Physics Solution: Kinematics of Linkages
1. Coordinate Setup and Geometry
Let the fixed joint A be at the origin $(0,0)$. Let the position of joint C be $(x, 0)$ as it moves along the x-axis. Since the rods form a parallelogram (specifically a rhombus because the rods are identical), the joint B will always be horizontally centered between A and C.
Let the length of each rod be $l$. The coordinates of B are $(x_B, y_B)$.
$$ x_B = \frac{x}{2} $$
From the right-angled triangle formed by the vertical line dropping from B to the diagonal AC:
$$ \left(\frac{x}{2}\right)^2 + y_B^2 = l^2 $$
$$ y_B = \sqrt{l^2 – \frac{x^2}{4}} $$
Let the angle the rod makes with the vertical at joint B be $\theta$. From geometry:
$$ \tan\theta = \frac{x/2}{y_B} = \frac{x}{2y_B} \quad \Rightarrow \quad x = 2y_B \tan\theta $$
2. Velocity Analysis
We are given that C starts from rest (where A and C were coincident, $x=0$) and moves with constant acceleration $a$.
$$ \ddot{x} = a $$
$$ v_x = \dot{x} = at \quad (\text{since } u=0) $$
$$ x = \frac{1}{2}at^2 $$
Now, let’s find the velocity components of B.
Differentiating $x_B = x/2$:
$$ v_{Bx} = \dot{x}_B = \frac{\dot{x}}{2} = \frac{at}{2} $$
Differentiating the constraint equation $\frac{x^2}{4} + y_B^2 = l^2$ with respect to time:
$$ \frac{1}{4}(2x\dot{x}) + 2y_B\dot{y}_B = 0 $$
$$ \frac{x\dot{x}}{2} + 2y_B\dot{y}_B = 0 \quad \Rightarrow \quad \dot{y}_B = -\frac{x\dot{x}}{4y_B} $$
Substituting $\frac{x}{2y_B} = \tan\theta$:
$$ \dot{y}_B = -\frac{1}{2} \dot{x} \tan\theta = -\frac{1}{2}(at)\tan\theta $$
3. Acceleration Analysis
The acceleration of B has two components: $\vec{a}_B = a_{Bx}\hat{i} + a_{By}\hat{j}$.
x-component:
$$ a_{Bx} = \frac{d}{dt}(v_{Bx}) = \frac{d}{dt}\left(\frac{at}{2}\right) = \frac{a}{2} $$
y-component:
We differentiate the velocity relation $\dot{y}_B = -\frac{x\dot{x}}{4y_B}$ or implicitly differentiate the constraint equation again:
$$ \frac{d}{dt} \left( \frac{x\dot{x}}{4} + y_B\dot{y}_B \right) = 0 $$
$$ \frac{1}{4}(\dot{x}^2 + x\ddot{x}) + (\dot{y}_B^2 + y_B\ddot{y}_B) = 0 $$
We need to solve for $\ddot{y}_B$:
$$ y_B\ddot{y}_B = – \left( \dot{y}_B^2 + \frac{\dot{x}^2}{4} + \frac{x\ddot{x}}{4} \right) $$
Substitute the known kinematic values ($\ddot{x}=a, \dot{x}^2 = (at)^2 = 2ax, \dot{y}_B = -\frac{x\dot{x}}{4y_B}$):
First, simplify the term in the bracket. Note that $v_x^2 = 2ax$.
$$ \dot{y}_B^2 = \left( -\frac{x \dot{x}}{4y_B} \right)^2 = \frac{x^2 (2ax)}{16y_B^2} = \frac{ax^3}{8y_B^2} $$
Now the bracketed term becomes:
$$ \left[ \frac{ax^3}{8y_B^2} + \frac{2ax}{4} + \frac{xa}{4} \right] = a \left[ \frac{x^3}{8y_B^2} + \frac{3x}{4} \right] $$
So,
$$ y_B\ddot{y}_B = -ax \left[ \frac{x^2}{8y_B^2} + \frac{3}{4} \right] $$
$$ \ddot{y}_B = -\frac{ax}{y_B} \left[ \frac{1}{2} \left(\frac{x}{2y_B}\right)^2 + \frac{3}{4} \right] $$
4. Final Substitution
Recall that $\frac{x}{2y_B} = \tan\theta$. Therefore $\frac{x}{y_B} = 2\tan\theta$.
$$ \ddot{y}_B = -a (2\tan\theta) \left[ \frac{1}{2}\tan^2\theta + \frac{3}{4} \right] $$
$$ \ddot{y}_B = -a \left[ \tan^3\theta + \frac{3}{2}\tan\theta \right] $$
The negative sign indicates the acceleration is directed downwards (as the parallelogram expands horizontally, the height decreases).
Conclusion:
The acceleration vector of joint B is:
$$ \vec{a}_B = \frac{a}{2}\hat{i} – a\left( \frac{3}{2}\tan\theta + \tan^3\theta \right)\hat{j} $$
(Note: While the magnitude direction for $y$ is negative in standard Cartesian coordinates, some answer keys might present the magnitude of the components or use a coordinate system where $y$ is positive downwards. The physical direction is strictly: x-component to the right, y-component downwards).
