Finding the apparent acceleration due to gravity versus latitude

[pandafish]

Homework Statement

A mass at rest on the surface of the earth at latitude λ , experiences a reaction from the earth that consists of a normal component S per unit mass and a tangential frictional component F per unit mass, directed towards a point vertically above the north pole. The earth is assumed to be a uniform sphere of radius R and mass M rotating with an angular velocity ω about its axis. How is the magnitude and direction of the apparent ’acceleration due to gravity’ g , acting on the mass at latitude , related to these forces?

Show that:

$$g^2 = [\frac{GM}{R^2}-R\omega^2cos^2\lambda]^2 + [\frac{1}{2}\omega^2Rsin2\lambda]^2$$

Relevant Equations

$$g=-\frac{GMm}{R^2}$$

I began by drawing a diagram and resolving the forces. Since the question asked for 'apparent gravity' I tried to find the normal force.

I started with the equations:

$$\\(\frac{GM}{R^2}-N)sin\lambda-Fsin\lambda=m\omega^2Rcos\lambda$$
$$\\(\frac{GM}{R^2}-N)sin\lambda-Fcos\lambda=0$$

Solving simultaneously, I ended up with:

$$\\N=\frac{GM}{R^2}-R\omega^2cos^2\lambda$$

 

[haruspex]

$$\\(\frac{GM}{R^2}-N)sin\lambda-Fsin\lambda=m\omega^2Rcos\lambda$$

The terms on the left cannot both be sine. F is tangential while the other force sum is radial. I don’t think having trig terms both sides is right either.
Which direction do you intend this force balance to be in?

 

[pandafish]

My bad, its suppose to be:

$$\\(\frac{GM}{R^2}-N)cos\lambda+Fsin\lambda=m\omega^2Rcos\lambda$$

This force balance is suppose to be perpendicular to the axis of Earth's rotation.

 

[haruspex]

My bad, its suppose to be:

$$\\(\frac{GM}{R^2}-N)cos\lambda+Fsin\lambda=m\omega^2Rcos\lambda$$

This force balance is suppose to be perpendicular to the axis of Earth's rotation.

Then why the cos on the right?

 

[pandafish]

Then why the cos on the right?

The right side is the centripetal acceleration,

$$a = \frac{v^2}{r} = \omega^2r$$

I believe r is distance from the mass to the earth's axis of rotation, so

$$a = \omega^2(Rcos\lambda)$$

I made a mistake by including mass on the right.

 

[haruspex]

The right side is the centripetal acceleration,

$$a = \frac{v^2}{r} = \omega^2r$$

I believe r is distance from the mass to the earth's axis of rotation, so

$$a = \omega^2(Rcos\lambda)$$

I made a mistake by including mass on the right.

Sorry, I read R as though it was r.

You need an expression for g. Note how the problem statement defines g. It is not as used in your "relevant equation".

 

[pandafish]

I see what you mean. Working backwards from the solution, I realised that the 'apparent gravitational acceleration' was made up from the normal force , and the component of the centrifugal force tangential to Earth's surface. However, I'm don't quite understand why this is the case.

 

[haruspex]

I see what you mean. Working backwards from the solution, I realised that the 'apparent gravitational acceleration' was made up from the normal force , and the component of the centrifugal force tangential to Earth's surface. However, I'm don't quite understand why this is the case.

Consider a mass placed on the surface. Apparent gravity is the net force per unit mass. This corresponds to the weight that would be recorded by a conventional weighing machine.

 

[pandafish]

Thank you, that explanation makes it a lot clearer what the apparent force is.