Question

The rod AB, of length 1 m, shown in the figure is connected to two soliders at each end through pins. The sliders can slide along QP and QR. If the velocity $V_A$ of the slider at A is 2 m.s, the velocity of the midpoint of the rod at this instant is m/s.

Answer 1

${\omega }_3={\omega }_{AB}=\frac{V_A}{I_{13A}}=\frac{V_B}{I_{13B}}=\frac{V_C}{I_{13C}}$
$I_{13A}=$ Distance between $I_{13}$ and point A
$I_{13B}=$ Distance between $I_{13}$ and point B
$I_{13C}=$ Distance between $I_{13}$ and point C
$\frac{V_A}{I_{13A}}=\frac{V_C}{I_{13C}};\frac{2}{I_{13A}}=\frac{V_C}{I_{13C}}$
Applying sine law in triangle $AB{I}_{13}$

$\frac{1}{sin{120}^o}=\frac{I_{13A}}{sin{30}^o}$
$I_{13A}=\frac{sin3{=}^o}{sin{120}^o}$
Applying sine lau in $\Delta AC{I}_{13}=0.5774$
$\frac{I_{13C}}{sin{30}^o}=\frac{I_{13A}}{sin{90}^o}$
$I_{13C}=\frac{I_{13A}sin{30}^o}{sin{90}^o}$
$=0.5774\times 0.5=0.2887m$
$V_C=\frac{V_A.I_{13C}}{I_{13a}}=\frac{2\times 0.2887}{0.5774}=1$m/s