Question

A body of mass `10kg` is acted upon by two per pendicular forces `6N` and `8N`. The resultant ac-celeration of the body is .
A. `1 ms^(-2)` at angle of `tan^(-1)((4)/(3)) 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 `8N` force .
D. `0.2 ms^(-2)` at an angle of `tan^(-1)((4)/(3))` w.r.t `8N` force .

Answer 1

Correct Answer - A
Consider the adjacent diagram
Given, mass ` = m= 10kg`
`F_(1) = 6N, F_(2) = 8N`
Resultant
`=F =sqrt(F_(1)^(2)+F_(2)^(2)) =sqrt(36+64)`
force `=10N`
`a = (F)/(m) = (10)/(10) =1m//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.tF_(1)=6N`
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.tF_(2) =8N` .