The resultant force is 4.12N, 75.96° North of East.
Why?
To calculate the resultant force, we need to calculate the resultant forces for each direction (vertical and horizontal) and then, find the direction, considering the signs of the resultant forces.
From the statement, we know that the forces are non-parallel forces, but there is no information about the angles, so, to solve the problem I'll assume that they are parallel to the x-axis (West and East) and the y-axis (North and South). Please note that there would be totally different results if the directions were given.
So, calculating we have:
[tex]Fr_{x}=5N-4N=1N\\\\\\Fr_{y}=10N-6N=4N\\\\ResultantForce=\sqrt{(Fr_{x})^{2} +(Fr_{y})^{2}}\\\\ResultantForce=\sqrt{(1N)^{2} +(4N)^{2}}=\sqrt{1N^{2}+16N^{2}}=4.123N[/tex]
The resultant force is equal to 4.12N
Now, calculating the direction, we have:
[tex]\alpha =Arctan(\frac{Fr_{y}}{Fr_{x}})=75.96\°[/tex]
[tex]\alpha =Arctan(\frac{4}{1})=75.96\°[/tex]
Hence, we have that the resultant force is 4.12N, 75.96° North of East.
Have a nice day!
