Is the force exerted on the rope 10 KN or 20?

[Caterell]

I have a (hopefully) quick question.
If I have a set-up like the one below, with a cable attached to a rope, and a cable puller pulling the rope through a pulley thereby pulling the cable - and I have a measuring device on the pulley saying that 20KN is being exerted on the pulley - is the force that is being exerted on the cable 20KN? Or is it only 10KN, because half of the 20KN measured force is on the rope on the other side of the pulley?

 

[rcgldr]

The pulley exerts an upwards force on the rope of 20KN. The horizontal force is not specified. The angle of the rope to the retractor isn't mentioned. The tension in the rope exerts a downwards component of force on both sides of the pulley, and also has a horizontal component of force on the left side. If you know the angle of the rope on the left you should be able to figure out all the forces.

 

[Caterell]

Thanks for that...I suppose my question was not specific enough.

What I was really trying to ask was: if the force measured at the measuring device was 20KN and given no additional loses occurred through the pully system then what was the approximate force exerted on the cable?

People are saying to me that the force was divided as a pulley system was used and the actual force on the cable was approx. 10kn. They tried to confuse me by saying that using force vector pricinples the force was shared between the cable being pulled and the pulling device.....what are your thoughts?

 

[rcgldr]

If the pulling device was located directly under the pulley, so the two sections of rope were vertical, then you'd have 10KN of tension in the rope, 10KN on the left side of the pulley, 10KN on the right side of the pulley, so the pulley would experience 20KN of downwards force.

Is the measuring device hinged so that it can measure the total force or does it only measure the vertical force? If the measuring device is hinged, then it simply opposes the vector sum of forces due to the tension in the two sections of rope on both sides of the pulley.

If the measuring device is hinged, and the pulling device is located to the side so that the left side of the rope is horizontal, then you have a vector of x KN pointed to the left, and another vector of x KN pointed straight down. The vector sum would be sqrt(2) x KN, pointed 45 degrees to the left of straight down. For a reading of 20KN, x would be 20/sqrt(1) ~= 14.142 KN.

 

[Caterell]

Oh... sorry! Misunderstanding: it's not vertical, it's horiztonal!
The picture was an overhead view...