Question

A Force of 6N acts on a body due east and the other force 8N acts due north. what is resultant of the forces

Answer 1

Force in north direction=8N

force in east direction=6N

Since the 2 forces are perpendicular

Fnet=√(8²+6²)

=√100

=10N

Hope this helps

Answer 2

As given that the

Two force are acting perpendicular to each other ....means angle between them is 90°.

And also

$R= \sqrt{ {A}^{2} + {B}^{2} + 2AB \cos \: x}$

where

R is the resultant of two vectors

A is the magnitude of one vector

B Is the magnitude of another vector

And

x is the angle between two given vectors

So......Here,

A=6N

B=8N

x=90°

So...,

$= > R = \sqrt{ {6}^{2} + {8}^{2} + 2(6)(8) \cos(90) } \\ = > R = \sqrt{36 + 64} \\ = > R= \sqrt{100} = 10$

So the resultant is 10N