Collar B moves up a fixed vertical rod with a constant velocity of 3 m/s. Rod AB, 2 m long, is pinned to the collar at B and makes 45° with the horizontal at B, running down to the left to end A. End A slides on a fixed surface inclined at 30° to the…

Answer: \(\omega_{AB}=1.345\) rad/s (anticlockwise); \(v_A=2.20\) m/s up the incline (at 30° to the horizontal); \(v_G=2.26\) m/s.

Directions of velocities: \(v_B\) is vertical (upward); \(v_A\) is along the incline (at 30° to the horizontal).

Locating I: draw a line through B perpendicular to \(v_B\) (a horizontal line through B), and a line through A perpendicular to the incline (at 120° to the horizontal, i.e. leaning up and to the left). They meet at I, to the left of B on the horizontal through B.

Angles of triangle IAB:

• At B: between BI (horizontal) and BA (45° below the horizontal) \(=45^\circ\).
• At A: AB points up-right at 45°, AI points up-left at 120° from the horizontal, so angle \(=120^\circ-45^\circ=75^\circ\).
• At I: \(180^\circ-45^\circ-75^\circ=60^\circ\).

Sine rule: \(\dfrac{IB}{\sin75^\circ}=\dfrac{IA}{\sin45^\circ}=\dfrac{AB}{\sin60^\circ}=\dfrac{2}{0.8660}=2.309\)

\(IB=2.309\times0.9659=2.231\) m; \(IA=2.309\times0.7071=1.633\) m.

Angular velocity: \(\omega_{AB}=\dfrac{v_B}{IB}=\dfrac{3}{2.231}=1.345\) rad/s. B (to the right of I) moves up, so the rod turns anticlockwise.

Velocity of A: \(v_A=\omega\times IA=1.345\times1.633=2.20\) m/s, directed up the incline (anticlockwise about I, with A below I, moves A to the right and up along the slope).

Velocity of G: in triangle IBG, \(BG=1\) m, angle at B \(=45^\circ\):

\(IG^2=IB^2+BG^2-2(IB)(BG)\cos45^\circ=4.976+1-3.155=2.821\Rightarrow IG=1.680\) m.

\(v_G=\omega\times IG=1.345\times1.680=2.26\) m/s, perpendicular to IG.