E-Store
linear momentum.............
|
| Forum Index -> Mechanics -> View Full Question |
|
| Author | Message | |||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
hey i am using the normal reaction offered by the floor on the chain which is equal to the force exerted by the chain on the floor N=Nwt+Nthrust Nwt=(m/l)xg Nthrust=Vrelative(dm/dt) Vrelative=u-v = v dm/dt=(dm/dy)(dy/dt)=m/l(v) Nthrust=v2(m/l) v=(2gx)1/2 Nthrust=2gx(m/l) N=(m/l)gx+2gx(m/l)=3(m/l)gx
|
|||||||||||||
| Like 0 people liked this | ||||||||||||||
|
|
||||||||||||||
Free Sign Up!
Preparing for IIT-JEE ?
Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor
@ INR 1,995/-
For Quick Info
|
|
|
|
|
| 35,229,755 Engineering Questions Attempted |
6,69,530 Visitors Monthly |
79 Engg. Exam Covered |
190 Students Currently Online |
Connect with Us  |
 |
Vriti Communities
Privacy Policy © Vriti 2011