IIT JEE , AIEEE Forum - Physics , Mechanics

karthik2007

a)

As the train is moving up the incline with an acceleration of g/2, you will have to consider a pseudo force of mg/2 downwards parallel to the incline.

Let # be the angle made by the string with the normal to the ceiling, and @ be the inclination of the plane.

Balancing the components horizontally (relative to the train)

Tcos# = mgcos30

Tsin# = mg/2 + mgsin30

dividing, we get tan# = (mg/2+mg/2)/mgcos30

which gives tan# = 2/

3.

Now putting the value of # either of the two equations, we can get the value of T which works out to be 5

7 N.

2) As the train is travelling with a constant velocity, we do not need to consider a pseudo force. So, balancing the horizontal components (horizontal wrt the train)

Tsin# = mgsin30

Tcos# = mgcos30.

dividing, we get Tan# = tan30

or # = 30 degrees.

Putting this value in the first equation, we get

T = mgsin30/sin30

T = 1x10x1

thus T = 10N.

Hence the result.

Cheers :)

Ask & Discuss Questions with Community & Experts