${\vec{a}}_W=A\hat{i}$. We need to find its component in the $y^{\prime }$ direction. From geometry, the angle between the horizontal axis ($\hat{i}$) and the normal axis ($y^{\prime }$) is $90^{\circ }-\theta$. Therefore, the projection is:
     $a_{W,y^{\prime }}=|{\vec{a}}_W|\cos (90^{\circ }-\theta )=A\sin \theta$
     (This is exactly the result a_y = a sin(theta) derived in the video at 13:20, where their a is our A).

   • Block: The block's acceleration is ${\vec{a}}_B$. We can also express it as the sum of the wedge's acceleration and the block's acceleration relative to the wedge, ${\vec{a}}_{B/W}$.
     ${\vec{a}}_B={\vec{a}}_W+{\vec{a}}_{B/W}$
     The key insight is that the relative motion ${\vec{a}}_{B/W}$ is constrained to be parallel to the incline (in the $x^{\prime }$ direction). It has no component in the normal ($y^{\prime }$) direction.

     Therefore, when we look at the components of ${\vec{a}}_B$ in the $y^{\prime }$ direction, the relative motion part contributes nothing. The entire $y^{\prime }$ component of the block's acceleration comes from the motion of the wedge it's sitting on.
     $a_{B,y^{\prime }}=a_{W,y^{\prime }}+a_{B/W,y^{\prime }}=a_{W,y^{\prime }}+0=a_{W,y^{\prime }}$

This confirms our constraint equation. The component of the block's acceleration perpendicular to the incline is equal to $A\sin \theta$. This relation between the block's motion and the wedge's acceleration $A$ is the geometric constraint we need to solve the full dynamics problem. The video you watched continues in the next section ([LINK](https://www.youtube.com/watch?v=C1LfDh33wj0)-3) to use this very constraint to solve for the unknown accelerations.

3. Method 2: A More Elegant View from a Non-Inertial Frame

Your strong mathematical foundation makes you well-equipped to appreciate a more abstract and often more efficient method: analyzing the system from a non-inertial (accelerating) reference frame.

Let's place ourselves on the accelerating wedge and see how the block moves from our new perspective.

The resource you just read jumps right into the solution. Let's explore the reasoning behind this powerful technique with another video.

Let's summarize the kinematic argument for this method:

1. Ground Frame (Inertial): The block's acceleration vector ${\vec{a}}_B$ points in a complicated direction, being the sum of the wedge's horizontal motion and its own slide downwards. The constraint is a relationship between components of acceleration.
2. Wedge Frame (Non-Inertial): An observer on the wedge sees the wedge as stationary. From this perspective, the block's motion is simple: it accelerates purely along the surface of the incline. Let's call this relative acceleration ${\vec{a}}_{rel}$. The kinematic constraint is embedded in the very description of motion in this frame. The trajectory is one-dimensional.

The challenge is that Newton's second law ($\vec{F}=m\vec{a}$) doesn't hold in its simple form in an accelerating frame. One must add "fictitious forces," as mentioned in the video. While the full dynamics of this are a future topic, the kinematic simplification itself is a powerful tool to understand now. For any wedge-block problem, you can always state that in the reference frame of the wedge, the block's acceleration is purely parallel to the surface of contact.

Conclusion

Today, we've extended our understanding of constraints from simple 1D string systems to 2D geometric constraints for surfaces in contact.

Key Takeaways:

• The fundamental principle for bodies in contact is that their components of acceleration normal (perpendicular) to the contact surface must be equal: ${\vec{a}}_A\cdot \hat{n}={\vec{a}}_B\cdot \hat{n}$.
• Method 1 (Inertial Frame): We can apply this rule by decomposing the acceleration vectors of each body in a fixed coordinate system and equating their components normal to the contact surface.
• Method 2 (Non-Inertial Frame): Often, it's kinematically simpler to analyze the motion from the reference frame of one of the bodies (e.g., the wedge). In this frame, the other body's motion is simplified (e.g., purely along the incline), which directly represents the constraint.

Mastering both perspectives will provide you with a versatile and powerful toolkit for tackling complex dynamics problems.

Preview of the Next Lesson:

We have now studied string constraints and surface constraints in isolation. The next logical step is to tackle problems where both are present. In the next lesson, we will learn to solve problems involving complex pulley systems with multiple strings and moving pulleys using systematic methods. This will often involve combining the length-differentiation method with the geometric constraints we learned today, preparing you for true JEE Advanced and Olympiad-level challenges.