Question

A heavy box is to be dragged along a rough horizontal floor . To do so ,person A pushes it an angle 30 0 from the horizontal and requires a minimum force FA ,while person B pulls the box at an angle 60 0 from the horizontal and needs minimum force FB. If the coefficient of friction between the box and the floor is 3 1/2 / 5 ,the ratio FA/ FB is

Answer 1

Given:
case 1:person pushes the box

Facos30=(mg+Fa sin30)μ
Fa=μmg/(cos30-μsin30)-----(1)

Case 2:Person pulls the box

Fbcos60=(mg-Fbsin60)μ
Fb=μmg/(Cos60+μsin60) -----(2)
Dividing equation 1 by 2, we get:
=(cos60+μsin60)/(cos30-μsin30)

Fa/Fb=2/√3

Answer 2

Answer:

Your answer is in the given attachment... hope it will help you.....

Given:

case 1:person pushes the box

Facos30=(mg+Fa sin30)μ

Fa=μmg/(cos30-μsin30)-----(1)

Case 2:Person pulls the box

Fbcos60=(mg-Fbsin60)μ

Fb=μmg/(Cos60+μsin60) -----(2)

Dividing equation 1 by 2, we get:

=(cos60+μsin60)/(cos30-μsin30)

Fa/Fb=2/√3

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