# 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?

• You should think of the fact, that the kid has to pull the double length of rope than someone standing on ground had to pull. – Georg Nov 6 '11 at 14:10
• So since someone else would have to pull $6/5 \cdot mg$, the kid would have to pull half of that, being $3/5 \cdot mg$? Then the force on either side of the rope is equal, it that correct? – Martin Ueding Nov 6 '11 at 15:15
• Is there any way to include a crude sketch here. I am having trouble visualizing the problem. – ja72 Nov 6 '11 at 21:55
• A diagram would be very helpful. – Doresoom Nov 7 '11 at 18:07
• Added a sketch based on Doresoom's answer. – Martin Ueding Nov 8 '11 at 17:43

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.

• I went ahead and gave a full solution to this since you had already posted the correct answer yourself. – Doresoom Nov 7 '11 at 18:25
• Thanks for mentioning the tension, that clears it up a lot. I created a diagram with your forces in it. – Martin Ueding Nov 8 '11 at 17:44
• To the "no friction"-part: Yes, and the inertia (rotation) of the roll can be neglected as well. – Martin Ueding Nov 8 '11 at 17:44
• You still need to add F acting upward on the kid, since he's exerting F down on the other end of the rope. – Doresoom Nov 8 '11 at 19:15
• I thought that the tension on the rope on the right side would be the opposing force against the pull of the kid. – Martin Ueding Nov 9 '11 at 16:51

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

• We do have MathJax enabled on this site, so you can use Latex-like mark-up to produce equations. – Kyle Kanos Sep 18 '15 at 21:10