Will this rope break due to the tidal forces or not? While I was thinking about how tidal forces can make objects float at the surface of a planet orbiting a massive object like a black hole, the fact that any material on the Earth isn't held together by gravity only, but also by chemical bonds which give it its tensile strength came into my mind. 
The scenario I thought of is as following:
1- A 10 kg metal sphere is tied to a 1 meter long thin rope and this rope is nailed into the surface of the Earth. 
2- The tensile strength of the rope is 10 N. (So the metal sphere has to accelerate at 1 $m/s^2$ to break the rope)
3- The Earth is orbiting a black hole at its Roche radius, and so our metal sphere on the surface of the Earth is effectively weightless and floating, but it is still held by the rope.
Here is a simple picture to summarize:

Now, if we move the Earth to orbit the black hole even closer until the tidal forces on Earth due to the black hole become $\Delta a$ = 10.8 $m/s^2$ and so the metal sphere is pulled by the difference between $\Delta a$ and the Earth's gravitational acceleration (9.8 $m/s^2$) which is 1 $m/s^2$ towards the BH, will the rope break or not ?
In other words, if a is the gravitational acceleration towards the BH, in order to break the rope, which one do we need ? :
1- $a_{Point A}$ - $a_{Point B}$ = 1 $m/s^2$
OR
2- $a_{Point A}$ - $a_{Point C}$ = 1 $m/s^2$ 
 A: Between points A and C.  The effect of having the Earth there is that it is providing a link between the center of mass of the Earth (C) and the test sphere (A).  Effectively, there is an infinitely strong, rigid rod going from B to C, which is then connected to the rope.  The tensions in that rod (the Earth) must match that of the rope.
Another way to think about it, is that if you removed the Earth entirely from the problem, and wanted to see if the rope was strong enough to resist the tidal forces --- then you would consider the difference in acceleration between points A and B.
A: Between points A and C
The earth can be considered as a wall and thus to break off the 'wall' a relative acceleration with respect to the wall is a must , If the tensile strength is 10N and mass 10 kg then the rope must move at a relative acceleration of 1m/s^2 or 
Acceleration(a) - Acceleration(c) = 1
A: The acceleration of the points $B$ and $C$ will be same assuming that the Earth is not breaking apart, i.e., the acceleration of all points on the Earth will be same. If you think of the Earth as a very strong rod, only the Tension in the rod varies so that the entire rod accelerates with the same acceleration as that of the center of mass of the rod.
So, for the rope to break, we need - 
${ a }_{ A } - { a }_{ B } = 1 m/s^2$
or 
${ a }_{ A }-{ a }_{ C } = 1 m/s^2$ 
As ${ a }_{ B }={ a }_{ C }$
A: You can't add forces on different objects to get an acceleration of one object. All the forces should act on the center of the sphere to have effect on breaking the sphere. All the three forces: the tidal force $F_{T}$, the gravitational force $F_{g}$ and the Tension force because of the rope pulling the sphere $T$ act at the center of the sphere for the sphere to move intact without breaking apart. 
Mathematically, 
   $$F_{T} - F_{g} - T = m_{s}a_{s}$$
where $m_{s}$ is mass of the sphere and $a_{s}$ is acceleration of the sphere towards the  Blackhole. 
$$\Rightarrow m_{s}a_{T} - m_{s}g - T =m_{s}a_{s}$$
where $a_{T}$ is the tidal acceleration on the sphere. 
   which gives us,
$$a_{T} - a_{s} = \frac{m_{s}g + T}{m_{s}}$$
$a_{T}$ and $a_{s}$ act at the center of the sphere. g is acceleration due to gravity and acts again at the center of the sphere not at points A, B and C.  
