Answer to Question #239978 in Quantum Mechanics for nic

Question

A rocket fires two engines simultaneously. One produces a thrust

of 725 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 which these

engines exert on the rocket.

Answer 1

First resolve the 513N thrust into two components of the right angles.

In the forward side, we have 513N cos(32.4°)

At right angles, we have 513N sin(32.4°)

So...

513N cos(32.4°) = 433.14 N

513N sin(32.4°) = 274.88 N

Next get the total forward force of the rocket.

725 N + 433.14 N = 1,158.14 N

And the total force at right angles:

0 + 274.88 N = 274.88 N

Next solve the Resultant Magnitude (F) through Pythagorean theorem. 

F² = a² + b²

F² = (1158.14 N)² + (274.88 N)²

F² = 1,341,288.26 + 75,559.01 

F² = 1,416,847.27

F = √1,416,847.27

F = 1,190.3139

F = 1,190.31 

Now that we have resultant magnitude, find the direction by dividing total force exerted at right angles by the total force exerted at the forward side. 

tanC = 274.88 N / 1,158.14 N

tanC = 0.237346089419241

C = tan⁻¹ 0.237346089419241

C = 13.35°