1.71 A rocket fires two engines simultaneously. One produces a thrust of 480 N directly forward, while the other gives a 513-N thrust at 32.4 degrees above the forward direction.

1.71 A rocket fires two engines simultaneously. One produces a thrust of 480 N directly forward, while the other gives a 513-N thrust at $32.4^o$ above the forward direction. Find the magnitude and direction (relative to the forward direction) of the resultant force that these engines exert on the rocket.

SOLUTION

Figure a shows the given vectors. Make a diagram like that of Figure b which shows vector-addition.

Let's solve for the $x$- and $y$-components of the given vectors, where the angles are measured from the +x-axis rotates toward +y-axis,

$A_x=(480 \ N) \ cos \ 0^o=480 \ N$

$A_y=(480 \ N) \ sin \ 0^o=0$

$B_x=(513 \ N) \ cos \ 32.4^o=433.1 \ N$

$B_y=(513 \ N) \ sin\ 32.4^o=274.9 \ N$

Then the components of the resultant, $\vec{R}$ are

$R_x=A_x+B_x=913.1 \ N$

$R_y=A_y+B_y=274.9 \ N$

Using the Pythagorean theorem and trigonometry, we have the magnitude and direction,

$R=\sqrt{(R_x)^2+(R_y)^2}=954 \ N$

$tan \ \theta=\frac{R_y}{R_x}$

$\theta=16.8^o$ above the forward direction. (Since we have to write direction relative to the forward direction as the problem says.)

Figure a.

Figure b.

You can check your answer using the law of cosines.