Our weight is the normal force of the Earth pushing back on us du to the effect

Question

Our weight is the normal force of the Earth pushing back on us du to the effect of gravity (i.e. mg). however, we actually all are moving in large circles at constant speeds on the surface of the rotating Earth. If the effect of uniform circular motion of an object on the Earth's surface were include in its free body diagram, what would the vertical force balance equation be? Use the gravitational force as simply mg. Assume the Earth has a constant equatorial radius of 6,378 km. you may refer to the figure at the right which portrays the rotating Earth as seen from above the North pole and an Object on the equator. What would be the length of day if the rate of revolution of the Earth were such that a person had no weight at the equator? The captain of a small private airplane is planning to fly from Binghamton, NY to Albany, NY. There are 116,92 miles between Albany, NY and Binghamton, NY. On the map of NY State, the pilot determines that the direct compass heading from Binghamton to Albany is 70.2 degree east of north. The pilot has a current weather report indicating that at the flight altitude the wind speed is 72 mph coming from a direction of 51 degree west of north. The cruise sped of the airplane is 140 mph. In what direction should the plane head so that it arrives in Albany taking into account the wind speed? How long will the flight take from Binghamton to Albany? Two masses on either side of an triangle with angles of 40degree, m_1 = 5 kg, m_2 is adjustable by the additional or withdrawal of sand. The coefficient of friction is 0.7 between each mass and the incline and the coefficient of kinetic friction is 0.35 between each mass and the incline. What is the maximum mass of m_2 so that the system will remain at rest? What is the minimum mass of m_2 so that the system will remain at rest?

Explanation / Answer

a) Here, Gravitational Force = centripetal Force

=> GMm/R2   =   mv2/R

=>    mg = mv2/R

where,    GM/R2   =   g   = acceleration due to gravity

where, R = 6378 km

b)   Here , v = sqrt(6378 * 1000 * 9.8)

                       = 7906 m/sec

=> length of day = (2 * 3.14 * 6378 * 1000)/7906

                            =   1.407 hours .

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