Let at any moment the man be at a depth h below the horizontal. As both his friends exert equal forces, the strings make equal angles with the vertical. let at this instant the string makes an angle with the vertical.
The forces on the man are the two tensions T making an angle with the vertical and his weight mg acting down. Its stated in the question that the friends are slowly pulling him out. So his acceleration in the vertical direction may be assumed to be zero. Then, 2T cos = mg T = mg / 2cos
As the man moves up increases. Hence cos decreases and so the value of T increases. So force exerted by the friends increases as the man comes up.
At a depth d, T = mg / 2cos Now, cos = d/(d/2)2 + d2 = 2/ 5 T = 5mg/4d
Moderation Team
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with the vertical.
with the vertical and his weight mg acting down. Its stated in the question that the friends are slowly pulling him out. So his acceleration in the vertical direction may be assumed to be zero.
(d/2)2 + d2 = 2/
5
