Calculating actual resistance in plank rowing I hope this is the correct TA here, but please consider the below rudimentary diagram, which depicts bodyweight plank rows using a particular device.  
The image shows the setup of the device, and the exercise:

I have not studied physics for years, and have no idea where to start with this, so I am asking here:
1) Given a particular angle Y and distance X, what will be the force on point A/B when carrying out the movement?  I ask this because I am concerned about damaging the door or the house structure.
2) Given a particular angle Y and distance X, how much of the 200lb bodyweight will actually be being pulled?  That is, what is the resistance needed to be overcome to complete the pull?  I ask this because I need to know how much easier moving further from the door makes the exercise, and how much easier moving closer to the door makes it!
(Sorry if the tags are inappropriate; I don't know what to tag this question with!)
Follow up based on answer to confirm correct understanding of angles
Is the use of the formula given correct?

 A: The answer to both questions is
$$
F= \frac{\cos{(Z)}}{\sin{(Y+Z)}}\times \text{(body weight)}
$$
where I called $Z$ the angle between the body and the floor.
So the force that on the door and the pull is $F$.
They are equal because, in simple words, the force on the door is the opposite reaction to the pull. "For every action, there is an equal and opposite reaction."
If you really want a formula using the distance $X$ and not the angle $Z$, it is possible, but the formula will look more complicated. I think that the angle $Z$ is easy to measure anyway. In any case, if the distance $X$ is long enough, and the guy is tall (compared with the door), then it should be possible to pull in a direction almost perpendicular to the body. In this case the angle Z + Y become almost $90$ degrees, and you can approximate $\sin(Y+Z)\approx1$ and also $\cos Z\approx \sin Y$. In this case the formula becomes simpler:
$$
F\approx {\sin{(Y)}} \times \text{(body weight)}
$$
If you are interested in the derivation, here it is.
So you want to have a static balance of all forces involved, so you need to have
$$
F\sin Y+N-mg=0
$$
$$
F\cos Y-A=0
$$
$$
N\cos Z - A \sin Z=0
$$
Here, $N$ is the normal force from the ground, $A$ is the friction on your shoes, and $mg$ is the body weight.
The first 2 equations are the balance of the forces, the last equations is the balance of the torques. Now after some algebra you get
$$
F\times\Big(\sin Y+\frac{\cos Y \sin Z}{\cos Z}\Big)=mg=\text{bodyweight}
$$
Now using some trigonometry you get the formula that I wrote in the beginning.
Edit:
the angles Y and Z can be calculated using trigonometry:
$$
Y=\arccos(X/L)
$$
$$
Z=\arcsin\Big( \frac{H-X \tan(Y)}{h} \Big)
$$
where $L$ is the length of the rope, $h$ is the height of the guy, and $H$ the height of the door.
If you really want the full formula, here it is
$$F=
\frac{h L \sqrt{1-\frac{\left(H-L \sqrt{1-\frac{X^2}{L^2}}\right)^2}{h^2}}}{L \sqrt{1-\frac{X^2}{L^2}} \left(h \sqrt{\frac{h^2-H^2+2 H L \sqrt{1-\frac{X^2}{L^2}}-L^2+X^2}{h^2}}-X\right)+H X}
\times\text{bodyweight}
$$
A: To understand what happens a bit better I decided to follow a dynamic
approach, not just a static one, as it makes a difference.

I
am going to assume that the feet of the person are fixed
throughout the exercise. The horizontal distance between the feet
of the person and the vertical axis, along the closet door, is $X$
and the exercise rope is attached to the closet door (the vertical
axis) at height $H$ from the ground.  Furthermore, I am going to
represent the human as a straight-line segment of height $h$ and
mass $m$. The feet of the person are the lower end of the segment,
that has contact with the ground. The upper end of the segment is
where the shoulders of the person are, or the upper chest,
whichever is more appropriate. For simplicity I am going to assume
that the force with which the person pull themselves along the
rope is always applied to the upper end of the segment and it is
aligned with the line from the upper end of the segment (the
shoulders/upper chest) to the point where the rope is attached on
the vertical axis/closet door. Moreover, I am going to denote the
distance from the feet to the center of gravity of the person/the
line segment by $l$. By assumption, the distance from the feet to
the upper end of the segment is $h$. If we denote by $L$ the
length of the rope and by $s$ the distance from the shoulder/upper
chest to the end of the rope, held by the person (i.e. $s$ is the
stretch of the arm stretch so to say), we get that the distance
between the shoulders/upper chest to attachment point of the rope
and the closet door is $L + s$.
