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#1 Re: Help Me ! » Mechanics - Friction » 2011-03-01 02:28:49

I was thinking a couple of things should be mentioned.
The solution is based on the 30N angled up force on P, being the least force that prevents P from sliding. If the coefficient of friction were appreciably larger, a smaller force than 30N would still not allow P to slide. From the wording of the problem, 30N is obviously the least. 30N is in "exact balance" with the frictional force, allowing the coefficient to be calculated. This consideration is somewhat trivial.
Also, as a general rule, static friction is greater than kinetic friction. In other words, the frictional force, when starting something sliding from a dead stop, is greater than the frictional force when it is sliding.

An example is automotive anti-lock brakes stopping a lot quicker than having all tires completely locked up and skidding. Both cars are moving until they stop, but with anti-lock brakes the tires aren't sliding on the pavement the entire time, therefore they are "static" relative to the pavement some of the time.

You might also experience this when trying to push something really heavy. That initial push can be a lot harder than the pushing needed to keep it going.

Again, from the wording of the problem, the same coefficient for the static and the kinetic condition is used. That's okay for solving a problem like this, but in the real world, that might not be the case.

#2 Re: Help Me ! » Mechanics - Friction » 2011-02-28 15:41:27

A Newton of force accelerates 1 kg at 1 m/sec^2. The acceleration of gravity is 9.81 m/sec^2. So the force of gravity is 9.81 Newtons on 1kg of mass. Simply looking it up will confirm this.

Define down as vertical + and from P toward the pulley as horizontal +.
Due to gravity pulling down on Q, it produces a horizontal force on P of +(3kg*9.81N/kg)=29.43N.   (1)

Due to gravity, P has a vertical force down on itself of +(2kg*9.81N/kg)=19.62N.   (2)

The 30N angled up force, produces
-(30N*cos(30°)) = -30N*0.866 = -25.98N horizontally   (3)
and -(30N*sin(30°))=-30N*0.5=-15N vertically.      (4)

So P has a horizontal force of
29.43N - 25.98N = 3.45N,   From (1)&(3) : (5)
and a vertical force of 19.62N - 15N = 4.62N. From (2)&(4) : (6)

The coefficient of friction is the horizontal force divided by the vertical force which is
3.45N/4.62N = 0.747 ≈ 0.75. From (5)&(6) : (7)
You can treat P's and Q's mass together and determine the acceleration of P and Q with the total forces acting on them. Then the force needed to accelerate just P plus the friction would be the tension in the string.

Once the 30N force is removed, the force down on P is just due to gravity acting on its mass. Using the coefficient of friction this produces a horizontal force of
-(19.62N*0.747)=-14.65N. From(2)&(7) : (8)
The sign is negative since it resists being dragged toward the pulley.

Due to gravity, Q is producing a force of 29.43N. From (1).

The total force on P and Q together is
29.43N - 14.65N = 14.78N. From (1)&(8) : (9)
F = ma so a = F/m = 14.78N/(3kg+2kg)
= (14.78kg-m/sec^2)/5kg = 2.956 m/sec^2. From(8) : (10)

Since P is 2kg and F = ma = 2kg * 2.956m/sec^2
= 5.912kg-m/sec^2 = 5.912N. From(10) : (11)

So with the friction on P and the force required to accelerate P, the tension in the string is
14.65N + 5.912N = 20.56N. From(8)&(11) : (12)
We can double check by calculating the force due to gravity on Q that is left over based on it's acceleration.

If there were no tension in the string, in other words, nothing pulling up on Q it would accelerate down at 9.81m/sec^2. It is accelerating down at only 2.956m/sec^2, which is obviously the same as for P and Q together.

The resistance to being accelerated down is
3kg * 2.956m/sec^2 = 8.868N. From (10) : (13)

The tension in the string is the force of gravity on Q minus Q's resistance to being accelerated, which is
29.43N - 8.868N = 20.56N. From(1)&(13) : (14).

The double check is good, since both (12) and (14) produced a result of 20.56N.

#3 Re: Help Me ! » Compound Interest calculation on savings... » 2011-02-28 13:34:19

I stand corrected, the compound interest link gives a good explanation of annualized/effective annual/APY. I missed it when I first scanned over it. It is called APR at that link.

So many different terms for the same thing, do those behind it make it intentionally confusing? wink

But those links really are excellent, good job!

#4 Re: Help Me ! » Compound Interest calculation on savings... » 2011-02-28 12:54:44


Hmmm ..... except it sounded like the monthly interest rate was 0.5%, not 6%, while £100 was indeed being added monthly.

That is also showing compounding monthly. Compounding was not specified, but the explanation made it sound like it was not, which would be unusual, but is possible. Compounding is much more common.

The links gave a good explanation of compounding, but there is also a question as to whether the 6% is the effective annual, also just called annualized, rate, sometimes abbreviated APY (Annual Percentage Yield). If it is the APY, then the monthly rate, if it is compounded monthly, would be slightly lower than 0.5%. A monthly interest rate of 0.5% would give an apparent yield of more than 6% when compounded monthly over a year.

