Determine the magnitude and direction of the effective value of g

In summary, at a latitude of 45 degrees on the Earth, the effective value of g is the combination of the gravitational force towards the center of the Earth and the centrifugal force away from the axis of rotation. This results in a deflection of the weight vector away from "straight down" to the center of the Earth, and a net acceleration that produces the object's effective weight.
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Homework Statement


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Determine the magnitude and direction of the effective value of g at a latitude of 45 degrees on the Earth. Assume the Earth is a rotating sphere.

Homework Equations



a = v2/r

The Attempt at a Solution



This question requires a vector diagram no doubt.
What I know is that there is gravity force towards to center of the Earth and normal force exactly opposite to it, and also centripetal force along the x axis. But centripetal force here confuses me as I don't know how to draw it. In my early problems with circular uniform motion, centripetal force was always component of some other force -like normal force, tension force, or friction force. I don't recall drawing an Fcentripetal in the vector diagram for questions like these:

fetch.php?media=phy141%3Alectures%3Aconicalpendulum.png
carbank.gif



But in this gravity question, I think I need to show that there is an Fcentripetal along the x axis? Or is it nothing more than a component of gravity force?
 
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An object on the surface of the Earth moves in a circle about the axis of rotation of the Earth, so there must be something providing the centripetal force. You're right that it is a portion of the gravitational force that is co-opted for this purpose. The net weight of objects are decreased accordingly, and the net direction for the weight vector is deflected a small amount away from "straight down" to the center of the Earth when the object in question is not located on the equator.

If you choose as your frame of reference the rotating Earth (a non-inertial frame of reference since it's rotating) it allows you to speak of a centrifugal pseudo force without getting too many vocal objections from physicists :) In that case you can draw a free-body diagram for an object and pencil in the gravitational acceleration and centrifugal acceleration. Sum them to find the net acceleration. That net acceleration is what produces the effective weight of the object.
 
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Related to Determine the magnitude and direction of the effective value of g

1. What is the effective value of g?

The effective value of g is the magnitude and direction of the gravitational acceleration experienced by an object at a specific location. It takes into account the gravitational force from all nearby objects, including the Earth and other celestial bodies.

2. How is the effective value of g determined?

The effective value of g can be determined through experiments and calculations using the principles of Newton's laws of motion and universal gravitation. It can also be measured using specialized instruments, such as accelerometers and gravimeters.

3. What factors can affect the effective value of g?

The effective value of g can be affected by several factors, including the mass and distance of surrounding objects, the shape and rotation of the Earth, and the altitude and latitude of the location where it is being measured.

4. Why is it important to know the effective value of g?

Knowing the effective value of g is important in many scientific fields, such as physics and astronomy, as it helps in understanding the behavior of objects under the influence of gravity. It also has practical applications in areas such as navigation and space exploration.

5. How does the effective value of g differ from the standard value of g?

The standard value of g is the average value of gravitational acceleration at sea level on Earth, while the effective value of g takes into account variations due to different factors. The effective value of g can be higher or lower than the standard value depending on the location and other influencing factors.

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