Saved Bookmarks
| 1. |
In the figure acceleration of bodies A,B and C are shown with directions. Values b and c are w.r.t. ground. Whereas a is acceleration of block A w.r.t. wedge C. accelerationof block A w.r.t. ground is `sqrt(beta) m//s^(2)`. Find `beta` (use b=c=`1 m//s^(2), theta=60^(@)`) |
|
Answer» `a=b+c` Net acceleration of `A=sqrt(a^(2)+c^(2)+2ac cos(pi-theta))=sqrt((b+c)^(2)+c^(2)-2(b+c)c costheta)=sqrt(3)` |
|