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Question-30994

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Question Number 30994 by ajfour last updated on 01/Mar/18
Commented by ajfour last updated on 01/Mar/18
Commented by ajfour last updated on 01/Mar/18
Please solve. Free body diagrams i have attached alongwith.
$${Please}\:{solve}.\:{Free}\:{body}\:{diagrams} \\ $$$${i}\:{have}\:{attached}\:{alongwith}. \\ $$
Commented by ajfour last updated on 02/Mar/18
mgsin α+mAcos α−μ_1 N=ma_(rel) N+mAsin α=mgcos α R=Ncos α+Mg+μ_1 Nsin α Nsin α−μ_1 Ncos α−μ_2 R=MA ⇒ N=mgcos α−mAsin α & Nsin α−μ_1 Ncos α −μ_2 (Ncos α+Mg+μ_1 Nsin α)=MA So N(sin α−μ_1 cos α−μ_2 cos α−μ_1 μ_2 sin α) −μ_2 Mg = MA (mgcos α−mAsin α)(sin α−μ_1 cos α−μ_2 cos α−μ_1 μ_2 sin α) = μ_2 Mg+MA A=((mgcos α(sin α−μ_1 cos α−μ_2 cos α−μ_1 μ_2 sin α)−μ_2 Mg)/(M+msin α(sin α−μ_1 cos α−μ_2 sin α−μ_1 μ_2 sin α))) =((20×(4/5)((3/5)−0.35×(4/5)−0.025×(3/5))−0.1×80)/(8+2×(3/5)((3/5)−0.35×(4/5)−0.025×(3/5)))) =((16(0.6−0.28−0.015)−8)/(8+1.2(0.6−0.28−0.015))) = ((16×0.305−8)/(8+1.2×0.305)) < 0 ⇒ A=0 .
$${mg}\mathrm{sin}\:\alpha+{mA}\mathrm{cos}\:\alpha−\mu_{\mathrm{1}} {N}={ma}_{{rel}} \\ $$$${N}+{mA}\mathrm{sin}\:\alpha={mg}\mathrm{cos}\:\alpha \\ $$$${R}={N}\mathrm{cos}\:\alpha+{Mg}+\mu_{\mathrm{1}} {N}\mathrm{sin}\:\alpha \\ $$$${N}\mathrm{sin}\:\alpha−\mu_{\mathrm{1}} {N}\mathrm{cos}\:\alpha−\mu_{\mathrm{2}} {R}={MA} \\ $$$$\Rightarrow\:{N}={mg}\mathrm{cos}\:\alpha−{mA}\mathrm{sin}\:\alpha \\ $$$$\&\:\:{N}\mathrm{sin}\:\alpha−\mu_{\mathrm{1}} {N}\mathrm{cos}\:\alpha \\ $$$$−\mu_{\mathrm{2}} \left({N}\mathrm{cos}\:\alpha+{Mg}+\mu_{\mathrm{1}} {N}\mathrm{sin}\:\alpha\right)={MA} \\ $$$${So} \\ $$$${N}\left(\mathrm{sin}\:\alpha−\mu_{\mathrm{1}} \mathrm{cos}\:\alpha−\mu_{\mathrm{2}} \mathrm{cos}\:\alpha−\mu_{\mathrm{1}} \mu_{\mathrm{2}} \mathrm{sin}\:\alpha\right) \\ $$$$\:\:\:\:\:−\mu_{\mathrm{2}} {Mg}\:=\:{MA} \\ $$$$\left({mg}\mathrm{cos}\:\alpha−{mA}\mathrm{sin}\:\alpha\right)\left(\mathrm{sin}\:\alpha−\mu_{\mathrm{1}} \mathrm{cos}\:\alpha−\mu_{\mathrm{2}} \mathrm{cos}\:\alpha−\mu_{\mathrm{1}} \mu_{\mathrm{2}} \mathrm{sin}\:\alpha\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\mu_{\mathrm{2}} {Mg}+{MA} \\ $$$${A}=\frac{{mg}\mathrm{cos}\:\alpha\left(\mathrm{sin}\:\alpha−\mu_{\mathrm{1}} \mathrm{cos}\:\alpha−\mu_{\mathrm{2}} \mathrm{cos}\:\alpha−\mu_{\mathrm{1}} \mu_{\mathrm{2}} \mathrm{sin}\:\alpha\right)−\mu_{\mathrm{2}} {Mg}}{{M}+{m}\mathrm{sin}\:\alpha\left(\mathrm{sin}\:\alpha−\mu_{\mathrm{1}} \mathrm{cos}\:\alpha−\mu_{\mathrm{2}} \mathrm{sin}\:\alpha−\mu_{\mathrm{1}} \mu_{\mathrm{2}} \mathrm{sin}\:\alpha\right)}\:\: \\ $$$$\:\:\:=\frac{\mathrm{20}×\frac{\mathrm{4}}{\mathrm{5}}\left(\frac{\mathrm{3}}{\mathrm{5}}−\mathrm{0}.\mathrm{35}×\frac{\mathrm{4}}{\mathrm{5}}−\mathrm{0}.\mathrm{025}×\frac{\mathrm{3}}{\mathrm{5}}\right)−\mathrm{0}.\mathrm{1}×\mathrm{80}}{\mathrm{8}+\mathrm{2}×\frac{\mathrm{3}}{\mathrm{5}}\left(\frac{\mathrm{3}}{\mathrm{5}}−\mathrm{0}.\mathrm{35}×\frac{\mathrm{4}}{\mathrm{5}}−\mathrm{0}.\mathrm{025}×\frac{\mathrm{3}}{\mathrm{5}}\right)} \\ $$$$\:\:=\frac{\mathrm{16}\left(\mathrm{0}.\mathrm{6}−\mathrm{0}.\mathrm{28}−\mathrm{0}.\mathrm{015}\right)−\mathrm{8}}{\mathrm{8}+\mathrm{1}.\mathrm{2}\left(\mathrm{0}.\mathrm{6}−\mathrm{0}.\mathrm{28}−\mathrm{0}.\mathrm{015}\right)} \\ $$$$\:\:=\:\frac{\mathrm{16}×\mathrm{0}.\mathrm{305}−\mathrm{8}}{\mathrm{8}+\mathrm{1}.\mathrm{2}×\mathrm{0}.\mathrm{305}}\:\:<\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:{A}=\mathrm{0}\:\:. \\ $$

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