Question

Question #e4d20

Answer 1

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 x-axis.

`sin(theta)="opposite"/"hypotenuse"`

`sin(theta)=F_(2y)/F_2`

Solving for `F_(2y)`:

`=>color(blue)(F_(2y)=F_2sin(theta))`

Similarly, using the cosine, we find that `color(blue)(F_(2x)=F_2cos(theta))`. Note that because the parallel component is to the left of the origin, we will say that this is negative.

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!