Four identical rods are hinged at their ends to make a parallelogram ABCD. The hinged joint A is rig

Q: Why do we differentiate the constraint equation twice?

The first derivative relates the velocities of the joints. Since we need to find the acceleration, we must differentiate a second time to bring $\ddot{x}$ and $\ddot{y}$ into the equation.

Q: How do we know $v_C^2 = 2a x_C$?

The problem states that initially A and C were coincident, meaning the system started from rest ($v=0$ at $x=0$). Since C is pulled with a constant acceleration $a$, we can use the third equation of motion: $v^2 = u^2 + 2as$.

Q: Why is the horizontal acceleration of B exactly $a/2$?

Because the rods form a rhombus, the diagonals bisect each other. The horizontal distance to B is always exactly half the horizontal distance to C. Therefore, its velocity and acceleration are also exactly half.

Q: Why does the answer use $\tan\theta$ instead of $\cot\theta$?

This depends on how $\theta$ is defined. If $\theta$ is the angle the rod makes with the vertical y-axis, then $x_B/y_B = \tan\theta$. If it were the angle with the horizontal, it would be $\cot\theta$.

Q: Why does the given answer have a positive $\hat{j}$ component?

The upper joint B is accelerating downwards, so its true vector has a negative $\hat{j}$ component. The positive sign in the answer corresponds to the lower joint D, which accelerates upwards symmetrically.

Question

Four identical rods are hinged at their ends to make a parallelogram ABCD. The hinged joint A is rigidly attached to a wall and the opposite joint C is pulled away from the wall with a constant acceleration $a$ as shown in the figure. Initially, the joints A and C were coincident. Find the acceleration vector of the joint B at the instant shown.

Sigma Percentile · Accepted Answer

Imagine a scissor lift or a folding mirror extending outwards. The motion of the joints is entirely dictated by the rigid rods connecting them. This problem is a beautiful exploration of constrained motion, where the geometry of a rhombus forces the velocities and accelerations of its vertices to dance in perfect synchronization.

### Setting the Stage: The Coordinate Method

When dealing with complex linkages, relying purely on angles and chain rules can quickly lead to a tangled mess of trigonometric identities. Instead, we anchor our analysis in a robust Cartesian coordinate system.

Let the fixed joint A be our origin $(0,0)$. Since joint C is pulled horizontally away from the wall, it moves strictly along the x-axis. We can denote its position as $(x_C, 0)$. Because the four identical rods form a rhombus, the diagonals bisect each other perfectly. This geometric symmetry guarantees that joint B will always be exactly halfway between A and C horizontally. Therefore, the coordinates of B are $(x_B, y_B)$, where $x_C = 2x_B$.

### The Master Constraint Equation

The fundamental physics of this system is governed by a single, unbreakable rule: the length of the rod AB is constant. We can express this using the Pythagorean theorem:

$$x_B^2 + y_B^2 = l^2$$

To uncover how the joints move, we differentiate this constraint equation with respect to time. The first derivative links the velocities:

$$2x_B \dot{x}_B + 2y_B \dot{y}_B = 0 \implies x_B \dot{x}_B + y_B \dot{y}_B = 0$$

But we need accelerations. So, we take a deep breath and differentiate one more time, carefully applying the product rule:

$$\dot{x}_B^2 + x_B \ddot{x}_B + \dot{y}_B^2 + y_B \ddot{y}_B = 0$$

This elegant equation is the key to unlocking the vertical acceleration of joint B.

### Translating the Motion of C to B

The problem states that initially, joints A and C were coincident. This means the system started from rest, completely folded along the y-axis. Joint C is then pulled with a constant horizontal acceleration $a$. Using basic kinematics ($v^2 = u^2 + 2as$), the square of C's velocity is directly proportional to its displacement:

$$v_C^2 = 2a x_C$$

Because $x_B$ is always exactly half of $x_C$, the horizontal velocity and acceleration of B are simply half of C's:

$$\dot{x}_B = \frac{1}{2} v_C \implies \dot{x}_B^2 = \frac{1}{4} v_C^2 = \frac{1}{4} (2a x_C) = a x_B$$

$$\ddot{x}_B = \frac{1}{2} a_C = \frac{a}{2}$$

### The Final Algebraic Assembly

Now, we substitute these horizontal components back into our acceleration constraint. First, we express $\dot{y}_B^2$ using the velocity relation:

$$\dot{y}_B = -\frac{x_B \dot{x}_B}{y_B} \implies \dot{y}_B^2 = \frac{x_B^2 (a x_B)}{y_B^2} = a \frac{x_B^3}{y_B^2}$$

Plugging everything into the second derivative equation yields:

$$a x_B + x_B \left(\frac{a}{2}ight) + a \frac{x_B^3}{y_B^2} + y_B \ddot{y}_B = 0$$

Combining the terms and isolating $\ddot{y}_B$, we get:

$$y_B \ddot{y}_B = -a \left( \frac{3}{2} x_B + \frac{x_B^3}{y_B^2} ight)$$

$$\ddot{y}_B = -a \left( \frac{3}{2} \frac{x_B}{y_B} + \left(\frac{x_B}{y_B}ight)^3 ight)$$

Notice how the ratio $\frac{x_B}{y_B}$ naturally emerges. If we define $	heta$ as the angle the rod makes with the vertical y-axis, then from the right triangle formed by the coordinates of B, we have $	an	heta = \frac{x_B}{y_B}$.

Substituting this trigonometric ratio gives us the final vertical acceleration:

$$\ddot{y}_B = -a \left( \frac{3}{2} 	an	heta + 	an^3	heta ight)$$

The negative sign perfectly captures the physical reality: as the rhombus is pulled horizontally, it flattens out, forcing the upper joint B to accelerate downwards. The given answer in the text features a positive $\hat{j}$ component, which corresponds to the symmetric lower joint D accelerating upwards, or implies a coordinate system where downwards is taken as positive.

Visualized Solution

Coordinate Setup

Geometric Constraint

Velocity Relation

Acceleration Relation

Kinematics of C

Kinematics of B (x-axis)

Vertical Acceleration of B

Simplifying $\overset{y}{¨}_{B}$

Final Acceleration Vector

The Sigma Insight: Motion in a Plane

Setting the Stage

The Master Constraint Equation

Translating the Motion of C to B

The Final Algebraic Assembly

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