Question

What is the effective weight of a person of mass 60 kg carried vertically up in a
rocket with an acceleration of 2g? Draw a properly labelled free-body diagram.
b) An aeroplane flies due east along the equator with a speed of 300 ms−1. Determine the
magnitude and direction of the Coriolis acceleration.

Answer 1

Answer on Question #61702-Physics-Mechanics

What is the effective weight of a person of mass $60\mathrm{kg}$ carried vertically up in a rocket with an acceleration of $2\mathrm{g}$ ? Draw a properly labeled free-body diagram.

Solution

Write Newton's second law in the vector form

$$\vec {F} = \vec {N} + \vec {W}$$

The equation will look like in the projection on the axis of $+\Upsilon$

$$m a = N - m g$$ $$N = m a + m g$$

Where, $a = 2g$

$$N = 2 m g + m g$$

Finally,

$$N = 3 m g = 3 \cdot 6 0 k g \cdot 9. 8 \frac {m}{s ^ {2}} = 1 7 6 4 N$$

Answer: 1764 N.

b) An aeroplane flies due east along the equator with a speed of 300 ms-1. Determine the magnitude and direction of the Coriolis acceleration.

Solution

The Coriolis acceleration is

$$\boldsymbol {a} _ {C} = 2 \boldsymbol {\Omega} \times \boldsymbol {v}$$

Where

$$\Omega = \frac {2 \pi}{2 4 \cdot 6 0 \cdot 6 0}.$$

The magnitude is

$$a _ {C} = 2 \frac {2 \pi}{24 \cdot 60 \cdot 60}300 \sin 90 = 0.044 \frac {m}{s ^ {2}}.$$

The direction is towards the center of Earth.

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