Question

Find the resultant of the following forces acting at a point:
1) 100√2 N along north-east,
2) 980√2 N along north-west,
3)1960 N along south​
Pls answer this asap!!! ​

Answer 1

Answer:

Resultant force due to North East force and North-west force will be along North .

This resultant force is given by

Explanation:

$Fn = \sqrt{( \sqrt{2} \times 980) {}^{2} + ( \sqrt{2} \times 100) } \\ = 1393 \: dyne$

Hence resultant due to the force FN and force along south is 1960-1393 = 596 dyne along south.