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A weightless ladder 6m long rests agains...

A weightless ladder `6m` long rests against a frictionless wall at an angle of `60^(@)` from the horizontal. A `60kg` man standing on it is `2m` from the top of the ladder. A horizontal force is applied at the lower end to keep it from slipping. The magnitude of the force is

A

`(10)/(sqrt(3))N`

B

`(20)/(sqrt(3))N`

C

`(sqrt(3))/(20)N`

D

`20sqrt(3)N`

Text Solution

Verified by Experts

The correct Answer is:
B

Taking moments about `A`
`F(OA)=W(CD)`
`F[6sin60^(@)]=60[2cos60^(@)]`
`impliesF=(20)/(sqrt(3))N`
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