N=840 RPM
P=5.25KW
(a) Forces for bevel gears
=2N/60= (2*840)/60= 87.96 rad/s
T=P*= 5.25*87.96= 461.79Nm
F(smaller gear)=T/r1= 461.79/52.5= 8.796 N
F(larger gear)= T/r2=461.79/157.5= 2.932 N
(b) Work done on each bearing of the larger gear
W=F(larger)* s = 2.932* [(126+84)/2-51]= 158.328 J
Work done on smaller gear's bearing
W=F(smaller)* s= 8.796* (105/2)= 461.79 J
(c) After checking the SKF catalogue, roller and ball bearings are most suitable ones.
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