Question

A body of mass `10kg` is acted upon by two perpendcular force, `6N` . The resultant acceleration of the body is.
A. `1ms^(-2)` at an angle of` tan^(-1) (4//3) w. r. t. 6 N`
B. `0.2ms^(-2)` at angle of `tan^(-1) (4//3) w.r.t. 6N`
C. `1ms^(-2)` at an angle of `tan^(-1) (3//4) w.r.t 8N`
D. `0.2ms^(-2)` at an angle of `tan^(-1) (3//4) w.r.t. 8N` force

Answer 1

Correct Answer - a,c
Here, `m = 10kg`
`F_(1) = 6 N, F_(2) = 8 N`
Resultant force `R =sqrt(F_(1)^(2)+ F_(2)^(2)) = sqrt(6^(2) + 8^(2)) = 10N`
`a = (R)/(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.t. `F_(1) = 6 N` force
Let `theta_(2)` be angle between `R` and `F_(2) tan theta_(2) = (6)/(8) = (3)/(4)`
`theta_(2) = tan^(-1) ((3)/(4))` w.r.t `F_(2) = 8 N` Force
.