be specific abt the situation

Q : two bodies of masses 0.03 kag, and 0.04 kg are tied to the ends fo a massless string. This string passes over the frictionless pulley. the tension in the string is???

sorry for not writing the question properly..

i am new to all these questions.. might be this is the simplest to you.. but i am new to this...

And also remeber that tension always acts away from the body.

Practise free body diagrams and remember this, it'll be very easy.

Hii Chandana ...well ....this is natural ..I also used to face such kind of difficulty ..see...suppose u have mass m1 and mass m2 ..and they are tied to the string which passes through a frictionless pulley then ...

the tension in the string is given by  T = (2g x  m1 x m2 ) / ( m1 + m2 )

the acceleration is given by        a = (m2 - m1 ) g / ( m2 + m1 )     ( assuming m2 > m1 )

( if the answer comes negetive ..its all right ...the thing is ..just the supposition of direction is opposite ...your teacher  may have a different supposition ....{so take the mod value } )

thrust acting on the pulley ..i.e. reaction at the pulley is given by  R  = (4 g x m1 x m2 ) / ( m1 + m2 )

when you will put a weight on a string the string will become taut

and there willl be an internal tension force setup in the wire

at the point of contact of the string with mass1 since the rope is massless no matter what the accelaration of the point be the net force on it

F  =  M * A   = 0   because M = 0

so net resultant force at the point of contact is zero

therefore the tension force is always opposite to the applied force

as far as the magnitude of tension force is concerned it will be equal to the external applied force

now you must be wondering that there are 2 different external forces acting on the rope so what will be the tension force

see the answer to this is that on the either side of the pulley tension forces of difeernet magnitude will be acting and this will be responsible for the rotation of the pulley