Pulley question + extras

I have this pulley question which I'm stuck on. I can prove the acceleration result, but after that I have no idea. Also, some equations I don't know how to solve are at the bottom - any ideas please.

"A planed is inclined at an angle arctan ¾ to the horizontal and a small, smooth, light pulley P is fixed to the top of the place. A string, APB, passes over the pulley. A particle of mass m1 is attached to the string at A and tests on the inclined place with AP parallel to a line of greatest slope in the place. A particle of mass m2, where m2 > m1, is attached to the string at B and hangs freely with BP vertical. The coefficient of friction between the particle at A and the plane is ½.

The system is released from rest with the string taut. Show that the acceleration of the particles is ((m2 – m1)g)/(m2 + m1).

At a time T after release, the string breaks. Given that the particle at A does not reach the pulley at any point in its motion, find an expression in terms of T for the time after release at which the particle at A reaches its maximum height. It is found that, regardless of when the string broke, this time is equal to the time taken by the particle at A to descend from its point of maximum height to the point at which it was released. Find the ratio m1 : m2."

Equations:
y = x^x
2^x = x^2


Cheers.

Quote:

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Originally Posted by Thomas154321

Also, some equations I don't know how to solve are at the bottom - any ideas please.

Equations:
y = x^x
2^x = x^2

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Are these suppose to be 2 simultaneous equations with 2 unknowns? If so, there must be a typo.

Regarding the mass on incline, what you have to do is break it into 2 separate problems:

(1) M2 falling and pulling M1 up the incline. This has already been solved and you know the acceleration. From this you can determine the velocity that M1 has at time = T.

(2) With the string broken, there is no force pulling M1 up the incline but it does have a velocity that you can determine from problem 1. Basically, the velocity of M1 at time = T from the 1st problem (i.e. when the string breaks) is the initial condition for problem #2.

Think of it this way. A block of mass M1 has an initial velocity V1 up an incline acted upon by gravity and friction. Find the maximum height that M1 reaches before stopping and sliding back down the incline. V1 comes from the first problem (and is a function of T).

Good luck. :)

Quote:

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Originally Posted by VBAhack

Are these suppose to be 2 simultaneous equations with 2 unknowns? If so, there must be a typo.

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Sorry, I want to differentiate them. I forgot to put that, and no, they are separate equations. Thanks for the pulley answer.

Differentiate. OK, that's important to know. For y = x^x suggest you take the log of both sides, then differentiate. The other one still doesn't look right (i.e. no y). :(

Incidentally, to make the pulley problem easier to think about, try considering the direction parallel to the slope and resolve your forces, velocities etc in this direction. At time T, the string breaks and m1 starts decelerating due to the frictional force and the component of m1g down the slope. This will sooner or later cause the velocity to reduce to zero and m1 will start to slide back down. So that is the maximum height.

To do this bit just involves using v=u+at and F=ma.

Then go from there to work out how long it takes to come back down again.


zaza

I have the time to reach the maximum height as t = 2(m2)^2T/(m2-m1). Is that correct?

How do you take logs of y = x^x??

log_x (y) = log_x (x) ??

Is it okay having a base to the logs that isn't a constant?

Quote:

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Originally Posted by Mattywoo2

How do you take logs of y = x^x??

log_x (y) = log_x (x) ??

Is it okay having a base to the logs that isn't a constant?

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I guess it's okay... though I've never seen it done! Taking bases for the above equation would yield:

ln(y) = ln(x^x)

giving:

ln(y) = xln(x)

(where all "ln"s are natural logarithms)

Ah yes, I suppose it would :p Easy to miss a trick when you over complicate things!!

so...

ln(y) = xln(x)

differentiating w.r.t. x

1/y . dy/dx = ln(x) + 1

dy/dx = x^x (ln(x) + 1)

I think you just did Thomas154321's homework. ;)

They weren't my homework, I was just interested in them. Homework was the pulley part. Which I'm still not done with. :rolleyes:

Quote:

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Originally Posted by Thomas154321

I have the time to reach the maximum height as t = 2(m2)^2T/(m2-m1). Is that correct?

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Hmmm, I got a different answer. What did you get for the velocity of M1 at time = T (i.e. when the string breaks)?

I also get a different answer.

I struggled a bit but believe I finally got it. A tricky part is realizing that the time durations are from the point at which M1 is initially released. In the end I treated like 3 separate problems:

1. M1 connected to M2 which is in free fall from start to when string breaks
2. String breaks until M1 reaches max height. The time being requested is the sum of T (time until string breaks) plus the time it takes to reach the max height after the string breaks
3. M1 goes from max height back to original starting point

One trick is to realize that the frictional force in #3 is in the opposite direction as in #2. Also, the distance traveled in #3 is the sum of the distance traveled in #1 and #2.

This is a tricky problem.