Question

Four rods each of length $I$ have been hinged to form a rhombus. Vertex $A$ is fixed to a rigid support, vertex $C$ is being moved along the $x-$axis with a constant velocity $v$ as shown in the figure. The rate at which vertex $B$ is approaching the $x-$axis at the moment the rhombus is in the form of a square is

Answer 1

Step 1: Given Data

$AB=l$

From the figure, let $AC=x$ and $BE=y$

$x=2y$, when the rhombus is a square.

Let the velocity of vertex $B$ and $C$ be $v_B$ and $v_C$ respectively.

Step 2: Formula Used

According to Pythagoras' theorem,

${\left(Penpendicular\right)}^2+{\left(Base\right)}^2={\left(Hypotenuse\right)}^2$

Step 3: Calculate the velocity of vertex $B$

From the figure, we can see that $△ABE$ is a right-angled triangle.

Since, $ABCD$ is a rhombus, the diagonals $AC$ and $BD$ will intersect each other.

$\therefore AE=\frac{AC}{2}=\frac{x}{2}$

According to Pythagoras' theorem,

$$\begin{gathered} B{E}^2+A{E}^2=A{B}^2=l^2 \\ \Rightarrow y^2+{\left(\frac{x}{2}\right)}^2=l^2 \end{gathered}$$

Upon differentiating with respect to $t$ we get,

$$\begin{gathered} 2y\frac{dy}{dt}+\frac{x}{2}\frac{dx}{dt}=0\left[\because l=const\right] \\ \Rightarrow -\frac{dy}{dt}=\frac{1}{2}\frac{x}{2y}\frac{dx}{dt} \\ \Rightarrow v_B=\frac{1}{2}{v}_C=\frac{v}{2} \end{gathered}$$

Hence, the velocity of vertex $B$ is $\frac{v}{2}$.