Let us set an inertial coordinate system where the origin $O$ is
the point where the closet door meets the ground. The horizontal
axis $Ox$ is aligned with the ground, pointing from $O$ to the
feet of the person, while the vertical axis $Oy$ is aligned with
the closet door. Denote by $\vec{e}_x, \, \vec{e}_y$ be the pair
of orthogonal unit vectors aligned with the axes $Ox$ and $Oy$ of
the coordinate system and denote $\vec{e}_z$ the vector, pointing
perpendicularly from the picture towards us.
During the exercise, the person/segment is tilted at a
time-varying angle $\theta = \theta(t)$ relative to the ground,
i.e. that $45^{\circ}$ angle on your picture is an example of
$\theta$. If we denote by $\vec{h}$ and $\vec{l}$ the vectors
along the segment pointing from the feet to the shoulders and from
the feet to the center of mass respectively, then $\theta =
\angle\, (\vec{e}_x, \, \vec{h}) = \angle\, (\vec{e}_x, \,
\vec{l})$ (because $\vec{l}$ and $\vec{h}$ are aligned vectors).
$\vec{G} = -\, mg \, \vec{e}_y$ is the force of gravity acting on
the center of mass of the person and $\vec{F}$ is the force with
which the person pulls along the rope, where we have assumed that
at each moment of time, (or in each position of the segment)
$\vec{F}$ is aligned with the segment from the upper end of the
segment to the point where the rope is attached to the door.
Then the equations of motion of the person can be derived from the
law that the moment of inertia times the angular acceleration of the segment is equal to the
sum of the torques of all forces acting on the segment relative to
the point of where the segment meets the ground (the feet).
\begin{align}
I\,\frac{d^2\theta}{dt^2}\, \vec{e}_z = \vec{h} \times \vec{F} \,
+ \, \vec{l} \times \vec{G}
\end{align}
Let us decompose all the vectors along the axes $Ox$ and
$Oy$ (i.e. along the vectors $\vec{e}_x$ and $\vec{e}_y$)
\begin{align}
& \vec{h} = h\, \cos(\theta)\, \vec{e}_x + h \, \sin(\theta)\, \vec{e}_y\\
& \vec{l} = l\, \cos(\theta)\, \vec{e}_x + l \, \sin(\theta)\, \vec{e}_y\\
& \vec{G} = - \, mg\, \vec{e}_y\\
& \vec{F} = F\, \left(\, \frac{\,- \,\big(\, X + h\, \cos(\theta)
\, \big)\, \vec{e}_x \, + \, \big(\, H - h\, \sin(\theta)\,
\big)\, \vec{e}_y \,}{\,\sqrt{\big(\, X + h\, \cos(\theta) \,
\big)^2 \, +
\, \big(\, H - h\, \sin(\theta)\, \big)^2\,}\,}\,\right)\\
&L \, + \, s = \sqrt{\big(\, X + h\, \cos(\theta) \, \big)^2 \, +
\, \big(\, H - h\, \sin(\theta)\, \big)^2\,}
\end{align}
Calculate the cross-products, having in mind that $$\vec{e}_x
\times \vec{e}_y = - \, \vec{e}_y \times \vec{e}_x = \vec{e}_z
\,\,\text{ and } \,\,\vec{e}_x \times \vec{e}_x = \vec{e}_y \times
\vec{e}_y = \vec{0}$$
\begin{align}
