Just a raw estimation to be added to the other answers:
At takeoff, engine's thrust is used to accelerate the airplane and to overcome aerodynamic drag plus rolling resistance of the landing gears. Or, in the case of our sliding airplane, the sliding friction.
In order to understand if making the airplane slide on its belly instead of rolling on its tires is doable, let's ballpark estimate these four terms (acceleration, aerodynamic drag, rolling resistance and sliding friction). No worries, just a couple of simple multiplications.
Acceleration: since the airplane has to reach the same takeoff speed within the same airfield's length both with and without landing gears, then it has to accelerate in the same way for both cases; from basic physics, acceleration is $a=V²/2L$ and for Newton's second law the relevant force to be applied is $F=ma=mV²/2L$ where $m$ is the airplane's mass, $V$ the takeoff speed and $L$ the field's length. Let's use a B747 to do the maths (SI units):
$F=(370'000\times85²)/(2\times3'300)=405kN$
Drag: let's evaluate it just before takeoff which is where the highest value is reached before (any kind of) landing gears become useless; we use the usual drag equation $D=½\rho V²SC_d$ at sea level with $C_d=0.02$ as given in the B747's polar:
$D=½\times1.125 \times85² \times511 \times0.02=40kN$
Rolling resistance; this is simply the weight times a coefficient which depends on the materials; for tire on asphalt this is some 0.03:
$F_{roll}=370'000 \times 0.03=11kN$
Sliding friction; also in this case it is simply the weight times a friction coefficient that for steel (I couldn't find aluminium) on asphalt is more or less 1:
$F_{slide}=370'000 \times 1=370kN$
So, let's do the sum to get the needed thrust $T$ for the case with landing gears (1.+2.+3.) and without (1.+2.+4.) respectively:
$T=405+40+11=456kN$
vs.
$T=405+40+370=815kN$
The latter represents an increase of some 80% in respect to a takeoff with landing gears... most probably not feasible.