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A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60° with it. The other end rests on a rough horizontal floor.

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प्रश्न

A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60° with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is ______. (take g = 10 m/s2)

विकल्प

  • 100 N

  • `100sqrt3 N`

  • 200 N

  • `200sqrt3 N`

MCQ
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उत्तर

A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60° with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is `bbunderline(100sqrt3 N)`.

Explanation:

Since the rod is in equilibrium, 

ΣFY = 0 

⇒ N1 = mg = 200 N

ΣFX = 0 

⇒ N2 = f

ΣTA = 0 

`Mg(1/2 cos 30°) = N_2(L cos 60°)`

`(Mg)/2 sqrt3/2 = N_2/2`

`N_2 = sqrt3/2 xx 200`

= `100sqrt3` = f

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