Explanation:
let vertical distance from A to C be [tex]h=250mm[/tex] the constraint equation is:
[tex]l_{ac} +l_{ab} =L[/tex]
we want to find [tex]l_{bc}[/tex] distance can be written as [tex]l_{bc} =h_{1}+h_{2}[/tex] which we will find by using Pythagorean theorem h 1 is a constant and can be written as h:
[tex]h_{2} =\sqrt{l_{ab}^2- s_{A} ^2}[/tex]
[tex]l_{ac} =\sqrt{s_{A}^2+ h^2 }[/tex]
[tex]l_{ab} =L-\sqrt{s_{A} ^2+h^2 }[/tex]
taking derivative w.r.t time we get
[tex]h^._{2} =-v_{B =(l_{ab} l_{ab} ^.-s_{A} v_{A} )/h_{2}[/tex]
[tex]l_{ab} ^.=(s_{A} v_{A} )/l_{ab} -L[/tex]
given [tex]s_{A} =425mm[/tex] which gives us
[tex]l_{ab} =1050-\sqrt{450^2+250^2} =556.923mm\\\\h_{2} =\sqrt{556.923^2-425^2}=359.914mm[/tex]
[tex]l_{ab} ^.=(425.25)/(556.923-1050)=-21.548mm\\\\-v_{B} =(-556.923*21.548-425.25)/(359.914)\\\\v_{B} =62.864mm/s[/tex]
