coconutman2015
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Suppose three forces are acting on a model rocket launching into the air. The force of gravity is 1.2 N acting downward. The rocket engine has a force of 12.3 N acting at an angle of 65⁰ above the horizontal to the right. The wind pushes the rocket with a force of 2.5 N to the left. Find the magnitude and direction of the resultant force on the rocket.

Please answer!!! Offering 50 points for answer and explanation

Respuesta :

jaydensquires2323

Answer:

can you include a picture

LammettHash

Decompose the forces acting on the rocket in horizontal and vertical components.

• net horizontal force:

∑ F = (12.3 N) cos(65°) - 2.5 N ≈ 2.7 N

(notice we're taking "to the right" to be the positive direction)

• net vertical force:

∑ F = (12.3 N) sin(65°) - 1.2 N ≈ 9.9 N

The resultant force then has magnitude

√((2.7 N)² + (9.9 N)²) ≈ 10.3 N

Since the horizontal and vertical components of the resultant are both positive, it points at an angle between 0° and 90° from the positive horizontal, so that

tan(θ) ≈ (9.9 N) / (2.7 N) ≈ 3.69

===>   θ ≈ arctan(3.69) ≈ 75°

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