We’ve been given a rope with a breaking strength equal to two-thirds of the foreman sliding down on it; hence, the rope has a tension equal to two-thirds of the foreman’s weight in an upwards direction.
Let m represent a foreman’s mass.
Thus, we have W = mg (downward direction), T = 2/3mg (upwards direction)
Where W is the weight of a foreman and T is a tension
Now applying Newton’s Second Law of Motion on a foreman sliding downwards
F = ma
F is here is T – W (net force)
Substituting this in the above formula, we get 2/3mg−mg = ma
Cancelling out m and solving LHS we get a = −g/3
Therefore, the minimum acceleration is g3 in the downwards direction.
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