# A simple tension question

1. Aug 2, 2009

### Beamy

1. The problem statement, all variables and given/known data

Three equal weights are suspended from the midpoint of three spans of a near-horizontal negligible-mass wire supported at frictionless pulleys, and are counterbalanced by an adjustable spring set to exert an initial nominal force of 3W, this spring force being increased as the weights are 'pulled up'. The wire is fixed at point x. See attached gif.

I'm trying to determine the tensions t1, t2, t3, t4, t5 and t6.

2. Relevant equations

Taking the left-hand span, as a1 approaches zero, the vertical components (t1cosa1, t2cosa1) of t1 and t2 will approach t1 and t2 respectively, with t1 being equal to t2, since the attachment point of the weight to the wire does not move.

Similarly, in the other spans, t3 = t4, and t5 = t6.

For the vertical components (t1sina1, t2sina1) of t1 and t2, the sum of these will equal W:

t1sina1 + t2sina1 = W

Since t1 = t2, this vertical can be expressed as:

t2 = W/2sina1

(Similarly, in the other spans, t4 = W/2sina2, and t6 = W/2sina3.)

3. The attempt at a solution

As the angles approach zero, so does sina, and the vertical expressions indicate the tension needs to be increasingly large. If the tensions are considered bidirectional, then I think the sum t1 + t2 + t3 + t4 + t5 + t6 will be equal to the force needed to be exerted by the spring, but I don't have any relationship between the angles a1, a2 and a3, and therefore I can't prove that t1 = t2 = t3 = t4 = t5 = t6, which intuitively would seem to be the case in the near-zero angle case.

#### Attached Files:

• ###### 3-weights1.gif
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Last edited: Aug 2, 2009
2. Aug 2, 2009

### Lok

Try to ignore all weights and think about tension in wires. The distances are equal and the pulleys are frictionless so you are in a special case.

The weights seem to be fixed by pulleys as their distance from the middle does not change as they are pulled up.

Last edited: Aug 2, 2009
3. Aug 3, 2009

### kuruman

Can you calculate the net horizontal and vertical force at the two little circular thingies (pulleys?) that support the rope between the hanging masses? You should be able to get a relation between angles this way.

4. Aug 4, 2009

### Beamy

Responding to Lok and Kuruman's points:

If I ignore the weights, then the tension in the wire is the force being exerted by the coil spring.

For the vertical forces acting on the pulleys and the rest of the system, I've marked the reaction loads. The sum of these loads is 3W.

For the horizontal forces at the pulleys, in reality, the pulleys are not frictionless, so I've marked the friction loads on the attached new diagram, but they will apply only in the dynamic case when the weights are being lowered or raised. I don't see that gets me any closer to a relationship between the angles, if indeed that relationship could tell me anything about the tension(s) along the wire.

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5. Aug 5, 2009

### Lok

Not frictionless... this system is not stable, and even in this case one value still remanis the same as always. The total tension that is supported by the spring. t1=t2 because you said so ... But t2=t3 because tension gets transmitted from W1 to W2 through the wire, frictionless pulley or not.