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A heavy box is to be dragged along a rough horizontal floor. To do so, person A pushes it at an angle 300 from the horizontal and requires a minimum force FA, while person B pulls the box at an angle 600 from the horizontal and needs minimum force FB.If the coefficient of friction between the box and the floor is  \frac{\sqrt{3}}{5}  the ratio   \frac{F_{A}}{F_{B}}    is       

  • Option 1)

    \sqrt{3}

  • Option 2)

    \frac{5}{\sqrt{3}}

  • Option 3)

    \sqrt{\frac{3}{2}}

  • Option 4)

    \frac{2}{\sqrt{3}}

 

Answers (1)

best_answer

As learnt (Concept missing)

Given 1: person pushes box

Fa cos 30= (mg+Fa sin 30)M

Fa=Mmg \left | (\cos 30-\sin 30M) -------------- (1)

 

Case 2. Person pulls box

Fb \: \cos 60=Mmg \left | (\cos 60+M \sin 60) ------------------- (2)

\frac{(1)}{(2)}

\frac{Fa}{Fb}=\frac{2}{\sqrt{3}}


Option 1)

\sqrt{3}

This option is incorrect

Option 2)

\frac{5}{\sqrt{3}}

This option is incorrect

Option 3)

\sqrt{\frac{3}{2}}

This option is incorrect

Option 4)

\frac{2}{\sqrt{3}}

This option is correct

Posted by

SudhirSol

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