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Question Number 159675 by Tawa11 last updated on 19/Nov/21

Commented by mr W last updated on 20/Nov/21

$$irememberthisquestionisnotasked$$$$forthefirsttime.$$

Commented by Tawa11 last updated on 20/Nov/21

$$\mathrm{Wow},\mathrm{I}\mathrm{really}\mathrm{appreciate}\mathrm{sir}.$$$$\mathrm{I}\mathrm{understand}\mathrm{very}\mathrm{well}.$$

Answered by mr W last updated on 20/Nov/21

Commented by mr W last updated on 20/Nov/21

$$\left(a\right)$$$$N=mg\cos \theta$$$$F=mg\sin \theta -\mu N=mg(\sin \theta -\mu \cos \theta )$$

Commented by mr W last updated on 20/Nov/21

Commented by mr W last updated on 20/Nov/21

$$\left(b\right)$$$$N=mg\cos \theta$$$$F=mg\sin \theta +\mu N=mg(\sin \theta +\mu \cos \theta )$$

Commented by mr W last updated on 20/Nov/21

Commented by mr W last updated on 20/Nov/21

$$\left(c\right)$$$$N=mg\cos \theta +F\sin \theta$$$$F\cos \theta =mg\sin \theta +\mu N=mg\sin \theta +\mu (mg\cos \theta +F\sin \theta )$$$$F(\cos \theta -\mu \sin \theta )=mg(\sin \theta +\mu \cos \theta )$$$$F=\frac{mg(\sin \theta +\mu \cos \theta )}{\cos \theta -\mu \sin \theta }$$

Commented by mr W last updated on 20/Nov/21

Commented by mr W last updated on 20/Nov/21

$$\left(d\right)\left(notaskedinquestion\right)$$$$N=mg\cos \theta +F\sin \theta$$$$F\cos \theta =mg\sin \theta -\mu N=mg\sin \theta -\mu (mg\cos \theta +F\sin \theta )$$$$F(\cos \theta +\mu \sin \theta )=mg(\sin \theta -\mu \cos \theta )$$$$F=\frac{mg(\sin \theta -\mu \cos \theta )}{\cos \theta +\mu \sin \theta }$$

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