Final answer:
To find linkage configurations with M=1 using six links, we apply the mobility equation. We derive configurations by substituting L=6 and M=1 into the equation, resulting in several possible combinations of the number of links. Two examples of such combinations have been discussed.
Step-by-step explanation:
Deriving the two possible configurations of higher order linkages that produce M=1 using six total links involves using the equation L-3-M=T+2Q+3P+4H, with 'L' representing the total number of links, 'M' the mobility of the system, 'T' the number of lower pairs, 'Q' the number of quaternary links, 'P' the number of pentanary links, and 'H' the number of hexanary links. Given that we have six links (L=6) and want a mobility of 1 (M=1), substituting into the equation gives us 6-3-1=T+2Q+3P+4H.
From this, there are multiple potential combinations of T, Q, P, and H which will satisfy the equation. As an example, for a simple one degree of freedom system (M=1) with six links, we may have one ternary link and three binary links, resulting in T=3, Q=1, and P=H=0. Another example could be zero lower pairs, two quaternary links, and zero of the rest, giving T=0, Q=2, P=H=0. These combinations show the arrangements that satisfy the mobility equation, which can be further analyzed to show the specific linkage configurations.