A stick of mass density λ$$=8$$ $$kg/m$$ rests on a disc of radius R $$=$$ $$20cm$$ as shown in the figure. The stick makes an angle θ$$=37o$$ with the horizontal and is tangent to the disc at its upper end. Friction exists at all points of contact and assume that it is large enough to keep the system at rest. Find the friction force (in$$ $$Newton) between the ground and the disc. (take$$ g $$=$$ $$10$$ $$m/s2$$ $$)

Question

A stick of mass density λ$$=8$$ $$kg/m$$ rests on a disc of radius R $$=$$ $$20cm$$ as shown in the figure. The stick makes an angle θ$$=37o$$ with the horizontal and is tangent to the disc at its upper end. Friction exists at all points of contact and assume that it is large enough to keep the system at rest. Find the friction force (in$$ $$Newton) between the ground and the disc. (take$$ g  $$=$$ $$10$$  $$m/s2$$ $$)

Infinity Learn · Accepted Answer

Since, disc is in rotational equilibrium,friction due to rod and ground will be same.τ$$A=0At$$ point of contact of ground⇒$$NL=MgL2cos$$⁡θ⇒$$N=Mg2cos$$⁡θ For disc, Nsin⁡θ$$=f+fcos$$⁡θ∴$$f=Nsin$$⁡θ$$1+$$$$cos$$⁡θ$$=Mgsin$$⁡θ$$cos$$⁡θ$$2(1+$$$$cos$$⁡θ$$)M=$$λL and $$L=Rtan$$⁡θ$$2$$⇒$$f=12$$λRgcos⁡θ $$=12$$×$$8$$×$$10$$×$$210$$×$$45=325=6.40N$$

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