Question

A rigid link PQ is 2 m long and oriented at ${20}^o$ to the horizontal as shown in the figure. The magnitude and direction of velocity $V_Q$ , and the direction of velocity $V_P$ are given. The magnitude of $V_P$ (in m/s) at this instant is

• A. 2.14
• B. 1.89
• C. 1.21
• D. 0.96

Answer 1

The correct option is D 0.96

Method I :

$\therefore {\omega }_{PQ}=\frac{V_P}{P{I}_{PQ}}=\frac{V_Q}{Q{I}_{PQ}}$

$P.I_{PQ}=$ Distance between P and $I_{PQ}$

$=2cos{70}^o+2sin{70}^o\times tan{45}^o$

$=2.56m$

$Q.I_{PQ}=$ Distance between A and $I_{PQ}$

$=\sqrt{2}\times 2\times sin{70}^o=2.66m$

$\therefore \frac{V_P}{2.56}=\frac{1}{2.66}$

or $V_p=\frac{2.56}{2.66}=0.96m/s$
Method II:
As, according to the data, link PQ is rigid link, so the velocity at very along the link will be same
Therefore :

$V_{P}cos{20}^o=V_{Q}cos{25}^o$

$V_P=V_Q\times \frac{cos{25}^o}{cos{20}^o}=1\times 0.9644$

$V_P=0.96$m/s

Points to Remember:
In this problem, as we have seen that
$V_B=V_c$
but, velocity of B with respect to C is zero. i.e.
$V_{BC}=0$