Question

two cylinders of diameters 100mm and 50mm weighing 200N and 50N​

Answer 1

Answer:

Applying static conditions of equilibrium,

∴ΣFy(↑+ve)=0−200−200+RBcos10=0∴RBcos10=400∴RB=400cos10=406.17N(↑)cos10=50/x,∴x=50cos10=50.77mmtan10=y/50,∴y=50tan10=8.816mm

∴ Distance of RA from O=x+y=50.77+8.816=59.586mm

Now, z=1002−502−−−−−−−−−√………{By Pythagoras Theorem}

∴z=86.60mm

∴ Distance of RC from O=59.586+86.60=146.18mm Perpendicular distance of RB from O→

Consider a triangle ΔOAC: OC=OA2+AC2−−−−−−−−−−√=59.5862+502−−−−−−−−−−−√=77.78mm

∴ In right angled ΔOBC,OB=77.782−502−−−−−−−−−−√=59.586mm∴ΣFx(→+ve)=0∴RA−RC−RBsin10=0∴RA−RC=406.17sin10=70.53…………………(I)∴ΣMO≅0(+ve)∴−RA×59.586+RB×59.586−200×50−200×100+RC×146.18=0∴−59.586RA+146.18RC=200×50+200×100=406.17×59.586

Solving equations (I) and (II) simultaneously, we get

RA=186.02N,RC=115.488N