Question

A block of mass 2 kg is placed on the floor .The coefficient of static friction is 0.4.If a force of 2.8 N is applied
on the block parallel to the floor ,then find the force of friction between the block and the floor ?

Answer 1

$$\begin{gathered} \mathrm{Answer} \\ \mathrm{as}\mathrm{we}\mathrm{know} \\ \mathrm{limiting}\mathrm{force}\mathrm{of}\mathrm{friction}=\mathrm{\mu R} \\ =\mathrm{\mu }\times \mathrm{mg} \\ =0.4\times 2\times 10 \\ =8\mathrm{newton} \\ \mathrm{as}\mathrm{external}\mathrm{applied}\mathrm{force}\mathrm{of}2.8\mathrm{newton}\mathrm{is}\mathrm{less}\mathrm{than}\mathrm{the}\mathrm{limiting}\mathrm{friction},\mathrm{the} \\ \mathrm{actual}\mathrm{force}\mathrm{of}\mathrm{friction}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{external}\mathrm{force}\mathrm{of}2.8\mathrm{newton}\mathrm{but}\mathrm{opposite}\mathrm{in} \\ \mathrm{direction}. \end{gathered}$$