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PhysicsNLMFrictionMedium2 minPYQ_2014

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A heavy box is to be dragged along a rough horizontal floor. To do so, the person $A$ pushes it at an angle $30 °$ from the horizontal and requires a minimum force $F_{A}$ , while the person $B$ pulls the box at an angle $60 °$ from the horizontal and needs minimum force $F_{B}$ . If the coefficient of friction between the box and the floor is $\frac{\sqrt{3}}{5}$ , the ratio $\frac{F_{A}}{F_{B}}$ is

Options:

Answer:

Solution:

$F_{A} \cos 30 ° = μ (m g + F_{A} \sin 30 °)$

$F_{B} \cos 60 ° = μ (m g - F_{B} \sin 60 °)$

$F_{A} (\cos 30 ° - μ \sin 30 °) = μ m g$

$F_{B} (\cos 60 ° + μ \sin 60 °) = μ m g$

Divide the equation,

we get,

$\Rightarrow \frac{F_{A}}{F_{B}} = \frac{μ m g / (\cos 30 ° - μ \sin 30 °)}{μ m g / (\cos 60 ° + μ \sin 60 °)} = \frac{2}{\sqrt{3}}$

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Friction

⏱ 2mℹ️ Source: PYQ_2014

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