You know, these forces of friction are basically based on the force applied on the surfaces in contact. The force pressing the surface will be met by another force N (Normal force, always perpendicular to the pressing surfaces) according to the Action and Reaction law of Newton.
Enlarging the area of contact will not vary the force F but will decrease pressure. But it won't show in the computation. (Please refer to figure below)
...... W.................W
.____↓___........__↓__
.\............/......./.........\
...\_____/......./_______\
▒▒▒↑▒▒▒▒▒▒▒▒▒↑▒▒▒▒▒▒▒▒▒
.......N...................N
Irregardless of your contact area, Ff = µN
But if you have to specify the coefficient of static friction (µ), then your consideration of the larger area may fall on the characteristic of the surfaces in contact, whereby, you may specify a bigger coefficient.
...... W....................W
.____↓___...........__↓__
.\..P........./........./.p.......\
...\↓↓↓↓↓/........./↓↓↓↓↓↓↓↓\
▒▒▒↑▒▒▒▒▒▒▒▒▒↑▒▒▒▒▒▒▒▒▒
.......N....................N
If we have to analyze your point in terms of pressure, it will go this way:
................let a = smaller area
......................A = bigger area
the smaller area will give a pressure of
P = W/a
whereas the bigger area will give a pressure of
p = W/A
Now, the force F = pressure times area
...= P x A
So in the blocks, the Force will still come out the same, because
F = P x a <===for smaller area
F = p x A <====for bigger area
Smaller pressure is compensated by big area
Bigger pressure is compensated by small area
The result will be equal for force.