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# A Physics problem I can't explain properly

If a stick with dm/dl =const is hanging on three nonstretchable pieces of string (l1=l2=l3 and l is the distance between them) you can't find the tensions, because the problem is unsolvable if all the strings don't stretch. It is kind of obvious, because if a string happens to be 0,00000000001 mm longer than the others the tension will be zero. But how to prove this normally?

Sorry, I almost never use English for Physics conversation, so I don't know any proper words...

## Discussions

Best Answer 8 years ago

I think there are too many strings.

;-)

As a result I suspect this is one those problems with too much freedom.

It has not just one solution, but an infinite number of solutions.

Write the tensions in the strings as [T1, T2, T3]. The left one is T1. The middle one is T2. The right one is T3.

Try setting the tension T2 in the middle string to an arbitrary fraction of the weight of the stick. Then solve for T1 and T3. By that I mean choose a T2 in the range:

0 <= T2 <= W

Try T2=0. Solve for the other two strings. Get T1=W/2, T3=W/2

Similarly:

Try T2=0 ---> [T1,T2,T3] = [W/2, 0, W/2]

Try T2=W/3 ---> [T1,T2,T3] = [W/3,W/3, W/3]

Try T2=W/2 ---> [T1,T2,T3] = [W/4,W/2, W/4]

Try T2=W ---> [T1,T2,T3] = [0, W, 0]

You can do the same trick choosing T1, then solving for T2 and T3.

You can do the same trick choosing T3, then solving for T1 and T2.

Basically you get "freedom" to choose the tension in one string. Then the other two depend on what you picked for the one free string.

Answer 8 years ago

Yes, but you are only resolving vertically, and not taking moments.

Answer 8 years ago

Choose the center of the rod to be the center for all these moments.

Then the moments due to T2, upward on the center string, and W, the downward force due to the weight of the rod, these are both zero.

T2 x 0 = 0

W x 0 = 0

The other moments are (T1 x -L), and (T3 x L). All the moments have to sum to zero. The result is that T1 and T3 have to be equal. This is true of all the solutions I wrote in my previous answer. The moments balance, and the forces balance.

Try T2=0 ---> [T1,T2,T3] = [W/2, 0, W/2]

Try T2=W/3 ---> [T1,T2,T3] = [W/3,W/3, W/3]

Try T2=W/2 ---> [T1,T2,T3] = [W/4,W/2, W/4]

Try T2=W ---> [T1,T2,T3] = [0, W, 0]

8 years ago

In English, the definition of these problems, from distant memory would be.

"Three, light, inextensible strings of identical length support a rod, of uniform section, at positions l1,l2,l3. Find the tensions in each string"

The question, by saying identical MEANS ==, absolutely equal, yes, we all know you CAN'T in practice do that, but this is a mathematics exercise.

Don't sweat it.

Steve

8 years ago

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___ |___|___|____where the distance between the strings is the same, the force of tension in each of the strings will be the opposite of the force of gravity exerted on the stick.

I'm assuming that the length of the strings is the same.

So if the stick has a mass of 3(kg) and we round gravity off to 10(m/s^2), the force of gravity on the stick is:

3(kg)*10(m/s^2)=30(kg*m/s^2) = 30 newtons

Since the stick is not moving and the only force acting against gravity pulling the stick down is the force of tension in the strings pulling the stick up, the 30 N is shared out between all three strings, and the tension in each is then 10 N.