Question

A straight link AB 40cm long has, at a given instant, its end B moving along line OX at 0.8m/ s and acceleration at 4m/sec² and the other end A moving along OY, as shown in fig. Find the velocity and acceleration of the end A and of mid point C of the link when inclined at 300 with OX.

Answer 1

Let the length of link is L = 40cm and AD = Y, OB = X

X² + Y² = 1 ...(i)

Diff with respect to time, and -ive sign is taken for down word motion of A, when B is moving in +ive direction, we get

2Xdx/dt – 2Ydy/dt = 0XV_B – YV_A = 0

V_A = (X/Y)V_B = (Lcosθ/Lsinθ)V_B = VB/tanθV_A 

= 0.8/tan300 = 1.38m/sec ...(iii)

V_A = 1.38m/sec

Again differentiating equation (2), we get

Xd²x/dt² + (dx/dt)² – yd²y/dt² – (dy/dt)² = 0

X.a_B + (V_B)² – Y.a_A – (V_A)² = 0

0.4cos 300 X 0.4 – (0.8)² – 0.4sin 300 X a_A – (1.38)²= 0

1.38 + 0.64 – 0.2a_A – 1.9 = 0

a_A = 0.6.6m/sec²