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A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left end. The mass of the log is 200 kg. An engineer with a mass of 53.5 kg walks along the log from the left to the right until the log is balanced horizontally. How far from the left end of the log is the engineer when the log is horizontal?

Respuesta :

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

 ∑ M = 0

mass of the log will be concentrated at the center  

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

[tex]m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)[/tex]

[tex]200 \times 0.1 L = 53.5 \times (x - 0.6 L)[/tex]

[tex]0.374 L =x - 0.6 L[/tex]

       x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

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