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Two mechanics problems that are annoying me!

Got these two questions from my Cambridge Engineering book thing I need to do before I go there:

1)

Two identical beads A and E, each of mass m, are threaded on a rough straight horizontal wire. The beads are joined by a light inextensible string of length 4a. Particles B,C,D each of mass 2m, are attached to the spring so that AB=BC=CD=DE=a. The system hangs freely under gravity in a vertical plane. The coefficient of friction between each bead and the wire is

{\frac{1}{4}}

. Prove that, for the system in equilibrium in a symmetrical shape, the greatest possible value of the distance AE is:

Unparseable latex formula:

2a({\frac{1}{\sqrt{10}}} + \frac{1}{\sqrt{2}}})

I have no idea how to even start this one, there are hints in the back which talk about limiting equilibrium etc but I still have no clue, plus the answer looks to be more than a few basic sums.

2)

The greatest instantaneous acceleration a person can survive is 25g, where g is the acceleration of free fall. A climber's rope should be selected such that, if the climber falls when the rope is attached to a fixed point on a vertical rock, the fall will be survived.

A climber of mass m is attached to a rope which is attached firmly to a rock face at B. When at a point A, a distance L above B, the climber falls.

a) Assuming the rope obeys Hooke's Law up to breaking, use the principle of conservation of energy and the condition for greatest instantaneous acceleration to show the the part AB of the rope (of unstretched length L) must be able to stretch by more than

{\frac{L}{6}}

without breaking for the climber to survive.

I started this but always get it slightly wrong and don't see where.

I said that for the climber to survive:

{\frac{1}{2}}Fx > 2Lmg + xmg

since I figured the max distance the climber could fall before the rope became taut would be 2L (ie L below B) and then it would extend, thus losing more GPE. However, after I do all the sums and use F=ma and so substitute in F=25mg I get x must be greater than or equal to 4/23 which is just short of 1/6. I must have gone wrong somewhere.

b) A particular rope has a breaking strength of 25 times the weight of the climber, and obeys Hooke's law until breaking, when it has stretched by 20% of its length. Determine wheter this rope is suitable.

I haven't been able to do this because I keep ending up back where I got to for part a).

I know that those two questions are pretty long so I appreciate any help I receive.

Reply 1

v-zero

14

For 2.) a) I also get x is greater than 4L/23

This should be easy and I don't understand where either of us went wrong... I'll give it all some thought.

Reply 2

steve10

7

For Q1, Here is a diagram.= - can you follow it?

The system is in limiting equilibrium, which means that the friction force at the two end beads is the maximum (limiting) friction possible, and F = mu.R, where mu=1/4. There is a maximum value for AE if the system is to remain in equilibrium. If AE was too large, then the horizontal components of T1, acting on the two 1mg beads, would exceed the limiting friction force and the two beads would move together. If AE was less than the maximum value, then the horizontal components of T1 would be less than the maximum (limiting) friction force.

The problem is just basically a bit of trig and balance of forces at various nodes, e.g. C,D,E.
Also, I got T2=mg.sqrt(2), and T1 = mg.sqrt(10)

Edit: correction to value of T1.

v-zero

For 2.) a) I also get x is greater than 4L/23...

Well, that's the right answer, isn't it?

the rope (of unstretched length L) must be able to stretch by more than L/6, and 4/23 is greater than 1/6.

Reply 4

steve10

7

For Q2 b),
According to the information given, a rope must be able to stretch by more than 1/6, to avoid giving an accln greater than 25g.

1/6 = 16.6667%.
If a particular rope can then stretch by 20%, which is greater than 16.6667%, then this rope is suitable.

Reply 5

v-zero

14

steve10

Well, that's the right answer, isn't it?

the rope (of unstretched length L) must be able to stretch by more than L/6, and 4/23 is greater than 1/6.

I think it is implied in the question that you are supposed to show that L/6 should be part of the equation. If not then I think the question is worded badly. If they wanted 4L/23 surely they would have said it.... Anyway, bleh.

Reply 6

v-zero

14

steve10

For Q2 b),
According to the information given, a rope must be able to stretch by more than 1/6, to avoid giving an accln greater than 25g.

1/6 = 16.6667%.
If a particular rope can then stretch by 20%, which is greater than 16.6667%, then this rope is suitable.

This implies the natural length of the new rope is also L, which is not stated. These questions are quite vague at times it seems...

Reply 7

MarcdeV

OP

I think you are meant to get L/6 since in the answers it says extension is greater than or equal to L/6, then for the part b) it says something about how the extension for that cable is 18.7% so it doesnt snap

Reply 8

ukgea

6

For part a), I think you're supposed to take gravity into account: Because of gravity, even if the tension in the rope were 25mg, you would only be subject to a force of 24mg, so the tension in the rope can actually be 26mg. Plugging this into your equation yields the desired result of L/6.

Reply 9

v-zero

14

ukgea

For part a), I think you're supposed to take gravity into account: Because of gravity, even if the tension in the rope were 25mg, you would only be subject to a force of 24mg, so the tension in the rope can actually be 26mg. Plugging this into your equation yields the desired result of L/6.

Yes, I too came to this conclusion - not by energy calculation means as the question suggests though, but by calculus.

Reply 10

MarcdeV

OP

yep, that's the first part of that question solved, don't see where they get 18.7% for the second part though.

Also I still don't really get the first question (the one with the beads on the string). The diagram helps but I don't see how you can get the answer from there.

Reply 11

DFranklin

18

MarcdeV

Also I still don't really get the first question (the one with the beads on the string). The diagram helps but I don't see how you can get the answer from there.

It's just chasing around the various equations. First of all substitute the values of

\mu

and R into

T_1 \cos \beta = \mu R

, and then substitute through the other equations.

Spoiler

Probably worth noting that I don't agree with Steve10 about the relation between T1 and T2. But he's done all the hard work by doing the diagram and writing down the base equations.