Let the fixed joint A be at the origin $(0,0)$. Let the position of joint C be $(x, 0)$ as it moves along the x-axis. Since the rods form a parallelogram (specifically a rhombus because the rods are identical), the joint B will always be horizontally centered between A and C.
Let the length of each rod be $l$. The coordinates of B are $(x_B, y_B)$. $$ x_B = \frac{x}{2} $$ From the right-angled triangle formed by the vertical line dropping from B to the diagonal AC: $$ \left(\frac{x}{2}\right)^2 + y_B^2 = l^2 $$ $$ y_B = \sqrt{l^2 – \frac{x^2}{4}} $$ Let the angle the rod makes with the vertical at joint B be $\theta$. From geometry: $$ \tan\theta = \frac{x/2}{y_B} = \frac{x}{2y_B} \quad \Rightarrow \quad x = 2y_B \tan\theta $$
We are given that C starts from rest (where A and C were coincident, $x=0$) and moves with constant acceleration $a$. $$ \ddot{x} = a $$ $$ v_x = \dot{x} = at \quad (\text{since } u=0) $$ $$ x = \frac{1}{2}at^2 $$ Now, let’s find the velocity components of B. Differentiating $x_B = x/2$: $$ v_{Bx} = \dot{x}_B = \frac{\dot{x}}{2} = \frac{at}{2} $$ Differentiating the constraint equation $\frac{x^2}{4} + y_B^2 = l^2$ with respect to time: $$ \frac{1}{4}(2x\dot{x}) + 2y_B\dot{y}_B = 0 $$ $$ \frac{x\dot{x}}{2} + 2y_B\dot{y}_B = 0 \quad \Rightarrow \quad \dot{y}_B = -\frac{x\dot{x}}{4y_B} $$ Substituting $\frac{x}{2y_B} = \tan\theta$: $$ \dot{y}_B = -\frac{1}{2} \dot{x} \tan\theta = -\frac{1}{2}(at)\tan\theta $$
The acceleration of B has two components: $\vec{a}_B = a_{Bx}\hat{i} + a_{By}\hat{j}$. x-component: $$ a_{Bx} = \frac{d}{dt}(v_{Bx}) = \frac{d}{dt}\left(\frac{at}{2}\right) = \frac{a}{2} $$ y-component: We differentiate the velocity relation $\dot{y}_B = -\frac{x\dot{x}}{4y_B}$ or implicitly differentiate the constraint equation again: $$ \frac{d}{dt} \left( \frac{x\dot{x}}{4} + y_B\dot{y}_B \right) = 0 $$ $$ \frac{1}{4}(\dot{x}^2 + x\ddot{x}) + (\dot{y}_B^2 + y_B\ddot{y}_B) = 0 $$ We need to solve for $\ddot{y}_B$: $$ y_B\ddot{y}_B = – \left( \dot{y}_B^2 + \frac{\dot{x}^2}{4} + \frac{x\ddot{x}}{4} \right) $$ Substitute the known kinematic values ($\ddot{x}=a, \dot{x}^2 = (at)^2 = 2ax, \dot{y}_B = -\frac{x\dot{x}}{4y_B}$): First, simplify the term in the bracket. Note that $v_x^2 = 2ax$. $$ \dot{y}_B^2 = \left( -\frac{x \dot{x}}{4y_B} \right)^2 = \frac{x^2 (2ax)}{16y_B^2} = \frac{ax^3}{8y_B^2} $$ Now the bracketed term becomes: $$ \left[ \frac{ax^3}{8y_B^2} + \frac{2ax}{4} + \frac{xa}{4} \right] = a \left[ \frac{x^3}{8y_B^2} + \frac{3x}{4} \right] $$ So, $$ y_B\ddot{y}_B = -ax \left[ \frac{x^2}{8y_B^2} + \frac{3}{4} \right] $$ $$ \ddot{y}_B = -\frac{ax}{y_B} \left[ \frac{1}{2} \left(\frac{x}{2y_B}\right)^2 + \frac{3}{4} \right] $$
Recall that $\frac{x}{2y_B} = \tan\theta$. Therefore $\frac{x}{y_B} = 2\tan\theta$. $$ \ddot{y}_B = -a (2\tan\theta) \left[ \frac{1}{2}\tan^2\theta + \frac{3}{4} \right] $$ $$ \ddot{y}_B = -a \left[ \tan^3\theta + \frac{3}{2}\tan\theta \right] $$ The negative sign indicates the acceleration is directed downwards (as the parallelogram expands horizontally, the height decreases).
The acceleration vector of joint B is: $$ \vec{a}_B = \frac{a}{2}\hat{i} – a\left( \frac{3}{2}\tan\theta + \tan^3\theta \right)\hat{j} $$ (Note: While the magnitude direction for $y$ is negative in standard Cartesian coordinates, some answer keys might present the magnitude of the components or use a coordinate system where $y$ is positive downwards. The physical direction is strictly: x-component to the right, y-component downwards).