v_{A1} = v/2 ??
v_{A2}=5v/6 ?
v_{B2} = -v/6 ???
Traditionally been weak in this part of mechanics...pulley problems samajh aate they magar usse zyaada nahi.
Q. A hinged construction consists of three rhombuses with the ratio of sides 3:2:1 . Vertex A3 moves in the horizontal direction at a velocity v. Determine the velocities of vertices A1, A2, and B2 at the instant when the angles of the construction are 90 degrees.
How am I supposed to approach this? Kisi cheez ko constant lena hai...magar kisko?
take B_{3}A_{3} = r , B_{2}A_{2}=2r , B_{1}A_{1}=3r
now take angle B_{1}A_{1}A_{0} = B_{2}A_{2}A_{1}=B_{3}A_{3}A_{2}= a
v_{A1} = d(A_{0}A_{1})/dt = d/dt ( 2 x 3rcosa)
now
\frac{d(A_{1}A_{2})}{dt} = v_{A_{2}}-v_{A_{1}}
so v_{A_{2}}= \frac{d(A_{1}A_{2})}{dt}+ v_{A_{1}}
v_{A_{2}}= \frac{d(2\times 2rcosa)}{dt}+ v_{A_{1}}
\frac{d(A_{2}A_{3})}{dt} = v_{A_{3}}-v_{A_{2}} = v-v_{A_{2}}
v=\frac{d(A_{2}A_{3})}{dt}+v_{A_{2}}
v_{B}=\frac{d(2rsina)}{dt}
now find relation between v ,r , da/dt .
Nice! It sort of makes sense now :D
You took A_{0} as your stationary reference point and measured everything from there. To find the bases of those triangles you used trigo in a half triangle then multiplied by 2.
To find the velocity of A_{1} you again used A_{0} as a reference point, by subtracting those specific velocities. Nice one!