# What angle does the hanging mass make?

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1. Sep 28, 2015

### diredragon

1. The problem statement, all variables and given/known data
At a latitude of 50° north a mass is hanged by a massless string. By the means of an angle how much does the mass deviate from the earths radius due to rotation of the earth? Consider the Earth as a spherical body.

2. Relevant equations
Fcp=mv^2/r
w=dθ/dt
v=wr
3. The attempt at a solution
I have made a drawing show earth as a circle at whose latitude of 50
degrees a point on its circumference is connected by radius of a small circle of that latitude and R of earth. acceleration is directed along its r.
r=Rcosθ
How do i draw a free body diagram for thic situation. Three forces T,mg,Fcp are in play and i know only that the θ between mg and Fcp is 50°. Where is tension directed at?

Last edited: Sep 28, 2015
2. Sep 28, 2015

### Staff: Mentor

Tension is always along the string.
Be careful with the direction here. At the equator one is pointing inwards, the other is pointing directly outwards, so here you would have an angle of 180°. At 50° N the angle is different - but not 50°. It is still larger than 90°.

You can determine the magnitude of both Fcp and mg as function of the unknown mass.

3. Sep 28, 2015

### Mister T

Fcp doesn't belong on the diagram as a force. It's not a force, per se. It's the net force, or vector sum of T and Fg, where Fg is the gravitational force.

4. Sep 28, 2015

### diredragon

Last edited: Sep 28, 2015
5. Sep 28, 2015

### Mister T

The gravitational force should point towards the center of Earth.

6. Sep 29, 2015

### diredragon

Yes, but the whole diagram is tilted with respect to the earth one. Mg is drawn so that it acts towards the radius.

7. Sep 29, 2015

### Mister T

Note that some instructors and textbook authors present this scenario in a way that mg and T are equal in magnitude, opposite in direction. They will insist either that g be called the free fall acceleration or that mg be called the apparent weight, or perhaps both. Other instructors and textbook authors will instead say that mg equals the gravitational force or the true weight, or perhaps both. This is a confusion made worse when an instructor does it one way and the textbook another.

The best way to eliminate the confusion is refer only to Fg and T, where Fg is the gravitational force calculated using Newton's Law of Universal Gravitation. For an observer standing on the equator of a spherical Earth these two forces are opposite in direction, but they are not equal in magnitude. For the observer at a latitude of 50o they are neither equal in magnitude nor opposite in direction. But their vector sum has a magnitude of mv2/r.