The degrees of freedom of the linkage when link ABC is fixed are:
A) 0
B) 1
C) 2
D) 3
Correct Answer: A
Solution :
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, \[{{N}_{3}}=2\] \[{{P}_{1}}=\text{1/2}\text{.(2}\times 3+3\times 2\text{)=6,}\] \[N=5,\] \[{{P}_{2}}=0\] \[F=3(5-1)-2\times 6=0\]
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