[redsniper]

Alright, say you've got a man sitting in a chair which is hanging from a massless rope attached to a massless, frictionless pulley on the ceiling. The man is holding the other end of the rope so that he and the chair are just hanging there in the air. What would be the tension on the rope and what force is the chair exerting on the man? I thought at first that the tension might be the sum of the man's and chair's weights, then I thought it might be the difference, now I'm just confused. The man obviously weighs more than the chair and if they weren't in contact I could probably figure this out pretty easily, but since they're in contact I just can't quite wrap my brain around it. Help please. 

[Eishtmo]

The force of the chair on the man must be enough to hold the man up, and thus only requires his mass to figure out.  The tension on the rope, however, needs to include the mass of the chair and the mass of the man.  For the tension, you treat them as a single object.

[Herra Tohtori]

Let's consider the dilemma.

There is the rope. On other end there is a chair and on other end there is a person.

At first glance, we see that to acieve a status quo, the mass of the chair needs to be smaller than mass of the person. Otherwise the chair will lift the man from the chair, unless negative supportive force is applied.

So, m<M (m=chair mass, M=person mass).

There are three forces affecting the chair:

-Weight of the chair; G1=mg; points down (negative)
-Inverted normal force that affects between the chair and person's arse; -N; points down (negative)
-Tension of the rope; T; points upward (positive)

So, if we want the chair to be stationary, the sum of these forces needs be zero. Therefore,

G1+N=T

Similarly, there are three forces affecting the persin sitting in the chair:

-Weight of the person; G2=Mg; points down (negative)
-Normal force (or support force) that affects between the chair and person's arse; N; points upward (positive)
-Tension of the rope; T; points upward. (positive)

The sum of these also needs to be zero; therefore

G2= T+N


Now we get a nice little pair of equations in which we have two variables, therefore we can sove the dilemma.

Tension must be the same in each equation, same applies to normal supportive force. Therefore

{G1+N=T
{G2= T+N

G2 = G1+N+N = G1+2N

G2=G1+2N

2N=G2-G1

N=½(G2-G1)=½(M-m)g

There's the normal force which affects between the person's arse and the chair.

Tension is therefore

T=G1+N=G1+½G2-½G1 = ½G1+½G2 = ½(M+m)g

There.


Check if I missed something, but that would seem to be the answer.

Unsurprisingly, the tension is what it's supposed to be. Weigh of man and chair div. by two.

If one string keeps mass M up, the tension in string is Mg; if two strings keep the object up, the tension is MG/2, if three, MG/3 and soforth. In this case, the other forces do sum away and the chair-person system can be treated (carefully) as one body.

Ideally, this works when m<=M. When m=M, the normal and inverted normal forces are zero, and the tension is everything that keeps the system aloft, in which case the core of this dilemma becomes easier to reach.

The situation becomes more interesting if m>M... but, as you can guess, also in that case the tension is the same.

[Bobboau]

half his (+chair) weight.

...damn, that is a more complete answer.

[spartan_0214]

my solution:

There is no such thing as a massless rope, so no problem there.

If there was a massless rope: there's nothing holding the rope onto the pulley, so the man would be sitting in the chair on the ground...

 

[Herra Tohtori]

I find your lack of faith... disturbing. <Force Choke> There ya got massless...


Anyway, making things simpler than they are is a trade mark of physics excersizes. You will find out that even if you use a rope with mass, the tension is actually the equation I posted before - at the ends of the rope (or more accurately, the intersections of chair, rope and person's hand.

Going up the rope towards the pulley, more tension is applied to point that you evaluate - particularly the weight of the rope below the point you want to evaluate. And if you want to really have friction in the pulley, it doesn't matter in a static situation anyway.

[Davros]

keep in mind you have to use the following phrase in your answer
or you will fail
"There's the normal force which affects between the person's arse and the chair"