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

Options

A. `(sqrt(2)-1)a`
B. `(sqrt(2)+1)a`
C. `sqrt(2)a`
D. `a`

Answer

A

Explanation

Accelation ` a=(F )/(m )`

The resultant of `F_1` and `F_2 ` has magnitude `F.` given by
` F. = (F_(1)^(2) +F_(2)^(2) ) ^(1//2) = sqrt(2) F " " ( :. F_1 - F_2 =F)`
the direction of ` F.` is opposite to that of ` F_3`
` therefore ` Net force on body =` F. =F_3 = sqrt(3) = sqrt(2 ) F -F =( sqrt(2)-1) F`
` therefore ` Accelation `=(sqrt(2)-1) a` ( using (i))