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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
रिक्त स्थान भरें
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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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