Question Number 26113 by tawa tawa last updated on 19/Dec/17
Answered by ajfour last updated on 20/Dec/17
$$\mu_{{upper}} {mg}={kx} \\ $$$$\mu_{{upper}} {mg}+\mu_{{lower}} \left({M}+{m}\right){g}={F} \\ $$$${so}\:\:\:\boldsymbol{{x}}=\frac{\mathrm{0}.\mathrm{9}×\mathrm{15}×\mathrm{10}}{\mathrm{325}}\:=\frac{\mathrm{135}}{\mathrm{325}}\: \\ $$$$\:\:\:\:\:=\frac{\mathrm{27}}{\mathrm{65}}{metre}\:=\frac{\mathrm{5}.\mathrm{4}}{\mathrm{13}}\:=\mathrm{0}.\mathrm{415}\:{metre} \\ $$$${and}\:\:\boldsymbol{{F}}=\mathrm{0}.\mathrm{9}×\mathrm{15}×\mathrm{10}+\mathrm{0}.\mathrm{6}×\mathrm{45}×\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{135}+\mathrm{270}\:=\mathrm{405}{N}\:. \\ $$
Commented by tawa tawa last updated on 20/Dec/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$