Question-29727

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 ?

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 ?

Commented by Tinkutara last updated on 11/Feb/18

Answered by mrW2 last updated on 11/Feb/18

T o k e e p s l i d i n g t h e b o x :

F \cos θ = μ_{k} (m g - F \sin θ)

F = \frac{μ_{k} m g}{\cos θ + μ_{k} \sin θ}

F = \frac{μ_{k} m g}{\sqrt{1 + μ_{k}^{2}} (\frac{1}{\sqrt{1 + μ_{k}^{2}}} \cos θ + \frac{μ_{k}}{\sqrt{1 + μ_{k}^{2}}} \sin θ)}

F = \frac{μ_{k} m g}{\sqrt{1 + μ_{k}^{2}} (\cos α \cos θ + \sin α \sin θ)}

w i t h α = \tan^{- 1} μ_{k}

F = \frac{μ_{k} m g}{\sqrt{1 + μ_{k}^{2}} \cos (θ - α)}

F_{k, m i n} = \frac{μ_{k}}{\sqrt{1 + μ_{k}^{2}}} m g = \frac{0.2}{\sqrt{1 + 0 . 2^{2}}} \times 200 \times 10 = 392 N

a t θ = α_{k} = \tan^{- 1} 0.2 = 11.3 °

T o b r i n g t h e b o x t o m o v e :

F_{s, m i n} = \frac{μ_{s}}{\sqrt{1 + μ_{s}^{2}}} m g = \frac{0.25}{\sqrt{1 + 0 . 25^{2}}} \times 200 \times 10 = 485 N

W = F_{s, m i n} \cos α_{s} \times 0 + F_{k, m i n} \cos α_{k} \times l = 392 \times \cos 11.3 ° \times 20 \approx 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.