IIT JEE , AIEEE Forum - Physics , Mechanics

rhawks

There is a fixed inclined plane of angle $\theta$ =53 degrees.On the inclined plane a wedge(of mass M=1.2 kg) is kept at an angle of $\alpha$ =37 degrees with respect to the horizontal.On the wedge a small block of mass m=0.6 kg is kept.When the system is released , FIND THE ACCELARATION OF THE '0.6 KG' BLOCK IN THE HORIZONTAL DIRECTION

(hi the ANSWER is (144g/443),but i dont no how to solve it. plzzzz solve it)

celestine

I didnt understand how the inclination of wedge $\alpha$ is oriented. I assumed the usual case of a wedge and block and got the ans as .329 g . Could you upload the diagram of this question if available? I would be glad to solve it.

rhawks

tell me how to upload a daigram into my question.

celestine

you can scan the diagram and save it in a file and upload it or draw it in paint by yourself and send it

rhawks

I have drawn it in paint,may i know how to upload it,

rhawks

please give the equations celestine

celestine

could you mail it to me

rhawks

what is your email id

celestine

mastercelestinepreetham@yahoo.com

celestine

Considering FBD

The equations involved are

N- reaction btw wedge and blk

N=.6gcos37 = .48g

For acc of wedge, Aw along incline
Nsin(53-37) + 1.2gsin53 = 1.2Aw

Aw = .91 g

Now acc of blk wrt wedge , Ab = gsin37 ? Awcos16 = -0.275 g

Net Horizontal acc, A = Absin53 + Awsin37 = 0.326 _~ .3251_~ 144/443 g

Note: The actual ans is approximated bcos of the various initial approx that I made in the sin and cos values . But the method of approach is same and genuine.

celestine

, Ab = gsin37 ? Awcos16 = -0.275 g
is actually Ab = gsin37 - Awcos16 = -0.275 g

celestine

IT was a nice question
from where did u get it

rhawks

I Actually got it from an old olympaid book

shinee

can u also mail me, i have given it in ur nudge book

Ask & Discuss Questions with Community & Experts