Question

I answered part a and b already, need help with c and d.

Answer 1

Given the forces:

300 N at 0 degrees

400 N at 90 degrees

Let's solve for part C and D.

Part c.

The magnitude of the resultant force.

To find the magnitude of the resultant force, apply the formula:

$F_R=\sqrt[]{F^2_x+F^2_y}$

Where:

$\begin{gathered} F_x=A_x+B_x \\ \text{Where:} \\ A_x=A\cos \theta=300\cos 0=300 \\ B_x=B\cos \theta=400\cos 90=0 \\ \\ F_x=300+0=300N \end{gathered}$$\begin{gathered} F_y=A_y+B_y \\ \text{Where:} \\ A_y=A\sin \theta=300\sin 0=0N \\ B_y=B\sin \theta=400\sin 90=400\text{ N} \\ \\ F_y=0+400=400N_{} \end{gathered}$

Fx = 300

Fy = 400

Hence, we have:

$\begin{gathered} F_R=\sqrt[]{(300)^2+(400)^2} \\ \\ F_R=\sqrt[]{90000+160000} \\ \\ F_R=\sqrt[]{250000} \\ \\ F_R=500\text{ N} \end{gathered}$

Therefore, the magnitude of the resultant force is 500 N.

• Part d.

The direction of the resultant force.

To find the direction of the resulatnt force, apply the fomula(trig ratio for tan):

$\tan \theta=\frac{F_y}{F_x}$

Hence, we have:

$\begin{gathered} \tan \theta=\frac{400}{300} \\ \\ \tan \theta=1.33 \end{gathered}$

Take the tan inverse of both sides:

$\begin{gathered} \theta=\tan ^{-1}(1.33) \\ \\ \theta=53.13\degree \end{gathered}$

Therefore, the direction of the resultant force is 53.13° from the x-axis.

ANSWER:

(c) 500 N

(d) 53.13°