Force

Agnishom · 2011-05-20 21:41:44

1.In a tag of war between the teams A and B, The rope breaks at a point which is nearer to A. Who has applied more force? Why?
2. A stone is tied to the middle of a string. and suspended from one end. Here S is the stone and O is the point of suspension. If I give a jerk at P, the string will break below the stone. Why?
3.A stone is tied to the middle of a string. and suspended from one end. Here S is the stone and O is the point of suspension. If I gradually Increase the pull at the free end the string will break above the stone. Why?
4. Why does my book say that problems 2 and 3 are based on inertia?

Agnishom · 2011-05-20 22:05:29

I have another question:
A hammer weighing 3 Kg, moving with a velocity of 10 m/s, strikes against the head of a spike and drives it into a block of wood. if the hammer comes to rest in 0.025 s, the impulse would be -30 N s
Why

mathlover002 · 2011-05-21 00:02:58

hi, i think for your second question, it is just 30 Ns because you are looking at the change in momentum from before to when it comes to rest, i.e.

momentum before = mu = (3*10) = 30 Ns
momentum after = mv = (3*0) = 0 Ns

the difference is -30 Ns...

i am guessing your question is asking about the resistive forces the hammer encounters before coming to rest, then remember that

I = Ft (Impulse = Force*time)

so

-30 = F*0.025

re-arrange to find F, we get that the hammer encounters resistive forces of 1200N (the value you get is negative because it acts in the opposite direction)

i will try and look at your first question in a bit. forgive me if my answers are wrong

Bob · 2011-05-21 01:00:57

hi Agnishom,

and mathslover002

Nothing to forgive; that answer to the second question looks good to me.

EDIT: First question. I've thought about this for a long time. The rope breaks => it has reached its elastic limit => it is stretchable.

I am not at all confident about the following analysis but here goes:

If I assume that the work done in stretching the part of the rope from team A to the break point is all done by team A and the work done in stretching the longer section from the break to team B is all done by team B, then team B does the greater work.

But, I can think of no reason to allow this assumption; why should the work be divided in this way. Furthermore, the tension, extension, natural length and modulaus of elasticity are connected by the formula

We know the natural length of A's section is shorter than B's and lambda should be the same through the whole rope but the other variables are unkown; except I keep coming back to the idea that the forces are equal along the rope because that's how the tug of war works. If one team pulls harder they drag the other team along.

Hypothetical experiment:

Get two accurate strain gauges and point them towards each other. connect them by a stretchable string. Tie the other end of each gauge to two separate winches by stronger string.

Gradually wind up the winches.

I am confident that the stretching will cause equalisation of the strain as measured by the two gauges. If this wasn't the case the string would move until it was. At that moment the experiment would show the two 'teams' as represented by the strain gauges to be pulling equally. That's what a tug of war measures. Wind one winch some more and the string just stretches until both gauges show the same again.

So I think the answer is: it's irrelevent where the rope breaks; if the two teams were stationary at that moment, they were applying equal force.

But I may be wrong.

FURTHER EDIT.

The more I think about this, the more I'm coming to the conclusion it is a 'trick' question. If the teams are not moving when the rope breaks, then, by Newton's first law, the forces are balanced => both teams exert equal force.

In (2) where is P?

Inertia means 'mass', so try drawing a force diagram.

Bob

Last edited by Bob (2011-05-21 06:25:34)

Agnishom · 2011-05-21 14:19:35

P is a free point which is the free end point of the string

Bob · 2011-05-21 20:14:04

hi Agnishom,

Number 3.

Diagram below.

I have shown the forces acting at S and again at the bottom of the string.

TU is the tension in the upper part of the string and TL the lower tension.
The weight of the stone is Mg
F is the downwards force.

At S

TU = Mg + TL

At the bottom

TL = F

TU > TL so the string will break above the stone.

Number 2.

I'm still unsure about the diagram for this.

Maybe you can do it yourself, based on my answer to 3
or post a diagram
or, at least, see, from what I have done, what I need to know in order to do 2.

Bob

Last edited by Bob (2011-05-21 20:30:41)

Agnishom · 2011-05-22 20:43:37

The diagram is same for no 2 and 3 but the 2nd question is also very weird for me

Bob · 2011-05-23 10:17:16

hi Agnoshom,

Oh good. Not just me then?

I think that's as far as I can go with these. Sorry I cannot provide a fully satisfying set of answers.

Bob

Math Is Fun Forum

#1 2011-05-20 21:41:44

Force

#2 2011-05-20 22:05:29

Re: Force

#3 2011-05-21 00:02:58

Re: Force

#4 2011-05-21 01:00:57

Re: Force

#5 2011-05-21 14:19:35

Re: Force

#6 2011-05-21 20:14:04

Re: Force

#7 2011-05-22 20:43:37

Re: Force

#8 2011-05-23 10:17:16

Re: Force

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