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30 April, 15:42

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° above the forward

direction. Find the magnitude and direction of the resultant force acting on the rocket.

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  1. 30 April, 16:08
    0
    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°
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