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

Options

A. 1 `ms^(-2)` at an angle of `tan^(-1)(4/3)` w.r.t 6n force
B. 0.2 `ms^(-2)` at an angle of `tan^(-1)(4/3)` w.r.t 6n force
C. 1 `ms^(-2)` at an angle of `tan^(-1)(4/3)` w.r.t 6n force
D. 0.2 `ms^(-2)` at an angle of `tan^(-1)(4/3)` w.r.t 8n force

Answer

A, C

Explanation

Mass = m= 10 kg
`F_(1) =6N`
`f_(2) = 8N`
Resultant force `=f= sqrt(F_(1)^(2) + F_(2)^(2)) = sqrt(36+ 64)`
Let `theta _(1)` be angle and `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.t F_(2) = 8N`