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zetafunc.

Guest

M4 (Resolving Forces)

A particle P of mass m is attached to the mid-point of an elastic string, of natural length 2L and modulus of elasticity 2mk²L, where k is a positive constant. The ends of the string are attached to points A and B on a smooth horizontal surface, where AB = 3L. The particle is released from rest at the point C, where AC = 2L and ACB is a straight line. During the subsequent motion, P experiences air resistance of magnitude 2mkv, where v is the speed of P. At time t, AP = 1.5L + x.

Show that

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I'm getting the same answer but with a negative sign in front of the 4k²x. In my mind, I see this as someone pulling the particle towards B and plucking it, so it moves towards A, but which way are the forces pointing? I have:

-T[sub]1[/sub] (tension in AP) pointing towards A
-T[sub]2[/sub] (tension in PB) pointing towards B
-If the particle is moving towards A, then the air resistance acts in the direction AB
-Since this is SHM, the acceleration opposes the direction of displacement, so acceleration acts towards B

Yet, when you resolve using the above, you don't get the right signs. Can anyone help? Thanks.

zetafunc.

Guest

Re: M4 (Resolving Forces)

"A particle P of mass 4 kg moves along a horizontal straight line under the action of a
force directed towards a fixed point O on the line. At time t seconds, P is x metres from
O and the force towards O has magnitude 9x newtons. The particle P is also subject to air
resistance, which has magnitude 12v newtons when P is moving with speed v m s–1."

They've also done the same thing for this question! The force towards O and the air resistance are resolved as being in the same direction. Why?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: M4 (Resolving Forces)

hi zetafunc,

post #1 question

Yes, I see why this causes sign problems.

Firstly, does the string ever go slack, because there'll be no tension if that happens.

At point C, CB = the natural length of that half of the string so it is 'just not slack'.

T1 (tension due to AP) > T2 (tension due to PB) so motion is such as to stretch the right hand string just after release.

Without the air resistance the mass would oscillate symmetrically and return to C, so the string never goes slack.

That means we can use T = (ext) x (mod)/(nat length) without any worries.

I too will take T1 as acting towards A and T2 towards B.  That will not change during different amounts of stretching.

But the velocity and resistance will, so you need to be consistent about the signs for these.  That's where the problem lies.

I've got A on the left and B on the right so I'm going to take A --> B as my positive direction.

So I'll assume the acceleration is +ve when acting from left to right, similarly the velocity.

As the air resistance opposes the motion at all times I'll take it as an accelerating force of -2mkv acting towards the right.

so

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: M4 (Resolving Forces)

post #2 question.

I'm going to draw O in the middle and P to the right.  And I shall take O --> P as the positive direction.

So acceleration and velocity are positive in the direction OP.

The force is towards O, and so is negative.

The air resistance opposes motion and so is negative when v is positive.

So

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

zetafunc.

Guest

Re: M4 (Resolving Forces)

post #2 question.

I'm going to draw O in the middle and P to the right.  And I shall take O --> P as the positive direction.

So acceleration and velocity are positive in the direction OP.

The force is towards O, and so is negative.

The air resistance opposes motion and so is negative when v is positive.

So

Bob

Thanks for the replies. How do you know that the velocity and acceleration are positive in the direction OP? I thought that if the force (9x) acts towards O, then the acceleration is also towards O...

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: M4 (Resolving Forces)

the force (9x) acts towards O, then the acceleration is also towards O...

I'd say it acts away from O with value -9x.

Distance, velocity and acceleration will all have both positive and negative values during the motion.

So don't make any assumptions about them ... just declare that they will all be treated as positive in a certain direction.

If you are then asked to find, say, the acceleration, at a certain point and your calculations come out with a negative, then you know that the actual acceleration there is the other way.

Consider this:

the general quadratic is ax^2 + bx + c = 0

In a particular situation you might find b = -4

But you wouldn't start with ax^2 - bx + c = 0 would you?

(but if you did, then you would end up with -b = 4  ... in which case you'd swop it around to b = -4, and we are back to where we started.

Just be consistent about which direction you will define to be +, and remember that air resistence must act against the motion so will therefore be negative.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

zetafunc.

Guest

Re: M4 (Resolving Forces)

But how do you know if the particle is moving away from O or towards O? If it's towards O then 9x and air resistance have the same sign, but if it's moving away from O they don't. How do you know which is going on?

zetafunc.

Guest

Re: M4 (Resolving Forces)

Also, just to clarify, for the first problem, since AB is the positive direction and the particle is moving towards A, the resistive force is -2mkv because v is negative, right?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: M4 (Resolving Forces)

under the action of a
force directed towards a fixed point O on the line.

It doesn't say which way it is moving, just which way the force acts.  But you know about SHM, so you know that when it travels towards O it speeds up and overshoots, and then it is travelling away.  And when it travels away it gets slower, stops and reverses.

But how do you know if the particle is moving away from O or towards O?

You don't because it does both at some stage during the motion.

That's why you declare a positive direction and build the equations as if x, dx/dt, dv/dt are +ve in that direction.  If you choose the opposite you'll still get the same analysis, as long as you are consistent.

Let's look at a simpler case.

A particle moves along a straight line under the action of a centrally acting force, proportional to its distance, x, from a point O.  So I'd have

I've taken dv/dx as positive away from O, so the force acts negatively.

But you could have dv/dt acting towards O in which case the distance becomes negative and we get

Bob

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

zetafunc.

Guest

Re: M4 (Resolving Forces)

Okay, that makes sense, I think. So the key here is just that the particle has SHM. Is that implied by the force acting towards O? What is the key bit of info that tells us it moves with SHM?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: M4 (Resolving Forces)

hi zetafunc,

On the first question there's no direct clue, but as soon as you read elastic string, you should be thinking SHM.  Before SHM is introduced, M papers usually say inelastic.

On the second one it says

A particle P of mass 4 kg moves along a horizontal straight line under the action of a force directed towards a fixed point O on the line

The M4 syllabus says

Damped and/or forced harmonic motion. 
The damping to be proportional to the speed.
Solution of the relevant differential equations will be expected.

So you should be expecting this.  I'd assumed you have the syllabus but in case not here it is:

http://www.edexcel.com/migrationdocumen … 180510.pdf

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

zetafunc.

Guest

Re: M4 (Resolving Forces)

Oh, okay, thanks. So the key is that it's elastic...

16 days to go... uh oh.