Question

A 15,600 kg rocket in space fired two thrusters. One thruster propelled the rocket forward with a force of 77,500 N while the second thruster propelled the rocket to port (left) with a force of 27,200 N. Find its acceleration (angle and magnitude).

Answer 1

First, let us find the resultant force. Since both forces $\mathbf{F}_1$ and $\mathbf{F}_2$ (see the figure) are perpendicular to each other, the magnitude of the resultant force $\mathbf{F}$ can be found with Pythagorean theorem:

$$F = \sqrt{F_1^2 + F_2^2}$$

where $F_1 = 77500N$ and $F_2 = 27200N$ are magnitudes of the forces $\mathbf{F}_1$ and $\mathbf{F}_2$ respectively. Thus, obtain:

$$F = \sqrt{77500^2 +27200^2}\approx 82135N$$

The magnitude of the acceleration is (according to the second Newton's law):

$$a = \dfrac{F}{m} = \dfrac{82135N}{15600kg} \approx 5.3m/s^2$$

where $m = 15600kg$ is the mass of the rocket.

The angle of the acceleration with x-axis is equal to the corresponding angle $\theta$ (from the right triangle in the figure):

$$\theta= \arctan\left( \dfrac{-F_1}{F_2}\right)\\ \theta = 180\degree - \arctan\left( \dfrac{77500}{27200}\right) = 180\degree-70.7\degree = 109.3\degree$$

Answer. Magnitude: 5.3 m/s^2, angle: 109.3 degrees.