\vec{l} \times \vec{G} =& \big( \, l\, \cos(\theta)\, \vec{e}_x + l \, \sin(\theta)\, \vec{e}_y \, \big) \times \big(\,- \, mg \, \vec{e}_y\,\big)\\
=& -\, lmg \, \cos(\theta)\, \vec{e}_z
& \\
\vec{h} \times \vec{F} =&  \big( \, h\, \cos(\theta)\, \vec{e}_x + h \, \sin(\theta)\, \vec{e}_y \, \big) \\
&\times F\, \left(\, \frac{\,-\,\big(\, X + h\, \cos(\theta) \,
\big)\, \vec{e}_x \, + \, \big(\, H - h\, \sin(\theta)\, \big)\,
\vec{e}_y \,}{\,\sqrt{\big(\, X + h\, \cos(\theta) \, \big)^2 \, +
\, \big(\, H - h\, \sin(\theta)\, \big)^2\,}\,}\,\right)\\
=& F\left(\, \frac{\, h \, \sin(\theta)\,\big(\, X + h\,
\cos(\theta) \, \big) \, + \, h\, \cos(\theta)\,\big(\, H - h\,
\sin(\theta)\, \big) \,}{\,\sqrt{\big(\, X + h\, \cos(\theta) \,
\big)^2 \, + \, \big(\, H - h\, \sin(\theta)\,
\big)^2\,}\,}\,\right) \, \vec{e}_z\\
=& F\left(\, \frac{\, hX \, \sin(\theta) \, + \, hH\,
\cos(\theta)\,}{\,\sqrt{\big(\, X + h\, \cos(\theta) \, \big)^2 \,
+ \, \big(\, H - h\, \sin(\theta)\, \big)^2\,}\,}\,\right) \,
\vec{e}_z\\
=& F\left(\, \frac{\, hX \, \sin(\theta) \, + \, hH\,
\cos(\theta)\,}{\,\sqrt{X^2 + H^2 + h^2 + \, 2 h X \, \cos(\theta)
\,
 - \, 2 h H\, \sin(\theta)\,}\,}\,\right) \, \vec{e}_z
\end{align}
Hence, the equations of motion in vector form simplify to
\begin{align}
I \frac{d^2\theta}{dt}\, \vec{e}_z =& \, F\,\left(\, \frac{\, hX
\, \sin(\theta) \, + \, hH\, \cos(\theta)\,}{\,\sqrt{X^2 + H^2 +
h^2 + \, 2 h X \, \cos(\theta) \,
 - \, 2 h H\, \sin(\theta)\,}\,}\,\right) \, \vec{e}_z\\
& - \, lmg \, \cos(\theta) \, \vec{e}_z
\end{align}
and they are all along the same vector $\vec{e}_z$, so they reduce
to one equation of motion
\begin{align}
I \frac{d^2\theta}{dt}\, =& \, F\,\left(\, \frac{\, hX \,
\sin(\theta) \, + \, hH\, \cos(\theta)\,}{\,\sqrt{X^2 + H^2 + h^2
+ \, 2 h X \, \cos(\theta) \,
 - \, 2 h H\, \sin(\theta)\,}\,}\,\right) -  lmg \,
\cos(\theta)
\end{align}
We also have an equation that connects the angle variable $\theta
= \theta(t)$ (the tilt of the person relative to the ground) to
the variable $s = s(t)$ (the arm-stretch, i.e. the distance of the
upper chest of the person to the closest end of the rope)
$$s = \sqrt{X^2 + H^2 + h^2
+ \, 2 h X \, \cos(\theta) \,
 - \, 2 h H\, \sin(\theta)\,} - L$$
So we can express the magnitude of the force with which the person
is pulling, which is equal to the magnitude of the force applied
to the door:
\begin{align}
F = \frac{\left(\,I \frac{d^2\theta}{dt^2} + lmg \,
\cos(\theta)\,\right)\sqrt{X^2 + H^2 + h^2 + \, 2 h X \,
\cos(\theta) \,
 - \, 2 h H\, \sin(\theta)\,}}{\,hX \,
\sin(\theta) \, + \, hH\, \cos(\theta)\,}
\end{align}
Here $I = \frac{1}{3}mh^2$ is the moment of inertia of the person,
represented as a straight-line segment of height $h$. So what is
the meaning of the term $I \frac{d^2\theta}{dt^2}$?
