IIT JEE , AIEEE Forum - Physics , Mechanics

sulekha_hi

A man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleys. show that the force(assumed equal) exerted by each friend as the man moves up. Find the force when the man is at depth d.

catch_arnnie

i think ur question here is incomplete....plz check...

sulekha_hi

sorry catch arnie i had actually posted stoichiometry question in mechanics so i had to change. And yes i have asked about purity not impurity.

joyfrancis

Hi.

It is a kind of modified version of an HC verma problem 12 pg79.

We have to show that as the man in the ditch is pulled up the force exerted on the other two men increases...which is true.

See,

take the angle made by the rope connecting the man in the ditch and anyone of the other two and the horizontal be @.

Now as the man is pulled up the rope starts becoming horizontal(when it is perfectly horizontal the man would be out)....i.e the angle@ is DECREASING.

Now if we take tension along the rope connecting the man in the ditch and anyone of the other two to be T then the force on the other man is Tcos@.

Obviously if @ is decreasing cos@ would increase

.:the force exterted on the two men would increase as the man is pulled up.

As for the second part, i am getting the following answer.

Force exerted when he is at a depth d = 5mg/4d

elessar_iitkgp

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

iitkgp_bipin

First of all clearly draw a diagram for the situation.

Two ropes are at tension T and they are at an angle of

with the vertical.

Now balancing vertical forces :
2Tcos

= mg

Now simply apply trigonometry in right angled triangle formed by rope,half-width of ditch and vertical to find

:
cos

= d / {d² + (d/2)²}^1/2 = 2/

5

Hence T = mg/2cos

= (

5)mg/4

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