hii cud u plz solve this question............

 

 

a plumb bob of mass 1 kg is hung from the ceiling of a train compartment. the train moves on a inclined plane with an acceleration of a= g/2. if the angle of incline is 30⁰ .find the angle made by the string with the normal to the ceiling and also find the tension on the string..

ANS= tan^-1(2/3) , 57 N.

 

i tied to solve by resolving the two forces acting on the point which suspends teh bob but im not getting the answer.. cud sum1 plz help..

here train is moving in upward direction on an inclined plane having angle of inclination = 30 deg.

acceleration of the train = a = g/2

here component of acceleration in upward direction i.e. normal to the horizontal is (g/2)sin30 and along the horizontal it is (g/2)cos30

let string makes an angle  with the normal to the celing of train.

Tension in the string = T

Resolving the tension along two components as follows:

(a) Vertical to the ground

T cos( - 30) = m[g +(g sin30)/2]     (1)

(b) Along the ground that is in the horizontal direction

T sin( - 30) = mg (cos30)/2     (2)

Dividing eq. (2) by (1) we obtain,

tan( - 30) = [mg (cos30)]/m[g +(g sin30)/2] = 3 /5

or (tan - tan30)/(1+tantan30) = 3 /5

or tan = 2/3

or  = tan^-12/3 

Now to obtain tension in the string square and add then equate LHS and RHS of eq. (1) and (2)

by doing so we obtain and simplifing we obtain

T = 57