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I have a simple static mechanical system, but I reach a conclusion that seems to me counter-intuitive:

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There is a pulley fixed to the ceiling and there is a weight fixed to a rope which goes through the pulley and is fixed to a point on the floor. I denote by $T$ the tension in the rope and draw the forces applied to the weight (since the weight is in equilibrium, I should have $T=mg$), and to the pulley. Since the pulley is in equilibrium, the force exerted on it by the ceiling should be $2T$, and thus I arrive at the conclusion that in this setting, the force exerted on the ceiling (by Newton's 3rd law) is equal to $2mg$. Is my reasoning correct?

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The rope on the left pulls the weight and the pulley towards each other. The rope on the right pulls the floor and the pulley towards each other. Each of these pull downwards on the pulley with force $T$. The total downward force on the pulley is $2T$.

The pulley doesn't move because the ceiling pulls upward on it hard enough to keep it motionless. The pulley can only be motionless if the total force on the pulley is $0$. So the upward force from the ceiling is $2T$.

Perhaps the tension in the rope on the right is counter intuitive. The weight is motionless, so the total force on it is $0$. The two forces are gravity and the tension of the rope on the left. Both of these have magnitude $T$.

If the rope on the left is pulling upward on the weight, something must pull downward with force $T$ on the rope on the right. That is the floor, which pulls just hard enough to keep the rope from moving.

Another way you could arrange for a force $T$ to pull downward on the rope on the left is to hang another weight on the rope on the right. This would create the same tension in the rope on the right as the floor does.

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    $\begingroup$ "Another way you could arrange for a force T to pull downward on the rope on the left is to hang another weight on the rope on the right" - after this, it is not counter-intuitive to me anymore, thank you $\endgroup$
    – John Donn
    Commented Apr 7, 2014 at 16:06
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If there is no air drag, the pulley and the rope are frictionless and massless, the rope is not slacking anywhere, the rope is unstretchable and the rope is attached at exactly the centre of mass of the object so that no torque is produced, yes.

How is it counter-intuitive? It may seem like that it would require an exact $mg$ force on the top too but don't forget that one end of the rope is attached to the ground. So that extra $mg$ comes into the picture because there is an extra tension component on the other rope which means that the rope on the top with tension has to balance that by having tension $2T$.

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