Q.
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 )
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answer is 6.40.
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Detailed Solution
Since, disc is in rotational equilibrium,friction due to rod and ground will be same.τA=0At point of contact of ground⇒NL=MgL2cosθ⇒N=Mg2cosθ For disc, Nsinθ=f+fcosθ∴f=Nsinθ1+cosθ=Mgsinθcosθ2(1+cosθ)M=λL and L=Rtanθ2⇒f=12λRgcosθ =12×8×10×210×45=325=6.40N
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