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A ladder of mass 10 kg is resting with i...

A ladder of mass 10 kg is resting with its top end touching a vertical wall and the bottom end on the floor. If the angle between the ground and the ladder is 45°, find the horizontal force required at the bottom to prevent the ladder from sliding. Neglect friction and take `g = 10 ms^(-2)`.

Text Solution

Verified by Experts

Let `R_(1), R_(2)` be the reaction forces `F to ` horizontal force
`W to` weight
Since `theta = 45^(@), OA = OB =x`
To prevent sliding, net moment must be zero.

`R_(1) =F` and `R_(2) =W`, for translational equilibrium
Let us take moments about A.
`therefore` Moment of `R_(1)` is zero.
Total clockwise moment `=W xx (OB)/2 + F xx OA`
Total anticlockwise moment = `R_(2) xx OB = W_(x)`
For rotational equilibrium, total clockwise moment = anticlockwise moment or F = 50 N
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Knowledge Check

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