Saved Bookmarks
| 1. |
In the fig, find the acceleration of mass m_(2) |
|
Answer» SOLUTION :`l_(1)+2l_(2)=` constant on differentiating `v_(1)+2v_(2)=0` Again differentiating `a_(1)+2a_(2)=0 RARR a_(1)=-2a_(2)`  .-ve. sign indicates that the accelerations are in opposite direction. Suppose acceleration of `m_(2)` is `a_(2)` downward and then acceleration of `m_(1)` will be `a_(1)` UPWARDS. `T-m_(1)g=m_(1)a_(1)` `T=m_(1)g+m_(1)a_(1)` `m_(2)g-2T=m_(2)a_(2)`  `m_(2)g-2(m_(1)g+m_(1)a_(1))=m_(2)a_(2)` `m_(2)g-2m_(1)g=m_(2)a_(2)+4m_(1)a_(2)(THEREFORE a_(1)=2a_(2))` .-. sign should not be substituted `a_(2)=((m_(2)-2m_(1))g)/(4m_(1)+m_(2))MS^(-2)` |
|