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A uniform metal bar of length 6m and mass 100kg rest with its upper end against a smooth vertical wall and with its lower end on a rough surface of coefficient of friction 0.32.what is the maximum angle made with the horizontal to which the bar can be enclined without sliding?

Sagot :

Answer:

Let F1x = force of wall on ladder

F1x = F2x    the force of friction  (balances horizontal forces)

F2y = W      balance weight of ladder

F2x = μ F2y       force of friction on ladder

F1x L sin θ = W L / 2 cos θ     balance torque about point F2

tan θ  = W / (2 F1x)     previous equation

Now     μ W = μ F2y = F2x = F1x   from above

So tan θ  = W / (2 μ W) = 1 / (2 μ ) = 1 / .64

tan θ = 1.56

θ = 57.4 deg

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