• # question_answer A linkage is shown below in the figure in which links ABC and DBF are ternary links whereas AF, BE and CD are binary links. The degrees of freedom of the linkage when link ABC is fixed are: A) 0                                 B) 1C) 2                                 D) 3

Degrees of freedom, $F=3\,(N\,-1)-2{{P}_{1}}-{{P}_{2}},$ Where ${{P}_{1}}=\text{1/2(2}{{\text{N}}_{2}}\text{+3}{{\text{N}}_{3}}\text{)}\text{.}$ Number of binary links, ${{N}_{2}}=3,$ Number of ternary links, ${{N}_{3}}=2$ ${{P}_{1}}=\text{1/2}\text{.(2}\times 3+3\times 2\text{)=6,}$ $N=5,$ ${{P}_{2}}=0$ $F=3(5-1)-2\times 6=0$