Mechanics - Tension

1

Philip Calvert ·2009-01-07 08:07:53

can you give the angles

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Subash ·2009-01-07 08:09:41

angle made by the rope on either side with the vertical is theta

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Philip Calvert ·2009-01-07 08:13:39

t=mg/2cosÎ¸

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Subash ·2009-01-07 08:19:39

this is the correct figure

and the ans is mg/2tantheta

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prateek punj ·2009-01-07 08:22:03

i'm getting the ans mg/2tan(theta)

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Philip Calvert ·2009-01-07 08:23:02

you told the angle with the vertical [11]

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Philip Calvert ·2009-01-07 08:24:47

but how tanÎ¸ it should be sinÎ¸

2TsinÎ¸ = mg ??[7]

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prateek punj ·2009-01-07 08:28:04

no the ans is in tanÎ¸

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Subash ·2009-01-07 08:32:55

@philip i gave u the wrong angle sry

@prateek ur ans is correct can u post the method

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prateek punj ·2009-01-07 08:45:15

@ subhash
i don't think the Î¸ given by u earlier was wrong....

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Subash ·2009-01-07 08:47:24

the correct diagram is the second one

and the ans is mg/2tantheta

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prateek punj ·2009-01-07 09:04:25

if we consider the right portion of the wire....
the vertical component of tension at the point where the string is hanged by the wire is TsinÎ¸ and the horizontal component is TcosÎ¸(towards right).....
where T=tension on the wire due to wall.....

NOW,
Let, the tension at the middle point of wire will be T'....
and the direction of tension will be horizontal towards left....
since it is equilibrium....

TsinÎ¸=mg/2

TcosÎ¸=T'
Solving the two eqⁿ we can calculate T'.....

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prateek punj ·2009-01-17 08:24:02

check this out.....
is this not to be pinked....
guys many of my posts are not been pinked though they r useful....

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