Question #208089

An epicyclic reduction gear has a shaft A fixed to arm B.The arm B has a pin fixed to its outer end and two heard C and E which are rigidly fixed,revolve on this pin. Gear C meshes with annular wheel D and gear E with pinion F.G is the driver pulley and D is kept stationary.The number of teeth are:D=80,C=10,E=24 and F=18.If the pulley G runs at 200r.p .m,find the speed of shaft A.



Expert's answer

Speed ratio

NFND=NFNE∗NCND\frac{N_F}{N_D}=\frac{N_F}{N_E}*\frac{N_C}{N_D}


NFND=TETF∗TDTC\frac{N_F}{N_D}=\frac{T_E}{T_F}*\frac{T_D}{T_C}


The relative velocity of gears with respect to the arm

NF−NarmND−Narm=−TETF∗TDTC\frac{N_F -N_{arm}}{N_D-N_{arm}}=-\frac{T_E}{T_F}*\frac{T_D}{T_C}


200−NA0−NA=−2418∗8010\frac{200 -N_{A}}{0-N_{A}}=-\frac{24}{18}*\frac{80}{10}


200−NANA=2418∗8\frac{200 -N_{A}}{N_{A}}=\frac{24}{18}*8


200−NA=10.667NA200-N_A=10.667 N_A

NA=17.14rpmN_A=17.14 rpm same direction as gear F


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