Question

Question #31f43

Answer 1

`6.874 m/s^2` about 71.90 degrees w.r,t north

Explanation:

`15 N` force is acting along south and `50N` along northeast. So total force acting along north is `50cos 45` or,35.35 N,so net force acting along north is `(35.35-14)` or,`21.35N`

And force acting along east is `(50 cos 45+30)` or,`65.35N`.

so,let ,net force acting along `theta` degrees w.r,t North and its magnitude is `sqrt` `(21.35^2+65.35^2) N` or,`68.74 N`

`tan theta = 65.35/21.35` or,`theta= 71.90` degrees

so,magnitude of acceleration will be `68.74/10` or `6.874m/s^2` along the direction of net force.