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When a force F acs on a body of mass m the acceleration product in the body is a . If htree equal forces `F_(1)=F_(2)=F_(3)=F` act on the same body as shown in figure the accleration produced is A. `(sqrt(2)-1)a`B. `(sqrt(2)+1)a`C. `sqrt(2)a`D. a |
Answer» Correct Answer - A (a) Accelertion `a=(F)/(m)` Resultant of `F_(1)andF_(2)`will be `F_(12)=sqrt(2F)`(in opposite directiomn o f`F_(3)`) Now ,resultant of`F_(12)and F_(3)`will be , `F_("net")=(sqrt(2)-1)F` `thereforea=(F_("net"))/(m)=(sqrt(2)-1)(F)/(m)=(sqrt(2)-1)a` |
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