Force on rope with accelerating mass on pulley I have a pretty basic pulley problem where I lack the right start.
A child sits on a seat which is held by a rope going to a cable roll (attached to a tree) and back into the kid's hands.
Sketch http://wstaw.org/m/2011/11/08/m7.png
When it sits still, I believe that the force on either side of the rope must be equal to keep is static, therefore each rope holds $\frac{1}{2}mg$, the cable roll has to carry the full $mg$.
Now, the kid wants go up with $\frac{1}{5}g$. For the whole system to accelerate up, the cable roll has to support another $\frac{1}{5}mg$ resulting in $\frac{6}{5}mg$ of force.
The question that I cannot answer is:

How much force does the kid need to apply onto the rope in its hands?

As I said before, $F_k$ (kid) and $F_s$ (seat) have to be $\frac{6}{5}mg$. So I get this:
$$F_k + F_s = \frac{6}{5}mg $$
In order to solve for either one, I would need another equation. The forces cannot be equal, otherwise there would be no movement of the rope. So I just invented the condition, that the difference of the forces has to be the acceleration:
$$F_k - F_s = \frac{1}{5}mg $$
I can solve this giving me $F_s = \frac{5}{10}mg$ and $F_k = \frac{7}{10}mg$ which will sum up to the total force.
But is this the right approach at all?
 A: The solution is easier seen with a free body diagram. You'll need 2 equations, so use two points on the rope: one attached to the seat, and one where the kid holds the rope.
For the first, you've got the weight of the kid mg pulling downward, the force F the kid is exerting on the rope upward, (since he's lightening the load on the seat by effectively distributing his weight elsewhere) and the tension T in the rope pulling upward equal to ma:
$$T + F -mg = ma $$
The second FBD gives you the tension in the rope pulling upward, and the force the kid is exerting pulling downward. Since it's a weightless point, the ma portion is zero:
$$ T - F = 0$$
Combining and simplifying:
$$ 2F - mg = \frac{1}{5}mg $$
$$ 2F = \frac{6}{5}mg $$
$$ F = \frac{3}{5}mg $$
Of course, this is assuming the rope has no weight per unit length, and there is no friction in the drum.
A: The kid pulls himself, so there are two tensions 2T to support his body:
         2T-mg=(1/5)mg

so
                T=(3/5)mg
The force the kid applied equals the normal force n exerted by the seat, therefore
             T+n-mg=ma  

             n=mg-T+ma=mg-(3/5)mg+(1/5)mg=(3/5)mg

