Saved Bookmarks
| 1. |
Two blocks 'A' and 'B' having masses m_(A) and m_(a), respectively are connectd by an arrangement shown in the figure. Calculate the downward acceleration of the block B. Assume the pulleys tobe massless. Under what condition will block A have downward acceleration? |
|
Answer» SOLUTION :According to Newton `2^(nd)` Law `m_(B)g-T=m_(B)a_(B)""("From FBD of " m_(B))"" ...1` `2T-m_(A)g=m_(A)a_(A)""("From FBD of "m_(A))""...2` Constraint relation `a_(B) = 2a_(A)=a""...3` Solving EQUATION 1, 2 & 3 `a_(B)=a=(2g(2m_(B)-m_(A)))/((4m_(B)+m_(A))), a_(A)= (a)/(2)=(g(2m_(B)-m_(A)))/((4m_(B)+m_(A)))`. Block A will have downward motion if `a_(A) lt 0 rArr (2m_(B)-m_(A)) lt 0` `rArr m_(A) gt 2m_(B)` |
|