Sliding ladder by instantaneous centre

Given: $L = 4$ m, $\theta = 60^\circ$ with the floor, $v_A = 2$ m/s downward along the wall.

Locating the instantaneous centre

$v_A$ is vertical, so the perpendicular to it at A is the horizontal line through A. $v_B$ is horizontal, so the perpendicular at B is the vertical line through B. These meet at I.

   wall
    |A
    | \         I = instantaneous centre
  --+--\-----I     (horizontal from A,
    |   \G   |      vertical from B)
    |    \   |
    +-----\--B--- floor

Coordinates (origin at the wall-floor corner): A $(0,\ 3.464)$, B $(2,\ 0)$, so I $= (2,\ 3.464)$ m.

• $IA = L\cos60^\circ = 2.00$ m
• $IB = L\sin60^\circ = 3.464$ m
• $IG = L/2 = 2$ m (I, A, B and the corner form a rectangle, and G is the midpoint of the diagonal AB)

Angular velocity

$$\omega = \frac{v_A}{IA} = \frac{2}{2.00} = 1.00\ \text{rad/s (counter-clockwise)}$$

Velocities of B and G

$$v_B = \omega \times IB = 1.00 \times 3.464 = 3.464\ \text{m/s (horizontal, away from the wall)}$$ $$v_G = \omega \times IG = 1.00 \times 2 = 2.00\ \text{m/s}$$

$v_G$ is perpendicular to IG.

Answer: I is 2 m horizontally from A and 3.464 m vertically from B; ω = 1.00 rad/s; v_B = 3.46 m/s; v_G = 2.00 m/s.