I recently had a statics exam and was on the net looking up more information on pulleys and such so I could beef up my knowledge, and anyhow, I came across this question that someone else asked on another site that’s been bugging me now: http://answerboard.cramster.com/physics-topic-5-48087-0.aspx
When you add the forces up each mass vectorially (is that a word?), it does make sense that the tension measured by the device would be 0… However I know the tension in the cable must not equal zero. Could the device measure 0 and there still be tension in the device since it doesn’t have a static point from which to measure?
The vector sum of forces on a non-accelerating object always equal zero, but that doesn’t tell you anything about the tension at that point. Obviously, as you seem to have figured out intuitively, there has to be some tension there; if you cut the rope the weights would fall.
With frictionless pulleys, the tension in the rope will be the same along all points. If you look at just one of the vertical portions, the tension has to equal the weight of the hanging mass. Higher tension would cause the mass to accelerate upwards and lesser tension would allow it to fall. The tension measuring device would register a tension of mg.
Make a free body diagram of one of the weights. There must be a force in the rope equal to the weight of the 15 kg mass. Then make a free body diagram of a pulley. The rope pulls down by the weight of the 15kg mass, the pulley must push up on the rope by an equal amount. There is no horizontal force and thus no tension between the pulleys.
Now I’ve talked myself into the tension in the rope being equal to the weight of each mass. If one were to slide one of the pulleys toward the other, the problem doesn’t change. Keep sliding until they overlap and you have one pulley with a weight on each end. Tension in pulley = weight of 15kg.
If the weight on the left isn’t moving, the force upwards on it must be 15kg times* g*, or 147 N. So the tension on the rope just above it is 147N.
With frictionless pulleys, the tension on the rope is the same everywhere.
Therefore the tension at the center is 147 N
And when we follow the rope around to the right side, we find that the tension there should also be 147N to keep things from moving. That’s great, it means everything works out.
Remember, a rope can have tension, but not be moving.
There is most certaintly tension in the middle, as Quercus noted. Otherwise, the rope wouldn’t need to be there to support the weights. If you cut it, the weights would fall, as Baracus noted. Sure, the vector sum of forces where the measurement device is zero, with a 147N force from the left rope, ditto for the right rope. But just because that is zero, doesn’t mean the tension in the rope is zero.
The dude that answered the question on that site is totally wrong.
Ahh, that’s what I was thinking… But I still don’t completely get why the tension is 147 N instead of 2*147 N. Would you just consider one side of the diagram to be static and measure the tension due to one of the weights?
Think of it this way. Instead of two weights, just look at one with the rope attached to a wall after it goes horizontal. The tension in the rope would be 147N. Then imagine a second rope attached to the other side of the wall that goes over a pulley with a weight attached to it. The tension in that rope would also be 147N.
The net force on the wall is zero - not 2*147.
Now you can see that you don’t need the wall and you can just attach the two ropes together, the tension is still 147N
The question has already been correctly addressed by Quercus, but I think there might still be a little life in this horse yet. When drawing your free body diagram you have to account for the reaction forces on the pulley from its attachment to ground. So if you have a problem as stated in the O.P.'s link (and assuming the pulleys to be frictionless and the rope to be massless) the tension in the rope is 147N everywhere. The reaction components on the pulley are 147N down and 147N horizontal (which you can see by cutting both ends of the rope just away from the pulley and substituting in the rope tension) for a combined force vector of ~208N at 45 degrees.
This gets really interesting when you start looking at block and tackle machines; it looks like the “ground” is doing most of the work for you, because you can pick up a load that weights W with only W/n (n being the number of pulleys) for a simple pulley system, and even higher multiples for a more complex multi-spool pulley system. The tradeoff, however, is that you have to pull more rope length (in inverse proportion to your load reduction factor) to raise it by a given height, thus doing the same amount of work.
