Would it be possible to walk up a wall under right circumstances? For example, you are in a box that is connected a distance $R$ from a clockwise spinning centre. If I understand correctly, the spinning box is a result of the resulting centrifugal force $F_{centrifugal}$ = $\frac{mv^2}{R}$. The person then would be pushed against the wall opposite to the direction of $F_{centrifugal}$
See this picture in top view:

Where distance $R$, the direction of $F_{centrifugal}$ and v are indicated. The person in the box is looking in the opposite direction of the center and perpendicular to $v$ (speed).
If $R$ is small enough or $v$ large enough, then at a certain combination of $R$ and $v$, $F_{centrifugal}$ would be larger than $F_z$ = $mg$ (gravity).
Question
If $F_{centrifugal}$ > $F_z$ would it be possible to walk up the wall that is in front of him?
(The forces $F_{centrifugal}$ and $F_z$ are perpendicular on each other which mean they do not counteract each other. Similarly for example in a bus with the vertical gravity force and the horizontal acceleration of a person due to the acceleration of the bus.)
 A: Centrifugal force greater than gravity should be one of the prerequisites for walking up the wall. However, it is the friction of the wall which is the deciding factor.
Suppose the man lies down on the floor with his head towards the centre of the circle and the feet on the wall . Under normal circumstances, he would simply slip off the wall as the friction is negligible compared to weight. But due to the centrifugal force, he is now pushing against the wall with significant force. Now, it is possible for friction to overcome the gravity. But it depends upon the coefficient of friction. If it is k, then k times the centrifugal force must be greater than the person's weight. Only then can he walk up the wall
A: Partly agree with provided answers, but partly not, so here is my version.
TL;DR possible if speed and friction are big enough
Let's tie coordinate axis to box, which is a non-inertial reference frame. Man standing on the wall will be affected by four forces only:


*

*inertial force of $\frac{mv^2}{R}$ directed to the wall

*gravity

*normal force from the wall

*friction



Standing still (or moving without acceleration) means the following:
$$F_n = F_{inertia}$$
$$F_{friction} = mg $$
$F_{friction}$ can't extend $\mu F_n = \mu\frac{mv^2}{R}$, so the only condition for standing is $\mu\frac{mv^2}{R} \geq mg$. So the only requirement for climbing is that centrifugal force is greater than gravity divided by $\mu$.
Ok, once we met this condition, how would it look from climber's perspective?

Climber is affected by force (sum of gravity and inertial) which pushes him towards wall and along it, the other stuff (friction and normal force) is quite common. This situation is full equivalent of man climbing a hill on planet with gravity greater than on Earth. So @Lz4's advice to not remove both his legs while climbing is wrong. You are free to jump forward if you are strong enough to cope with "hill" angle and increased "gravity".
A: In simple words: Depends on your velocity.
Because if we look from the frame of reference of the box we find the the centrifugal force would provide the normal reaction (assuming cofficient of friction to be constant) and hence the friction force to balance gravitational force acting on the man.
u*(mv^2/r) ≥ m*g(this is the minimum condition for the man to stand on wall. If you want to be extra safe then you should solve the above equation with strict inequality to find v.) Once this condition is achieved the man can move on the wall. But remember while walking he must not remove both his legs at the same time because the frictional force is a contact force.It would only act when at least one the man's leg is in contact with the wall.
Where u is coefficient of friction between man's shoes and the wall
m is the mass of the man
v is the speed with which the box is rotating 
