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A 100 N uniform ladder of length 5.0 m rest against a smooth wall. The magnitude of the force of the wall, P, acting on the top of the ladder is 40 N. Find the angle 0, with the ground before the ladder begins to slip. (HINT: Use point O as axis of rotation and given that )
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Respuesta :

Answer:

Explanation:

Wall reactions are never constant as the ladder angle decreases from vertical.

I will ASSUME that the MAXIMUM wall reaction is 40 N before slippage occurs.

Let θ be the ladder angle to horizontal

Moments about the ladder foot will sum to zero

40[5sinθ) - 100[(5/2)cosθ = 0

40[5sinθ) = 100[(5/2)cosθ

 200sinθ = 250cosθ

sinθ/cosθ = 250/200

        tanθ = 1.25

             θ = 51.34019174...

            θ = 51°

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