$\frac{d^2\theta}{dt^2}$ is the angular acceleration of the person
at the moment of time $t$, i.e. this is how fast the angular
velocity of the person changes with time. It also means that
things are different when you are calculating $F$ while the person
is static and while the person is moving. If the person is static,
then $\theta$ doesn't change and therefore $I
\frac{d^2\theta}{dt^2} = 0$ and the equation simplifies to
\begin{align}
F = \frac{lmg \, \cos(\theta)\,\sqrt{X^2 + H^2 + h^2 + \, 2 h X \,
\cos(\theta) \,
 - \, 2 h H\, \sin(\theta)\,}}{\,hX \,
\sin(\theta) \, + \, hH\, \cos(\theta)\,}
\end{align}
However, when the person starts exercising actively, the angular
position of the person relative to the ground will start changing
and with it angular acceleration $\frac{d^2\theta}{dt^2}$ will
emerge, increasing the magnitude of the force. This explains why
if you are static, the door may hold the rope, but when the
exercise begins and the person starts moving, the magnitude $F$ of
the force may increase to a point of breaking the door.
I thought you should be aware of this effect.
So, the relevant equations you  may want to consider are
\begin{align}
&F = \frac{\left(\,\frac{1}{3}mh^2\, \frac{d^2\theta}{dt^2} + lmg
\, \cos(\theta)\,\right)\sqrt{X^2 + H^2 + h^2 + \, 2 h X \,
\cos(\theta) \,
 - \, 2 h H\, \sin(\theta)\,}}{\,hX \,
\sin(\theta) \, + \, hH\, \cos(\theta)\,}\\
& \tan(Y) = \frac{H - h\, \sin(\theta)}{X + h\, \cos(\theta)}\\
& \cos(Y) = \frac{X + h\, \cos(\theta)}{L + s}\\
& \sin(Y) = \frac{H - h\, \sin(\theta)}{L + s}\\
 &s = \sqrt{X^2 + H^2 + h^2 + \, 2 h X \, \cos(\theta) \,
 - \, 2 h H\, \sin(\theta)\,} - L
\end{align}
where the angle $Y$ is the one on your picture, in case you still
need it. If you want to calculate the static case, set the angular
acceleration $\frac{d^2\theta}{dt^2} = 0$. Otherwise, you can say
write a somewhat more realistic model, in which the person
exercises in a rhythmic way, between position angles $\theta_1$
and $\theta_2$ with period $T$ seconds starting from $\theta_1$,
reaching $\theta_2 > \theta_1$ and then returning back to
$\theta_1$. This is a periodic function $\theta = \theta(t)$ with
period $\theta(t + T) = \theta(t)$ so you can approximate it by
the trigonometric polynomial
$$\theta = \theta_1 \, \cos^2\left(\frac{2\pi}{T}\, t\right) + \theta_2 \, \sin^2\left(\frac{2\pi}{T}\, t\right)$$
and then differentiate twice this function and get the angular
acceleration
$$\frac{d^2\theta}{dt^2} = \frac{8\pi^2}{T^2}(\theta_2 - \theta_1)\cos\left(\frac{4\pi}{T}\, t\right)$$
so the angular acceleration can have a maximal value (for this
particular motion) $\frac{8\pi^2}{T^2}(\theta_2 - \theta_1)$, so
you can plug it in the formula above to get a slightly better
estimate for the force's magnitude (i.e. you could say that the
force's magnitude should be no greater than the approximate rough
estimate)
\begin{align}
&F = \frac{\left(\,\frac{1}{3}mh^2\, \frac{8\pi^2}{T^2}(\theta_2 -
\theta_1) + lmg \, \cos(\theta)\,\right)\sqrt{X^2 + H^2 + h^2 + \,
2 h X \, \cos(\theta) \,
 - \, 2 h H\, \sin(\theta)\,}}{\,hX \,
\sin(\theta) \, + \, hH\, \cos(\theta)\,}\\
& \tan(Y) = \frac{H - h\, \sin(\theta)}{X + h\, \cos(\theta)}\\
& \cos(Y) = \frac{X + h\, \cos(\theta)}{L + s}\\
& \sin(Y) = \frac{H - h\, \sin(\theta)}{L + s}\\
 &s = \sqrt{X^2 + H^2 + h^2 + \, 2 h X \, \cos(\theta) \,
 - \, 2 h H\, \sin(\theta)\,} - L
\end{align}
