How to calculate how much inward force is neccessary crush an object into a black hole I would you list some ways I could determine how much force is required to crush an object into a black hole, web sites that show this would also be helpful In fact moreso.
 A: You want to know the pressure, not the force. The way you do crush something into a black hole is to start with a massive star, at least 1.4 times as big as the Sun. Pressure is very high in the center, but ongoing fusion helps sustain it against collapse. So you wait for the star to exhaust its fuel. As the core cools, the weight of the star above it crushes it. 
Typically people don't talk about the pressure needed as much as the minimum size of the star. For example, that is what this site does. 
If a star isn't quite big enough to collapse to a black hole, it becomse a neutron star. People do talk about the pressure in the middle of a neutron star. If the pressure was any bigger than this (because the star was bigger), it would collapse to a black hole. So the answer, according to the Wikipedia Neutron Star article, is $1.6×10^{35} Pa$
You might also be interested in this site, even though it doesn't have numbers. 
A: As I wrote in my comment above, you must use the TOV equation: https://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_equation
Integrating this equation / solving the ODE, one obtains for the pressure of the spherically symmetric object at some distance $r$, :
$p(r) = \mu \left[  \frac{\sqrt{1-r_s /  R} - \sqrt{1 - r_s r^2 / R^3}}{ \sqrt{1 - r_s r^2 / R^3} - 3 \sqrt{1 - r_s / R}}\right]$
where $\mu$ is the mass density of the object, and $r_s$ I have denoted as the Schwarzschild radius: $r_s = 2G M$. 
Now, look at this result: it is only valid for when $r \leq R$. The pressure at $r=0$ which is what you're concerned with is:
$p(0) = \mu \left[ \frac{\sqrt{1 - r_s / R} - 1}{1 - 3 \sqrt{1 - r_s /R}}     \right]$. 
This becomes negative, $p(0) < 0$, when the denominator of this expression becomes negative, since $\mu > 0$ by assumption. So, this becomes negative when:
$1 - 3 \sqrt{1 - r_s / R} < 0$. 
This inequality is essentially the collapse condition. So, your object will collapse into a "black hole" as long as this inequality is satisfied. 
You essentially use G.R. to see when the pressure of the object you are trying to "collapse" becomes negative. If your object is spherically symmetric, you can use the TOV equation to show that this collapse into a "black hole" occurs for $R>9/8 r_s$, where $r_s$ is the Schwarzschild radius. The spherically symmetric case is nice because of symmetry, but for general geometries, it is slightly more difficult. This hopefully answers your question. (On a side note, one can always have black holes without any such notions of things collapsing: just take a Minkowski spacetime and cut out a hole, you'll get the Schwarzschild metric!) 
A: As per my back envelope calculation, we need this much "pressure" to crush anything that has mass $m$ into a black hole 1
$P = m^2 \frac {c^6} {h^2G}$,
where all constants hold its usual meaning.
This clearly shows the pressure needed is only function of mass $m$.
1 Kapil P. Chandra, Gravitational pressure of a black hole, DOI: 10.13140/RG.2.2.20473.34409
