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

Options

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

Answer

A, C

Explanation

Consider the adjacent diagram
Given , `" "` mass = m = 10 kg.
`F_(1) = 6N , F_(2) = 8N`
Resultant force `= F = sqrt(F_(1)^(2) + F_(2)^(2)) = sqrt(36 + 64)`
= 10 N
`a = (F)/(m) = (10)/(10) = 1 m//s^(2)` , along R .

Let `theta_(1)` be angle between R and `F_(1)`
tan `theta_(1) = (8)/(6) = (4)/(3)`
`theta_(1) = tan^(-1) (4 //3)` w.r.t. `F_(1) = 6 N`
Let `theta_(2)` be angle between F and `F_(2)`
tan `theta_(2) = (6)/(8) = (3)/(4)`
`theta_(2) = tan^(-1) ((3)/(4))` w.r.t. `F_(2) = 8N`