A body of mass 10 kg is acted upon by two perpendicular forces, 6N and 8N. The resultant acceleration of the body is

A body of mass 10 kg is acted upon by two perpendicular forces, 6N and

1.	A body of mass 10 kg is acted upon by two perpendicular forces, 6N and 8N. The resultant acceleration of the body is
Answer» 1 `MS^(-2)` at an angleof `tan^(-1)(4/3)` w.r.t 6n force 0.2 `ms^(-2)` at an angleof `tan^(-1)(4/3)` w.r.t 6n force 1 `ms^(-2)` at an angleof `tan^(-1)(4/3)` w.r.t 6n force 0.2 `ms^(-2)` at an angleof `tan^(-1)(4/3)` w.r.t 8n force Solution : Mass= m= 10 kg `F_(1)=6N` `f_(2) = 8N` Resultantforce`=f= sqrt(F_(1)^(2) + F_(2)^(2)) = sqrt(36+ 64)` Let`THETA _(1)` be angleand `vec(QR )` tan `theta _(1) = (F_(2))/(F_(1))= (8)/( 6) = (4)/(3)` `theta _(1) = tan^(-1)(4)/(3) w.r.t F_(1) = 6N` `tan theta_(2)= (F_(1))/( F_(2))= (6)/(8)= (3)/(4)` `theta _(2) = tan^(-1) ((3)/(4)) w.r.tF_(2) = 8N`