Saved Bookmarks
| 1. |
Referring to figure calculate the downward acceleration of mass m_(1). Assume the surfaces are frictionless and pullyes ar massless. |
|
Answer» Solution :Let a be the acceleration of mass M and `a_(1)"and" a_(2)`, the acceleration of `M_(1)"and " M_(2)` relative to fixed pulley , P. Then from shown in figure. `M_(1)G-T_(2)=M_(1)a_(1)`….(1) `M_(2)g-T_(2)=M_(2)a_(2)`////(2) and `T_(1) =Ma`....(3) Also `T_(1)=2T_(2)`....(4) Acceleration of `M_(1)` relative to movable pulley Q is `(a_(1)-a)`. Acceleration of `M_(2)` relative to pulley Q`=(a_(2)-a)` The acceleration of `M_(1) "and" M_(2)` relative to pulley Q are equal and opposite . `:. a_(1)-a=-(a_(2)-a)` or ` a=(a_(1)+a_(2))/2`....(5) substracting (2) from (1) `(M_(1)-M_(2)) g = M_(1)a_(1)-M_(2)a_(2)`....(6)Adding (1) and (2) , we get `(M_(1)+M_(2))g-2T_(2)=M_(1)a_(1)+M_(2)a_(2)` From (3) and (4) , 2T_(2)=T_(1)=Ma Using (5), `2T_(2)=(M(a_(1)+a_(2))/2`....(7) SUBSTITUTING this value in (7), we get `(M_(1)+M_(2))g-(M(a_(1)+a_(2)))/2=M_(1)a_(1)+M_(2)a_(2)` or` 2(m_(1)+M_(2))g-Ma_(1)-Ma_(2) =2M_(1)a_(1)+2M_(1)a_(1)+2M_(2)a_(2)` or `2(M_(1)+M_(2))g=(2M_(1)+M)a+(2M_(2)+M)a_(2)`....(8)Eliminating `a_(2)` from (6) and (8) , we get `a_(1)=[(4M_(1)M_(2)+M(M_(1)-M_(2))/(4M_(1)M_(2)+M(M_(1)+M_(2))]g` 
|
|