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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

$$\min \mathrm{force}550?$$

Commented by Tinkutara last updated on 11/Feb/18

No, it is ✖

Commented by 803jaideep@gmail.com last updated on 11/Feb/18

$$435\mathrm{thn}?$$

Commented by Tinkutara last updated on 11/Feb/18

❌

Answered by mrW2 last updated on 11/Feb/18

$$Tokeepslidingthebox:$$$$F\cos \theta ={\mu }_k(mg-F\sin \theta )$$$$F=\frac{{\mu }_{k}mg}{\cos \theta +{\mu }_k\sin \theta }$$$$F=\frac{{\mu }_{k}mg}{\sqrt{1+{\mu }_k^2}(\frac{1}{\sqrt{1+{\mu }_k^2}}\cos \theta +\frac{{\mu }_k}{\sqrt{1+{\mu }_k^2}}\sin \theta )}$$$$F=\frac{{\mu }_{k}mg}{\sqrt{1+{\mu }_k^2}(\cos \alpha \cos \theta +\sin \alpha \sin \theta )}$$$$with\alpha ={\tan }^{-1}{\mu }_k$$$$F=\frac{{\mu }_{k}mg}{\sqrt{1+{\mu }_k^2}\cos (\theta -\alpha )}$$$$F_{k,min}=\frac{{\mu }_k}{\sqrt{1+{\mu }_k^2}}mg=\frac{0.2}{\sqrt{1+0.2^2}}\times 200\times 10=392N$$$$at\theta ={\alpha }_k={\tan }^{-1}0.2=11.3°$$$$$$$$Tobringtheboxtomove:$$$$F_{s,min}=\frac{{\mu }_s}{\sqrt{1+{\mu }_s^2}}mg=\frac{0.25}{\sqrt{1+0.{25}^2}}\times 200\times 10=485N$$$$$$$$W=F_{s,min}\cos {\alpha }_s\times 0+F_{k,min}\cos {\alpha }_k\times l=392\times \cos 11.3°\times 20\approx 7688J$$

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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