# What is the force on the fixed pully if the other rope's end is fixed to the wall?

A fixed pulley is glued on the ceiling. A rope goes over it. The rope is fixed on the floor on one end. On the other end there is a weight.

I understand that the force that pulls at the wall (red) is the same as the weight. Let's assume that the red end of the rope actually goes straight down to the floor. What is the weight that pulls on the ceiling? (How strong does my glue has to be?)

If the setting was like

I understand that I would need a weight at the red end as heavy as the original weight to be in balance and that then glue would need to carry both weights.

However, in my case there is no second weight - the rope is fixed to the floor - which should not add weight, but should carry some. So I am confused how much weight/force my ceiling/glue has to deal with.

If the force pulling at the glue of the ceiling was A + B then I don't understand why the force suddenly doubles without any extra weight. The setup in picture 1 would be more stressful to the ceiling then if I glued the rope directly to the ceiling as in

That would seem rather unintuitive to me.

So, to rephrase my question: How is it that there is twice as much stress on the ceiling in picture 1 compare to picture 3?

Conceptually you can "replace" the force on the left side with a weight of 100 N, then the force downwards on the ceiling is 100 N plus 100 N $\times\cos \theta$, plus the weight of the pulley, where $\theta$ is the angle of the rope with the vertical.

The reason why the downward force on the ceiling is bigger than the weight of the object is that an extra 100N is needed to stop the pulley wheel from rotating.

Imagine there were two differently sized pulley wheels which rotate about the same axle. First of all we glue the string from the 100N weight on to the smaller pulley wheel, then we glue the string from a much smaller weight on to the pulley wheel with a much larger diameter. If the weights and diameters are in inverse proportion, we stop it from rotating, and we get a downward force on the ceiling which is smaller than 200N.

For example, if the large pulley wheel is twice as big as the first one, then to get equal torque you would only need 50N of force to stop the rotation, and then you would only have 150N of force acting on the ceiling due to the pulley.

Conversely the 100N could be attached to the larger wheel, then we would need a 200N weight on the other pulley, and we would get 300N total force downwards. That is to balance the torques on the pulley wheels.

And if the pulley bearing was seized, then you wouldn't need any extra weight at all. The extra force is to stop the rotation of the pulley wheel.

• How is it possible that the ceiling has to carry twice as much in this setting than if I would just glue the rope at the ceiling - that makes no sense to be intuitively. From what magical place is the twice as strong force coming from? I am confused. – Make42 Oct 1 '16 at 22:16
• If you glue the rope to the ceiling then the ceiling has a sideways force of 100 N plus the downward force of 100 N. The extra force in your example is the 100 N force downwards represented by the red arrow. You can always do the experiment and see. – Suzu Hirose Oct 1 '16 at 22:24
• What I meant was that I could get rid of the whole thing and just keep the rope and the weight, having the weight on the one end of the rope and the other end of the rope glued to the ceiling. Then the force on the ceiling's glue would be 100N downward. I think you misunderstood me? – Make42 Oct 1 '16 at 22:28
• @Make42, there was an error in my original answer which assumed the force was straight downwards. – Suzu Hirose Oct 1 '16 at 22:28
• The angle is such that it is nearly downwards. I just explained my previous reply with the direct glueing though... – Make42 Oct 1 '16 at 22:32

Consider everything to be stationary, so we can use Newton's second law:

$$\Sigma F_y=0$$ Or: $$F_3-F_1-F_2\cos\theta=0$$ And because $F_1=F_2$: $$\implies F_3=F_1(1+\cos\theta)$$ For $\theta \approx 0$ then $\cos\theta \approx 1$.

So:

$$F_3\approx 200\:\mathrm{N}$$

So, to rephrase my question: How is it that there is twice as much stress on the ceiling in picture 1 compare to picture 3?

This is simply not true. For one, you're comparing apples and oranges: in the first case you have a weight and a counterbalancing force acting on the pulley. In the third case you only have one weight acting on it.

The mistake you're making is that you should count all downward forces ($F_y$), not just the weights. Counting all $F_y$ always gives you the right answer re. the glue strength needed.

Method $1$ is definitively an ineffective way of suspending a weight from a ceiling because of the counterbalancing force $F_2$ needed to keep everything stationary.

• Sure, the formula is clear, but I don't get (intuitively) how picture 1 is supposed to put twice as much force on the ceiling than picture 3 (see my question). – Make42 Oct 1 '16 at 23:32
• Intuition is useless here. Just accept the math: it doesn't lie. – Gert Oct 1 '16 at 23:33
• You could do the experiment with a spring balance as well to check it out. – Suzu Hirose Oct 1 '16 at 23:38
• I don't agree that intuition is useless here; I've added an addendum to my answer to point out what the "extra" force is and how it could alter under different circumstances. – Suzu Hirose Oct 1 '16 at 23:47
• I've addressed your edit. – Gert Oct 1 '16 at 23:50