How large is the force that tightens the wire

[Daltohn]

Homework Statement

How large is the force that tightens the wire A (see picture), when the weight's mass is 670 kg? Guidance: Assume that the wires between the wheels are horizontal.

Homework Equations

G=mg?

The Attempt at a Solution

What seems weird to me is that none of the wheels look movable, wouldn't the tension force simply equal G in that case? I guess the weight is supported by three wires but the correct answer is 20 kN which seems very inexact if the answer is indeed G/3. Would appreciate a more thorough explanation :)

 

[Filip Larsen]

You may want to use the concept of virtual work (force times a small virtual displacement) and conservation of energy (work performed by virtual displacement of the weight equals work transmitted via A). You should arrive at the conclusion that the force is not mg/3 as you imply in your post but something else.

 

[NascentOxygen]

How large is the force that tightens the wire A (see picture), when the weight's mass is 670 kg? Guidance: Assume that the wires between the wheels are horizontal.

Not my forte, but you can see that the steel weight sets the tension in the rope. This end of arm A is acted on by 3 equal forces, each the tension in the rope.

You can sometimes see this sort of arrangement being used to maintain tension in high voltage power lines.

 

[Daltohn]

Not my forte, but you can see that the steel weight sets the tension in the rope. This end of arm A is acted on by 3 equal forces, each the tension in the rope.

You can sometimes see this sort of arrangement being used to maintain tension in high voltage power lines.

This scenario is about an electric train so you're exactly right :) The a) part of the problem asks why this sort of arrangement would be appropriate.

So is mg/3 the answer?

 

[Filip Larsen]

So is mg/3 the answer?

No, that is not the correct answer.

If you do not like to figure out how displacement relates to displacements at A (like I suggested before) you may also in this simple example look at the tension force in the string directly and make a free body diagram around the pulley at A so that the "cut" goes through the 3 strings in the middle. Since each time you "cut" a string in this system you must replace it with a force equal to the string tension you can directly count how many times more you need in order for a force at A to balance the forces from the string (this is a slightly different worded version of that NascentOxygen suggested).

 

[NascentOxygen]

So is mg/3 the answer?

What value do you calculate to be the tension in the rope?

 

[Daltohn]

I realize I missunserstood what the actual wire was, I understand now. The tension in the string is 670*9,81 and it affects wire A three times essentially. So 20kN. Thanks and sorry for the ignorance. :)

 

[NascentOxygen]

No problem. https://imageshack.com/a/img848/9251/cool0038.gif

 

[Daltohn]

But how does this work? How does one really draw the tension because if it's drawn like this, the tension in the lower part of the wire wouldn't be pulling on wire A? Tension is confusing me, I guess it tightens in both directions.

 

[NascentOxygen]

Under static equilibrium, the arrows all point to the right. Nothing moves, nothing turns.

 

[Daltohn]

How can the last (lower) arrow point to the right? Wouldn't that force have the same direction as the force of the weight?

 

[NascentOxygen]

As far as the pulley A is concerned, the load it supports pulls away on both ends of the rope looped over it. If the forces weren't balanced in opposition like that, the rope would unravel from the pulley.

Besides, you can't push rope.

Perhaps you should put together a couple of pulleys and string and observe how they work.