Question #e4d20
1 Answer
I would begin by drawing a force diagram.
We are given that:

#>vecF_1=30"N"# 
#>vecF_2=50"N"#
The net force is given by:
#color(blue)(abs(vecF_"net")=sqrt((F_x)^2+(F_y)^2))#
We will first need to find the parallel (x, horizontal) and perpendicular (y, vertical) components of the given forces.

For the northward force, there is no parallel component, as the force vector is entirely vertical, i.e. aligned with the y(north)axis.

For the northwest force, the force vector lies at an angle, and so this force has both parallel and perpendicular components. We can calculate these using basic trigonometry.

Because the force is said to act northwest and we are not given a specific angle, we assume it acts at exactly the midpoint between North and West, i.e. at
#45^o# above the negative xaxis.
#sin(theta)="opposite"/"hypotenuse"#
#sin(theta)=F_(2y)/F_2#
Solving for
#=>color(blue)(F_(2y)=F_2sin(theta))#
Similarly, using the cosine, we find that
Therefore, we have:
#F_"y net"=sumF_y=F_1+F_(2y)#
#F_"x net"=sumF_x=F_(2x)#
So, we know that:

#F_y=30+50sin(45^o)# 
#F_x=50cos(45^o)#
And so, the net force is:
#F_"net"=sqrt((50cos(45^o))^2+(30+50sin(45^o))^2)#
#=>color(blue)(F_"net"~~74"N")#
Which is closest to answer choice c. The slight difference in values is due to rounding (I used exact values).
Hope that helps!