You have to know where the center of gravity is. If $a$ is the % distance along the wheelbase for the center of gravity (50% = center, 0% = front, 100%=back), and $b$ the % distance along the track for the center of gravity (50% = center, 0% = left, 100% right) then the weight fractions for each wheel are:
$$ (\mbox{front-left}) = (\mbox{Weight}) \frac{3-2a-2b}{4} $$
$$ (\mbox{front-right}) = (\mbox{Weight}) \frac{1-2a+2b}{4} $$
$$ (\mbox{rear-left}) = (\mbox{Weight}) \frac{1+2a-2b}{4} $$
$$ (\mbox{rear-right}) = (\mbox{Weight}) \frac{2a+2b-1}{4} $$
There equations come from the balance of moments in two planes. Here is a top view of the balance.
Example:
I the front back-balance is $a=0.4$ and the left-right balance is $b=0.55$, with a chassis weight of $0.5\,{\rm kg}$ then the corner weights are:
$$ (\mbox{front-left}) = (0.5) \frac{3-2*0.4-2*0.55}{4} = 0.1365 {\rm kg}$$
$$ (\mbox{front-right}) = (0.5) \frac{1-2*0.4+2*0.55}{4} = 0.1625 {\rm kg}$$
$$ (\mbox{rear-left}) = (0.5) \frac{1+2*0.4-2*0.55}{4} =0.0865 {\rm kg}$$
$$ (\mbox{rear-right}) = (0.5) \frac{2*0.4+2*0.55-1}{4} =0.1125 {\rm kg}$$
Results Check
- Total weight on left wheels = $0.225 {\rm kg}$
- Total weight on right wheels = $0.275 {\rm kg}$
- Left-right balance = $b=0.275/0.5 = 0.55$
- Total weight on front wheels = $0.300 {\rm kg}$
- Total weight on rear wheels = $0.200 {\rm kg}$
- Front-back balance = $a=0.200/0.5 = 0.4$
- Total weight = $0.500 {\rm kg}$
A few thoughts to help you on your way.
When an elevator is moving, you have to do work against gravity. You are changing the potential energy of the system. The faster the elevator moves, the more work per unit time is needed (because power = work / time = force * velocity ). If you are changing the velocity of an object, you are changing its kinetic energy: if it's slowing down, it gives energy back to you; if it's speeding up, you need to give it more energy.
If the elevator car is not moving, no WORK needs to be done. You still need a FORCE - but you could tie a knot in the cable and turn off the power: the elevator car will stay in place, without electricity, without heat being generated...
The helicopter example is different. The only reason a helicopter can hover is because it is pushing air down. Every second that it hovers, it needs to move a volume of air at a certain speed. In this case, the helpful equation is
$$F\Delta t = \Delta p$$
The change in momentum of the air tells you the force that you can get. This can be done by moving a large volume of air a little bit, or moving a little bit of air by a lot. Both situations will give you the same momentum, but since energy goes as velocity squared, the larger blades will be more efficient (up to the point where the drag of the blades becomes an important factor).
To solve the problem you stated, you need to know the size of the blades of the helicopter. Making some really simplified assumptions (there is at least a factor 2 missing in this - but just to get the idea): if you have a helicopter blade that sweeps an area $A$ and pushes air of density $\rho$ down with velocity $v$ then the force is
$$F = m \cdot v = (A\rho v)\cdot v = A\rho v^2$$
and the power needed (kinetic energy of the air pushed down per second) is
$$E = \frac12 m v^2 = \frac12 (A \rho v) v^2 = \frac12 A \rho v^3$$
If we assume $A=3m^2$ (roughly 1 m radius), and $\rho=1 kg/m^3$, then for a force of $500 N$ we need a velocity
$$v = \sqrt{\frac{F}{A\rho}} = \sqrt{\frac{500}{3}} = 13 m/s$$
and the power required is
$$E = \frac12 A \rho v^3 = \frac12\cdot 3 \cdot 13^3 = 3.3 kW$$
This is a bit higher than the 2kW you have available. The solution would be to increase the size of the rotors - the larger the area, the lower the velocity of the air, and the better off you are.
As for the pulley and string, or fixed string - see the comments I made about the elevator. When nothing moves, no work is done. In the case of the helicopter, although the helicopter doesn't move, the wings (blades of the rotor) do - and so does the air that is being moved (and whose motion provides the force needed to keep the helicopter in the air).
I hope that clears up your understanding.
Best Answer
When the elevator of mass $m$ is pulled by the 'string' then after some time it will acquire some velocity, say $v$. At that point it will have a kinetic energy $K$:
$$K=\frac12 mv^2$$
Now you cut the power. Inertia demands that the elevator will continue to travel upwards for 'a bit'.
During this period, kinetic energy $K$ is converted to potential energy $U$, in such a way that:
$$\frac12 mv^2=mg\Delta h\tag{1}$$
where $\Delta h$ is the height it will travel before it comes to a standstill ($v=0$). After that, the elevator begins to free fall.
We can also find the deceleration $a$ during that first phase by taking the time derivative of $(1)$:
$$\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac12 mv^2\right)=\frac{\mathrm{d}}{\mathrm{d}t}\left(mg\Delta h\right)$$
which yields:
$$a=-g$$
(the sign is minus because $a$ is a deceleration)