Solution to Question 21
Method of Virtual Displacement (Constraint Analysis)
Let the position of the bead P be defined by the vector $\vec{r}$. The total length of the inextensible cord is constant. The cord is composed of segments connecting P to fixed points A, B, and C (via pulleys). Let $l_A, l_B, l_C$ be the lengths of the segments $PA$, $PB$, and $PC$ respectively. The segment $BC$ between pulleys is fixed and does not affect the derivative.
$$ L = l_A + l_B + l_C + \text{const} $$Differentiating with respect to time:
$$ \frac{dl_A}{dt} + \frac{dl_B}{dt} + \frac{dl_C}{dt} = 0 $$The rate of change of the distance between a moving point P and a fixed point is given by the projection of the velocity $\vec{v}$ of P onto the unit vector pointing from P towards the fixed point. Let $\hat{u}_A, \hat{u}_B, \hat{u}_C$ be the unit vectors pointing from P towards A, B, and C respectively.
$$ \frac{dl_A}{dt} = -\vec{v} \cdot \hat{u}_A $$ $$ \frac{dl_B}{dt} = -\vec{v} \cdot \hat{u}_B $$ $$ \frac{dl_C}{dt} = -\vec{v} \cdot \hat{u}_C $$Note: The negative sign appears because if $\vec{v}$ is in the direction of the unit vector (towards the pulley), the length of the string segment decreases.
Substituting these into the constraint equation:
$$ (-\vec{v} \cdot \hat{u}_A) + (-\vec{v} \cdot \hat{u}_B) + (-\vec{v} \cdot \hat{u}_C) = 0 $$ $$ \vec{v} \cdot (\hat{u}_A + \hat{u}_B + \hat{u}_C) = 0 $$This result is physically profound. It states that the velocity vector $\vec{v}$ is always perpendicular to the vector sum of the unit vectors along the strings. Let’s define this resultant vector as $\vec{S}$:
$$ \vec{S} = \hat{u}_A + \hat{u}_B + \hat{u}_C $$The bead is constrained to move in a direction perpendicular to $\vec{S}$. This is dynamically equivalent to a particle sliding on a smooth rigid surface whose normal vector is parallel to $\vec{S}$.
Calculating the Effective Incline Angle
We define a coordinate system with the x-axis horizontal (leftward positive to match the diagram’s geometry of A and B) and y-axis vertical (upward positive).
- $\hat{u}_A$: Points towards A. Angle $\alpha$ with horizontal. $\hat{u}_A = \cos\alpha \hat{i} + \sin\alpha \hat{j}$
- $\hat{u}_B$: Points towards B. Angle $\beta$ with horizontal. $\hat{u}_B = \cos\beta \hat{i} + \sin\beta \hat{j}$
- $\hat{u}_C$: Points towards C (Vertical). $\hat{u}_C = 0\hat{i} + 1\hat{j}$
The components of the resultant normal vector $\vec{S}$ are:
$$ S_x = \cos\alpha + \cos\beta $$ $$ S_y = \sin\alpha + \sin\beta + 1 $$Let $\phi$ be the angle this normal vector $\vec{S}$ makes with the vertical ($y$-axis).
$$ \tan\phi = \frac{S_x}{S_y} = \frac{\cos\alpha + \cos\beta}{1 + \sin\alpha + \sin\beta} $$Acceleration Calculation
The bead behaves like a block on a frictionless incline. The “plane” of motion is perpendicular to the normal $\vec{S}$. If the normal makes an angle $\phi$ with the vertical, the plane makes an angle $\phi$ with the horizontal.
The acceleration of a body sliding down a smooth incline of angle $\phi$ is:
$$ a = g \sin\phi $$From the geometry of the tangent/triangle associated with $\phi$:
$$ \theta = \phi = \tan^{-1} \left( \frac{\cos\alpha + \cos\beta}{1 + \sin\alpha + \sin\beta} \right) $$