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Question Number 29727 by Tinkutara last updated on 11/Feb/18
Commented by 803jaideep@gmail.com last updated on 11/Feb/18
min force 550?
$$\mathrm{min}\:\mathrm{force}\:\mathrm{550}? \\ $$
Commented by Tinkutara last updated on 11/Feb/18
No, it is ✖
Commented by 803jaideep@gmail.com last updated on 11/Feb/18
435 thn?
$$\mathrm{435}\:\mathrm{thn}? \\ $$
Commented by Tinkutara last updated on 11/Feb/18
❌
Answered by mrW2 last updated on 11/Feb/18
To keep sliding the box: F cos θ=μ_k (mg−F sin θ) F=((μ_k mg)/(cos θ+μ_k sin θ)) F=((μ_k mg)/( (√(1+μ_k ^2 )) ((1/( (√(1+μ_k ^2 ))))cos θ+(μ_k /( (√(1+μ_k ^2 ))))sin θ))) F=((μ_k mg)/( (√(1+μ_k ^2 )) (cos α cos θ+sin α sin θ))) with α=tan^(−1) μ_k F=((μ_k mg)/( (√(1+μ_k ^2 )) cos (θ−α))) F_(k,min) =(μ_k /( (√(1+μ_k ^2 )))) mg=((0.2)/( (√(1+0.2^2 ))))×200×10=392 N at θ=α_k =tan^(−1) 0.2=11.3° To bring the box to move: F_(s,min) =(μ_s /( (√(1+μ_s ^2 )))) mg=((0.25)/( (√(1+0.25^2 ))))×200×10=485 N W=F_(s,min) cos α_s ×0+F_(k,min) cos α_k ×l=392×cos 11.3°×20≈7688 J
$${To}\:{keep}\:{sliding}\:{the}\:{box}: \\ $$$${F}\:\mathrm{cos}\:\theta=\mu_{{k}} \left({mg}−{F}\:\mathrm{sin}\:\theta\right) \\ $$$${F}=\frac{\mu_{{k}} {mg}}{\mathrm{cos}\:\theta+\mu_{{k}} \mathrm{sin}\:\theta} \\ $$$${F}=\frac{\mu_{{k}} {mg}}{\:\sqrt{\mathrm{1}+\mu_{{k}} ^{\mathrm{2}} }\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mu_{{k}} ^{\mathrm{2}} }}\mathrm{cos}\:\theta+\frac{\mu_{{k}} }{\:\sqrt{\mathrm{1}+\mu_{{k}} ^{\mathrm{2}} }}\mathrm{sin}\:\theta\right)} \\ $$$${F}=\frac{\mu_{{k}} {mg}}{\:\sqrt{\mathrm{1}+\mu_{{k}} ^{\mathrm{2}} }\:\left(\mathrm{cos}\:\alpha\:\mathrm{cos}\:\theta+\mathrm{sin}\:\alpha\:\mathrm{sin}\:\theta\right)} \\ $$$${with}\:\alpha=\mathrm{tan}^{−\mathrm{1}} \mu_{{k}} \\ $$$${F}=\frac{\mu_{{k}} {mg}}{\:\sqrt{\mathrm{1}+\mu_{{k}} ^{\mathrm{2}} }\:\mathrm{cos}\:\left(\theta−\alpha\right)} \\ $$$${F}_{{k},{min}} =\frac{\mu_{{k}} }{\:\sqrt{\mathrm{1}+\mu_{{k}} ^{\mathrm{2}} }}\:{mg}=\frac{\mathrm{0}.\mathrm{2}}{\:\sqrt{\mathrm{1}+\mathrm{0}.\mathrm{2}^{\mathrm{2}} }}×\mathrm{200}×\mathrm{10}=\mathrm{392}\:{N} \\ $$$${at}\:\theta=\alpha_{{k}} =\mathrm{tan}^{−\mathrm{1}} \mathrm{0}.\mathrm{2}=\mathrm{11}.\mathrm{3}° \\ $$$$ \\ $$$${To}\:{bring}\:{the}\:{box}\:{to}\:{move}: \\ $$$${F}_{{s},{min}} =\frac{\mu_{{s}} }{\:\sqrt{\mathrm{1}+\mu_{{s}} ^{\mathrm{2}} }}\:{mg}=\frac{\mathrm{0}.\mathrm{25}}{\:\sqrt{\mathrm{1}+\mathrm{0}.\mathrm{25}^{\mathrm{2}} }}×\mathrm{200}×\mathrm{10}=\mathrm{485}\:{N} \\ $$$$ \\ $$$${W}={F}_{{s},{min}} \:\mathrm{cos}\:\alpha_{{s}} \:×\mathrm{0}+{F}_{{k},{min}} \:\mathrm{cos}\:\alpha_{{k}} \:×{l}=\mathrm{392}×\mathrm{cos}\:\mathrm{11}.\mathrm{3}°×\mathrm{20}\approx\mathrm{7688}\:{J} \\ $$
Commented by Tinkutara last updated on 12/Feb/18
Answers given are 7690 J, 470 N
Commented by Tinkutara last updated on 12/Feb/18
Thank you very much Sir! I got the answer.

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