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Question Number 208741 by Tawa11 last updated on 22/Jun/24
Answered by mr W last updated on 22/Jun/24
Commented by mr W last updated on 22/Jun/24
tan φ=μ=0.3 ((SC)/(BC))=((sin (θ+φ))/(sin ((π/2)−φ)))=((sin (θ+φ))/(cos φ)) ((SC)/(AC))=((sin ((π/2)−θ−φ))/(sin φ))=((cos (θ+φ))/(sin φ)) ((sin (θ+φ))/(cos φ))=((cos (θ+φ))/(sin φ)) tan (θ+φ)=(1/(tan φ)) ((tan θ+tan φ)/(1−tan θ tan φ))=(1/(tan φ)) ((tan θ+μ)/(1−μ tan θ))=(1/μ) ⇒tan θ=((1−μ^2 )/(2μ))=((1−0.3^2 )/(2×0.3))=((91)/(60)) ⇒θ≈56.6° (ans. iii) R_1 =mg cos φ N_1 =R_1 cos φ=mg cos^2 φ =30×10×(1^2 /(1+0.3^2 ))=275.2 N (ans. i) R_2 =mg sin φ F_2 =R_2 sin φ=mg sin^2 φ =30×10×((0.3^2 )/(1+0.3^2 ))=24.8 N (ans. ii)
$$\mathrm{tan}\:\phi=\mu=\mathrm{0}.\mathrm{3} \\ $$$$\frac{{SC}}{{BC}}=\frac{\mathrm{sin}\:\left(\theta+\phi\right)}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\phi\right)}=\frac{\mathrm{sin}\:\left(\theta+\phi\right)}{\mathrm{cos}\:\phi} \\ $$$$\frac{{SC}}{{AC}}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta−\phi\right)}{\mathrm{sin}\:\phi}=\frac{\mathrm{cos}\:\left(\theta+\phi\right)}{\mathrm{sin}\:\phi} \\ $$$$\frac{\mathrm{sin}\:\left(\theta+\phi\right)}{\mathrm{cos}\:\phi}=\frac{\mathrm{cos}\:\left(\theta+\phi\right)}{\mathrm{sin}\:\phi} \\ $$$$\mathrm{tan}\:\left(\theta+\phi\right)=\frac{\mathrm{1}}{\mathrm{tan}\:\phi} \\ $$$$\frac{\mathrm{tan}\:\theta+\mathrm{tan}\:\phi}{\mathrm{1}−\mathrm{tan}\:\theta\:\mathrm{tan}\:\phi}=\frac{\mathrm{1}}{\mathrm{tan}\:\phi} \\ $$$$\frac{\mathrm{tan}\:\theta+\mu}{\mathrm{1}−\mu\:\mathrm{tan}\:\theta}=\frac{\mathrm{1}}{\mu} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\mathrm{1}−\mu^{\mathrm{2}} }{\mathrm{2}\mu}=\frac{\mathrm{1}−\mathrm{0}.\mathrm{3}^{\mathrm{2}} }{\mathrm{2}×\mathrm{0}.\mathrm{3}}=\frac{\mathrm{91}}{\mathrm{60}}\: \\ $$$$\Rightarrow\theta\approx\mathrm{56}.\mathrm{6}°\:\:\:\:\left({ans}.\:{iii}\right) \\ $$$${R}_{\mathrm{1}} ={mg}\:\mathrm{cos}\:\phi \\ $$$${N}_{\mathrm{1}} ={R}_{\mathrm{1}} \:\mathrm{cos}\:\phi={mg}\:\mathrm{cos}^{\mathrm{2}} \:\phi \\ $$$$\:\:\:\:\:\:=\mathrm{30}×\mathrm{10}×\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{1}+\mathrm{0}.\mathrm{3}^{\mathrm{2}} }=\mathrm{275}.\mathrm{2}\:{N}\:\:\:\left({ans}.\:\:{i}\right) \\ $$$${R}_{\mathrm{2}} ={mg}\:\mathrm{sin}\:\phi \\ $$$${F}_{\mathrm{2}} ={R}_{\mathrm{2}} \mathrm{sin}\:\phi={mg}\:\mathrm{sin}^{\mathrm{2}} \:\phi \\ $$$$\:\:\:\:\:\:=\mathrm{30}×\mathrm{10}×\frac{\mathrm{0}.\mathrm{3}^{\mathrm{2}} }{\mathrm{1}+\mathrm{0}.\mathrm{3}^{\mathrm{2}} }=\mathrm{24}.\mathrm{8}\:{N}\:\left({ans}.\:{ii}\right) \\ $$
Commented by Tawa11 last updated on 22/Jun/24
Wow, have tried several time. God bless you sir.
$$\mathrm{Wow},\:\mathrm{have}\:\mathrm{tried}\:\mathrm{several}\:\mathrm{time}. \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 22/Jun/24
do you have same results?
$${do}\:{you}\:{have}\:{same}\:{results}? \\ $$
Commented by Tawa11 last updated on 22/Jun/24
Yes sir.
$$\mathrm{Yes}\:\mathrm{sir}. \\ $$

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