Hope I'm not complicating things too much. Compounding is much more usual, and using the effective annual rate isn't necesssary for understanding the basics of interest rates, but most investments and saving accounts list an effective rate, although some will list both.

#5 Re: Puzzles and Games » Lots of maths puzzles » 2011-02-28 12:02:44

Thanks. I was discussing that very problem with someone else and I think we found a more concise, but perhaps more subtle, way to explain it than the way I explained it before.

#6 Re: Puzzles and Games » Lots of maths puzzles » 2011-02-28 08:39:09

"2-Hell VS Heaven (*)" could be hard, if a person just doesn't think exactly how to word the question. Once the right question is realized, then it seems too simple.

Correct? That was a neat puzzle. It might be as easy, or as hard, for a middle school student to answer it as for a much older adult to answer it. Either way, I still thought that it was a good puzzle.

#7 Re: Help Me ! » Help please » 2011-02-27 10:06:47

Strange problem. What does "4 mark" mean? I can't do it "without further calculation" either and I'm not sure how the sum of w^2 helps. Maybe the sum of w^2 is an intentional red herring.

The change in the mean is obvious without further calculation since taking out a heavier and putting in a lower weight will always reduce the mean. The mean is the sum divided by the quantity of 15. The quantity doesn't change and the sum would obviously go down by taking out a heavier and putting in a lighter weight.

The change in standard deviation isn't at all obvious to me without further calculations. The orginal mean is 1145.3/15=76.35, but I get (1145.3-79.2+63.5)/15=75.31 for the reduced mean.

Still, since 63.5 is considerably farther from the reduced mean of 75.31, than 79.2 was from the original mean of 76.35 ((75.31-63.5)>(79.2-76.35)), the standard deviation would go up. Standard deviation is basically a measure of how far spread out weights are from the mean. Standard deviation is the square root of the mean of the squares of the difference between the weights and the mean.

The conclusion is clearly the same as bobbym's.

#8 Re: Help Me ! » recurrence » 2011-02-24 05:58:30

Nice, and perhaps obviously, to solve for a and b,

if T(0) and T(1) are givens.

If T(0)=0 and T(1)=1, then a=0 and b=0. However, if T(0) and T(1) are the initial conditions but are not given, then

If the course isn't this advanced, it might still be necessary to solve four simultaneous linear equations to solve for c, d, e, and f for n:even and n:odd, ignoring the a and b terms. The equations will be the same for n:even and n:odd if T(0)=0 and T(1)=1. If T(0) and T(1) are givens, but the values are not specified, T(0) and T(1) become terms in the solutions for n:even and n:odd, respectively, and the values of f are not the same for n:even and n:odd.

#9 Re: Puzzles and Games » Containers of marbles » 2011-02-21 09:16:17

The solution given in my previous message appears to be the only solution to wintersolstice's puzzle in the first message in this thread.

However, there is a qualification. In the original puzzle it does not state whether any of the individuals can see the earlier container labels, including Sally, in addition to their own, except for Sally. It doesn't matter. Whether they can, or not, changes nothing factual in the puzzle and changes nothing in the solution.

In wintersolstice's puzzle, there can be another solution if Sally can see everybody else's labels, at least one more solution, maybe more. For those of us solving the puzzle, we don't need to know what the labels are, we only need to know that the puzzle as stated is factual, and that Sally can see everybody else's labels.

As a funny trivial matter, the statement of the original puzzle is NOT factual,  or it is not complete if factual. If Harry has all the info that Sally has, then why doesn't he know the color of his third marble? Why doesn't he know everything that Sally knows? It's trivial but Harry should have come in late and not known the statements made by Tom and/or Rick[sic], for example. smile

That was enjoyable!

#10 Re: Puzzles and Games » Containers of marbles » 2011-02-21 08:58:11

The solution to the puzzle presented in the first message in this thread appears to be.......

The proof is....

Note: Had to use Tom, Rick, Harry, and Sally as the names instead of the names given in the original Marble Mix Up puzzle.

#12 Re: Puzzles and Games » chocolate bar - your solution blowed my mind - great! » 2011-02-03 03:01:51

Just like you, I did also. Our way is perhaps a more crude, but more direct, mathetmatical solution. It was the first thought that entered my mind and was the correct solution so it just stuck.

By the way, see
Click here to see message on hiding puzzle solutions in messages.
If you use the hide tag it will hide the solution that you just gave, just in case someone inadvertently sees the solution to the puzzle before they have tried to solve it for themselves. I did the same thing myself the first time I posted a message.

By the way, also, -ed is a common construction in English for the past tense, but "blowed" is not a word. The past tense of blow is "blew".


#13 Re: Formulas » Temperature » 2011-01-30 11:57:59

Celsius to Fahrenheit is not 9/5 of (C+32)  Nor (C+32)x9/5.

It is (9/5 of C)+32.

For example 0C is 32F, as the most obvious example. 9/5 of (0C+32) is not equal to 32F, but (9/5 of 0C)+32 is equal to 32F. The formual (9/5 of C)+32 also works on other perhaps obvious examples, like 100C is 212F and -40C is -40F.

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